id
int64
0
4.52k
code
stringlengths
142
31.2k
answer
stringclasses
7 values
prompt
stringlengths
1.64k
32.6k
test_prompt
stringlengths
1.7k
32.7k
token_length
int64
373
7.8k
__index_level_0__
int64
0
4.5k
3,927
import java.lang.*; import java.util.*; import java.io.*; public class Main { void solve() { int m=ni(); long a[][]=new long[m][]; HashMap<Long,Integer> mp=new HashMap<>(); long TS=0; long sm[]=new long[m]; for(int i=0;i<m;i++){ int sz=ni(); a[i]=new long[sz]; for(int j=0;j<sz;j++){ a[i][j]=nl(); mp.put(a[i][j],i); sm[i]+=a[i][j]; } TS+=sm[i]; } if(TS%m!=0){ pw.println("No"); return; } TS/=m; ArrayList<Node> ansForMask[]=new ArrayList[(1<<m)]; for(int i=0;i<(1<<m);i++) ansForMask[i]=new ArrayList<>(); ArrayList<Node> tempList=new ArrayList<>(); int vis[]=new int[m]; for(int i=0;i<m;i++){ out:for(int j=0;j<a[i].length;j++){ int mask=0; tempList.clear(); long val=a[i][j],req=Long.MAX_VALUE; int idx=i,idx2; Arrays.fill(vis,0); if(sm[i]==TS){ mask=1<<i; if(ansForMask[mask].size()==0) ansForMask[mask].add(new Node(val,i,i)); continue; } while(vis[idx]==0){ req=TS-(sm[idx]-val); if(!mp.containsKey(req)) continue out; idx2=mp.get(req); if(vis[idx]==1 || idx==idx2) continue out; if(vis[idx2]==1) break; // if(i==0 && j==1) pw.println(idx2+" "+req+" "+val); vis[idx]=1; mask+=(1<<idx); tempList.add(new Node(req,idx2,idx)); idx=idx2; val=req; } if(req!=a[i][j])continue out; mask+=(1<<idx); tempList.add(new Node(a[i][j],i,idx)); if(ansForMask[mask].size()==0){ // pw.println(Integer.toBinaryString(mask)); ansForMask[mask].addAll(tempList); } } } int dp[]=new int[1<<m]; dp[0]=1; out: for(int mask=1;mask <(1<<m);mask++){ if(ansForMask[mask].size()!=0){ dp[mask]=1; // pw.println(Integer.toBinaryString(mask)+" "+mask); continue; } for(int s=mask;s>0;s=(s-1)&mask){ if(dp[s]==1 && dp[mask^s]==1){ dp[mask]=1; ansForMask[mask].addAll(ansForMask[s]); ansForMask[mask].addAll(ansForMask[mask^s]); continue out; } } } if(dp[(1<<m)-1]==0){ pw.println("No"); return; } pw.println("Yes"); Pair ans[]=new Pair[m]; for(Node p : ansForMask[(1<<m)-1]){ ans[p.id1]=new Pair(p.c,p.id2); } for(int i=0;i<m;i++) pw.println(ans[i].c+" "+(ans[i].p+1)); } class Pair { long c; int p; public Pair(long c,int p){ this.c=c; this.p=p; } } class Node{ long c; int id1; int id2; public Node(long c,int id1,int id2){ this.c=c; this.id1=id1; this.id2=id2; } } long M = (long)1e9+7; // END PrintWriter pw; StringTokenizer st; BufferedReader br; void run() throws Exception { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); pw.flush(); } public static void main(String[] args) throws Exception { new Main().run(); } String ns() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() throws Exception { String str = ""; try { str = br.readLine(); } catch (IOException e) { throw new Exception(e.toString()); } return str; } int ni() { return Integer.parseInt(ns()); } long nl() { return Long.parseLong(ns()); } double nd() { return Double.parseDouble(ns()); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.lang.*; import java.util.*; import java.io.*; public class Main { void solve() { int m=ni(); long a[][]=new long[m][]; HashMap<Long,Integer> mp=new HashMap<>(); long TS=0; long sm[]=new long[m]; for(int i=0;i<m;i++){ int sz=ni(); a[i]=new long[sz]; for(int j=0;j<sz;j++){ a[i][j]=nl(); mp.put(a[i][j],i); sm[i]+=a[i][j]; } TS+=sm[i]; } if(TS%m!=0){ pw.println("No"); return; } TS/=m; ArrayList<Node> ansForMask[]=new ArrayList[(1<<m)]; for(int i=0;i<(1<<m);i++) ansForMask[i]=new ArrayList<>(); ArrayList<Node> tempList=new ArrayList<>(); int vis[]=new int[m]; for(int i=0;i<m;i++){ out:for(int j=0;j<a[i].length;j++){ int mask=0; tempList.clear(); long val=a[i][j],req=Long.MAX_VALUE; int idx=i,idx2; Arrays.fill(vis,0); if(sm[i]==TS){ mask=1<<i; if(ansForMask[mask].size()==0) ansForMask[mask].add(new Node(val,i,i)); continue; } while(vis[idx]==0){ req=TS-(sm[idx]-val); if(!mp.containsKey(req)) continue out; idx2=mp.get(req); if(vis[idx]==1 || idx==idx2) continue out; if(vis[idx2]==1) break; // if(i==0 && j==1) pw.println(idx2+" "+req+" "+val); vis[idx]=1; mask+=(1<<idx); tempList.add(new Node(req,idx2,idx)); idx=idx2; val=req; } if(req!=a[i][j])continue out; mask+=(1<<idx); tempList.add(new Node(a[i][j],i,idx)); if(ansForMask[mask].size()==0){ // pw.println(Integer.toBinaryString(mask)); ansForMask[mask].addAll(tempList); } } } int dp[]=new int[1<<m]; dp[0]=1; out: for(int mask=1;mask <(1<<m);mask++){ if(ansForMask[mask].size()!=0){ dp[mask]=1; // pw.println(Integer.toBinaryString(mask)+" "+mask); continue; } for(int s=mask;s>0;s=(s-1)&mask){ if(dp[s]==1 && dp[mask^s]==1){ dp[mask]=1; ansForMask[mask].addAll(ansForMask[s]); ansForMask[mask].addAll(ansForMask[mask^s]); continue out; } } } if(dp[(1<<m)-1]==0){ pw.println("No"); return; } pw.println("Yes"); Pair ans[]=new Pair[m]; for(Node p : ansForMask[(1<<m)-1]){ ans[p.id1]=new Pair(p.c,p.id2); } for(int i=0;i<m;i++) pw.println(ans[i].c+" "+(ans[i].p+1)); } class Pair { long c; int p; public Pair(long c,int p){ this.c=c; this.p=p; } } class Node{ long c; int id1; int id2; public Node(long c,int id1,int id2){ this.c=c; this.id1=id1; this.id2=id2; } } long M = (long)1e9+7; // END PrintWriter pw; StringTokenizer st; BufferedReader br; void run() throws Exception { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); pw.flush(); } public static void main(String[] args) throws Exception { new Main().run(); } String ns() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() throws Exception { String str = ""; try { str = br.readLine(); } catch (IOException e) { throw new Exception(e.toString()); } return str; } int ni() { return Integer.parseInt(ns()); } long nl() { return Long.parseLong(ns()); } double nd() { return Double.parseDouble(ns()); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.lang.*; import java.util.*; import java.io.*; public class Main { void solve() { int m=ni(); long a[][]=new long[m][]; HashMap<Long,Integer> mp=new HashMap<>(); long TS=0; long sm[]=new long[m]; for(int i=0;i<m;i++){ int sz=ni(); a[i]=new long[sz]; for(int j=0;j<sz;j++){ a[i][j]=nl(); mp.put(a[i][j],i); sm[i]+=a[i][j]; } TS+=sm[i]; } if(TS%m!=0){ pw.println("No"); return; } TS/=m; ArrayList<Node> ansForMask[]=new ArrayList[(1<<m)]; for(int i=0;i<(1<<m);i++) ansForMask[i]=new ArrayList<>(); ArrayList<Node> tempList=new ArrayList<>(); int vis[]=new int[m]; for(int i=0;i<m;i++){ out:for(int j=0;j<a[i].length;j++){ int mask=0; tempList.clear(); long val=a[i][j],req=Long.MAX_VALUE; int idx=i,idx2; Arrays.fill(vis,0); if(sm[i]==TS){ mask=1<<i; if(ansForMask[mask].size()==0) ansForMask[mask].add(new Node(val,i,i)); continue; } while(vis[idx]==0){ req=TS-(sm[idx]-val); if(!mp.containsKey(req)) continue out; idx2=mp.get(req); if(vis[idx]==1 || idx==idx2) continue out; if(vis[idx2]==1) break; // if(i==0 && j==1) pw.println(idx2+" "+req+" "+val); vis[idx]=1; mask+=(1<<idx); tempList.add(new Node(req,idx2,idx)); idx=idx2; val=req; } if(req!=a[i][j])continue out; mask+=(1<<idx); tempList.add(new Node(a[i][j],i,idx)); if(ansForMask[mask].size()==0){ // pw.println(Integer.toBinaryString(mask)); ansForMask[mask].addAll(tempList); } } } int dp[]=new int[1<<m]; dp[0]=1; out: for(int mask=1;mask <(1<<m);mask++){ if(ansForMask[mask].size()!=0){ dp[mask]=1; // pw.println(Integer.toBinaryString(mask)+" "+mask); continue; } for(int s=mask;s>0;s=(s-1)&mask){ if(dp[s]==1 && dp[mask^s]==1){ dp[mask]=1; ansForMask[mask].addAll(ansForMask[s]); ansForMask[mask].addAll(ansForMask[mask^s]); continue out; } } } if(dp[(1<<m)-1]==0){ pw.println("No"); return; } pw.println("Yes"); Pair ans[]=new Pair[m]; for(Node p : ansForMask[(1<<m)-1]){ ans[p.id1]=new Pair(p.c,p.id2); } for(int i=0;i<m;i++) pw.println(ans[i].c+" "+(ans[i].p+1)); } class Pair { long c; int p; public Pair(long c,int p){ this.c=c; this.p=p; } } class Node{ long c; int id1; int id2; public Node(long c,int id1,int id2){ this.c=c; this.id1=id1; this.id2=id2; } } long M = (long)1e9+7; // END PrintWriter pw; StringTokenizer st; BufferedReader br; void run() throws Exception { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); pw.flush(); } public static void main(String[] args) throws Exception { new Main().run(); } String ns() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() throws Exception { String str = ""; try { str = br.readLine(); } catch (IOException e) { throw new Exception(e.toString()); } return str; } int ni() { return Integer.parseInt(ns()); } long nl() { return Long.parseLong(ns()); } double nd() { return Double.parseDouble(ns()); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,408
3,917
815
import java.io.*; import java.util.*; public class Noldbach { public Scanner in = new Scanner(System.in); public PrintStream out = System.out; public boolean[] yes; public int n, k; public void main() { n = in.nextInt(); k = in.nextInt(); genPrime(); int i; yes = new boolean[n+1]; int x; for(i=0;i+1<prime.length;++i) { x = prime[i]+prime[i+1]+1; if(x <= n && fac[x] == x) yes[x] = true; } int count = 0; for(i=0;i<yes.length;++i) if(yes[i]) ++count; out.println((count>=k?"YES":"NO")); }//end public void main() //Generating Primes public int N = 100000+100; public int[] fac, rest; public int[] prime; public void genPrime() { ArrayList<Integer> ap = new ArrayList<Integer>(); fac = new int[N]; rest = new int[N]; int x,y; for(x=0;x<N;++x) { fac[x] = x; rest[x] = 1; } for(x=2;x<N;++x) if(fac[x]==x) { ap.add(x); for(y=x+x;y<N;y+=x) if(fac[y]==y) { fac[y] = x; rest[y] = y/x; } } prime = new int[ap.size()]; for(int i=0;i<prime.length;++i) prime[i] = ap.get(i); } public static void main(String[] args) { (new Noldbach()).main(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Noldbach { public Scanner in = new Scanner(System.in); public PrintStream out = System.out; public boolean[] yes; public int n, k; public void main() { n = in.nextInt(); k = in.nextInt(); genPrime(); int i; yes = new boolean[n+1]; int x; for(i=0;i+1<prime.length;++i) { x = prime[i]+prime[i+1]+1; if(x <= n && fac[x] == x) yes[x] = true; } int count = 0; for(i=0;i<yes.length;++i) if(yes[i]) ++count; out.println((count>=k?"YES":"NO")); }//end public void main() //Generating Primes public int N = 100000+100; public int[] fac, rest; public int[] prime; public void genPrime() { ArrayList<Integer> ap = new ArrayList<Integer>(); fac = new int[N]; rest = new int[N]; int x,y; for(x=0;x<N;++x) { fac[x] = x; rest[x] = 1; } for(x=2;x<N;++x) if(fac[x]==x) { ap.add(x); for(y=x+x;y<N;y+=x) if(fac[y]==y) { fac[y] = x; rest[y] = y/x; } } prime = new int[ap.size()]; for(int i=0;i<prime.length;++i) prime[i] = ap.get(i); } public static void main(String[] args) { (new Noldbach()).main(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Noldbach { public Scanner in = new Scanner(System.in); public PrintStream out = System.out; public boolean[] yes; public int n, k; public void main() { n = in.nextInt(); k = in.nextInt(); genPrime(); int i; yes = new boolean[n+1]; int x; for(i=0;i+1<prime.length;++i) { x = prime[i]+prime[i+1]+1; if(x <= n && fac[x] == x) yes[x] = true; } int count = 0; for(i=0;i<yes.length;++i) if(yes[i]) ++count; out.println((count>=k?"YES":"NO")); }//end public void main() //Generating Primes public int N = 100000+100; public int[] fac, rest; public int[] prime; public void genPrime() { ArrayList<Integer> ap = new ArrayList<Integer>(); fac = new int[N]; rest = new int[N]; int x,y; for(x=0;x<N;++x) { fac[x] = x; rest[x] = 1; } for(x=2;x<N;++x) if(fac[x]==x) { ap.add(x); for(y=x+x;y<N;y+=x) if(fac[y]==y) { fac[y] = x; rest[y] = y/x; } } prime = new int[ap.size()]; for(int i=0;i<prime.length;++i) prime[i] = ap.get(i); } public static void main(String[] args) { (new Noldbach()).main(); } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
772
814
3,764
// Don't place your source in a package import javax.swing.*; import java.lang.reflect.Array; import java.text.DecimalFormat; import java.util.*; import java.lang.*; import java.io.*; import java.math.*; import java.util.stream.Stream; // Please name your class Main public class Main { static FastScanner fs=new FastScanner(); static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); public String next() { while (!st.hasMoreElements()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int Int() { return Integer.parseInt(next()); } long Long() { return Long.parseLong(next()); } String Str(){ return next(); } } public static void main (String[] args) throws java.lang.Exception { PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int T=1; for(int t=0;t<T;t++){ int n=Int(),m=Int(),k=Int(); List<int[]>g[]=new ArrayList[n*m+1]; for(int i=0;i<g.length;i++){ g[i]=new ArrayList<>(); } for(int i=0;i<n;i++){ for(int j=0;j<m-1;j++){ int w=Int(); int u=i*m+j; int v=i*m+(j+1); g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } for(int i=0;i<n-1;i++){ for(int j=0;j<m;j++){ int w=Int(); int u=i*m+j; int v=(i+1)*m+j; g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } Solution sol=new Solution(out); sol.solution(n,m,k,g); } out.close(); } public static int Int(){ return fs.Int(); } public static long Long(){ return fs.Long(); } public static String Str(){ return fs.Str(); } } class Solution{ PrintWriter out; public Solution(PrintWriter out){ this.out=out; } List<int[]>g[]; int n,m; long INF=10000000000000000l; int curr=-1,curc=-1; long mn=Long.MAX_VALUE; long dp[][]; public void solution(int n,int m,int k,List<int[]>g[]){ //edge : 4 directions. this.n=n; this.m=m; long res[][]=new long[n][m]; if(k%2==1){ for(int i=0;i<n;i++){ Arrays.fill(res[i],-1); } print(res); return; } this.g=g; dp=new long[n*m+1][k/2+2]; for(int i=0;i<dp.length;i++){ Arrays.fill(dp[i],-1); } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; dfs(id,k/2); } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; res[i][j]=dp[id][k/2]; } } print(res); } public long dfs(int id,int cnt){ if(cnt==0){ return 0; } if(dp[id][cnt]!=-1)return dp[id][cnt]; int r=id/m; int c=id%m; long res=Long.MAX_VALUE; for(int p[]:g[id]){ int next=p[0],w=p[1]; res=Math.min(res,w*2+dfs(next,cnt-1)); } dp[id][cnt]=res; return res; } public int dis(int x1,int y1,int x2,int y2){ return Math.abs(x1-x2)+Math.abs(y1-y2); } public void print(long A[][]){ for(int i=0;i<A.length;i++){ for(int j=0;j<A[0].length;j++){ out.print(A[i][j]+" "); } out.println(); } } } class Solution1{ PrintWriter out; public Solution1(PrintWriter out){ this.out=out; } public void solution(int A[]){ } } /* ;\ |' \ _ ; : ; / `-. /: : | | ,-.`-. ,': : | \ : `. `. ,'-. : | \ ; ; `-.__,' `-.| \ ; ; ::: ,::'`:. `. \ `-. : ` :. `. \ \ \ , ; ,: (\ \ :., :. ,'o)): ` `-. ,/,' ;' ,::"'`.`---' `. `-._ ,/ : ; '" `;' ,--`. ;/ :; ; ,:' ( ,:) ,.,:. ; ,:., ,-._ `. \""'/ '::' `:'` ,'( \`._____.-'"' ;, ; `. `. `._`-. \\ ;:. ;: `-._`-.\ \`. '`:. : |' `. `\ ) \ -hrr- ` ;: | `--\__,' '` ,' ,-' free bug dog */
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> // Don't place your source in a package import javax.swing.*; import java.lang.reflect.Array; import java.text.DecimalFormat; import java.util.*; import java.lang.*; import java.io.*; import java.math.*; import java.util.stream.Stream; // Please name your class Main public class Main { static FastScanner fs=new FastScanner(); static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); public String next() { while (!st.hasMoreElements()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int Int() { return Integer.parseInt(next()); } long Long() { return Long.parseLong(next()); } String Str(){ return next(); } } public static void main (String[] args) throws java.lang.Exception { PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int T=1; for(int t=0;t<T;t++){ int n=Int(),m=Int(),k=Int(); List<int[]>g[]=new ArrayList[n*m+1]; for(int i=0;i<g.length;i++){ g[i]=new ArrayList<>(); } for(int i=0;i<n;i++){ for(int j=0;j<m-1;j++){ int w=Int(); int u=i*m+j; int v=i*m+(j+1); g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } for(int i=0;i<n-1;i++){ for(int j=0;j<m;j++){ int w=Int(); int u=i*m+j; int v=(i+1)*m+j; g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } Solution sol=new Solution(out); sol.solution(n,m,k,g); } out.close(); } public static int Int(){ return fs.Int(); } public static long Long(){ return fs.Long(); } public static String Str(){ return fs.Str(); } } class Solution{ PrintWriter out; public Solution(PrintWriter out){ this.out=out; } List<int[]>g[]; int n,m; long INF=10000000000000000l; int curr=-1,curc=-1; long mn=Long.MAX_VALUE; long dp[][]; public void solution(int n,int m,int k,List<int[]>g[]){ //edge : 4 directions. this.n=n; this.m=m; long res[][]=new long[n][m]; if(k%2==1){ for(int i=0;i<n;i++){ Arrays.fill(res[i],-1); } print(res); return; } this.g=g; dp=new long[n*m+1][k/2+2]; for(int i=0;i<dp.length;i++){ Arrays.fill(dp[i],-1); } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; dfs(id,k/2); } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; res[i][j]=dp[id][k/2]; } } print(res); } public long dfs(int id,int cnt){ if(cnt==0){ return 0; } if(dp[id][cnt]!=-1)return dp[id][cnt]; int r=id/m; int c=id%m; long res=Long.MAX_VALUE; for(int p[]:g[id]){ int next=p[0],w=p[1]; res=Math.min(res,w*2+dfs(next,cnt-1)); } dp[id][cnt]=res; return res; } public int dis(int x1,int y1,int x2,int y2){ return Math.abs(x1-x2)+Math.abs(y1-y2); } public void print(long A[][]){ for(int i=0;i<A.length;i++){ for(int j=0;j<A[0].length;j++){ out.print(A[i][j]+" "); } out.println(); } } } class Solution1{ PrintWriter out; public Solution1(PrintWriter out){ this.out=out; } public void solution(int A[]){ } } /* ;\ |' \ _ ; : ; / `-. /: : | | ,-.`-. ,': : | \ : `. `. ,'-. : | \ ; ; `-.__,' `-.| \ ; ; ::: ,::'`:. `. \ `-. : ` :. `. \ \ \ , ; ,: (\ \ :., :. ,'o)): ` `-. ,/,' ;' ,::"'`.`---' `. `-._ ,/ : ; '" `;' ,--`. ;/ :; ; ,:' ( ,:) ,.,:. ; ,:., ,-._ `. \""'/ '::' `:'` ,'( \`._____.-'"' ;, ; `. `. `._`-. \\ ;:. ;: `-._`-.\ \`. '`:. : |' `. `\ ) \ -hrr- ` ;: | `--\__,' '` ,' ,-' free bug dog */ </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> // Don't place your source in a package import javax.swing.*; import java.lang.reflect.Array; import java.text.DecimalFormat; import java.util.*; import java.lang.*; import java.io.*; import java.math.*; import java.util.stream.Stream; // Please name your class Main public class Main { static FastScanner fs=new FastScanner(); static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); public String next() { while (!st.hasMoreElements()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int Int() { return Integer.parseInt(next()); } long Long() { return Long.parseLong(next()); } String Str(){ return next(); } } public static void main (String[] args) throws java.lang.Exception { PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int T=1; for(int t=0;t<T;t++){ int n=Int(),m=Int(),k=Int(); List<int[]>g[]=new ArrayList[n*m+1]; for(int i=0;i<g.length;i++){ g[i]=new ArrayList<>(); } for(int i=0;i<n;i++){ for(int j=0;j<m-1;j++){ int w=Int(); int u=i*m+j; int v=i*m+(j+1); g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } for(int i=0;i<n-1;i++){ for(int j=0;j<m;j++){ int w=Int(); int u=i*m+j; int v=(i+1)*m+j; g[u].add(new int[]{v,w}); g[v].add(new int[]{u,w}); } } Solution sol=new Solution(out); sol.solution(n,m,k,g); } out.close(); } public static int Int(){ return fs.Int(); } public static long Long(){ return fs.Long(); } public static String Str(){ return fs.Str(); } } class Solution{ PrintWriter out; public Solution(PrintWriter out){ this.out=out; } List<int[]>g[]; int n,m; long INF=10000000000000000l; int curr=-1,curc=-1; long mn=Long.MAX_VALUE; long dp[][]; public void solution(int n,int m,int k,List<int[]>g[]){ //edge : 4 directions. this.n=n; this.m=m; long res[][]=new long[n][m]; if(k%2==1){ for(int i=0;i<n;i++){ Arrays.fill(res[i],-1); } print(res); return; } this.g=g; dp=new long[n*m+1][k/2+2]; for(int i=0;i<dp.length;i++){ Arrays.fill(dp[i],-1); } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; dfs(id,k/2); } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ int id=i*m+j; res[i][j]=dp[id][k/2]; } } print(res); } public long dfs(int id,int cnt){ if(cnt==0){ return 0; } if(dp[id][cnt]!=-1)return dp[id][cnt]; int r=id/m; int c=id%m; long res=Long.MAX_VALUE; for(int p[]:g[id]){ int next=p[0],w=p[1]; res=Math.min(res,w*2+dfs(next,cnt-1)); } dp[id][cnt]=res; return res; } public int dis(int x1,int y1,int x2,int y2){ return Math.abs(x1-x2)+Math.abs(y1-y2); } public void print(long A[][]){ for(int i=0;i<A.length;i++){ for(int j=0;j<A[0].length;j++){ out.print(A[i][j]+" "); } out.println(); } } } class Solution1{ PrintWriter out; public Solution1(PrintWriter out){ this.out=out; } public void solution(int A[]){ } } /* ;\ |' \ _ ; : ; / `-. /: : | | ,-.`-. ,': : | \ : `. `. ,'-. : | \ ; ; `-.__,' `-.| \ ; ; ::: ,::'`:. `. \ `-. : ` :. `. \ \ \ , ; ,: (\ \ :., :. ,'o)): ` `-. ,/,' ;' ,::"'`.`---' `. `-._ ,/ : ; '" `;' ,--`. ;/ :; ; ,:' ( ,:) ,.,:. ; ,:., ,-._ `. \""'/ '::' `:'` ,'( \`._____.-'"' ;, ; `. `. `._`-. \\ ;:. ;: `-._`-.\ \`. '`:. : |' `. `\ ) \ -hrr- ` ;: | `--\__,' '` ,' ,-' free bug dog */ </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,622
3,756
3,125
import java.io.*; import java.util.*; import java.math.*; public class Solution implements Runnable { private BufferedReader in; private PrintWriter out; private StringTokenizer st; private Random rnd; double Vend; private double calcFirstSegment(double a, double Vmax, double Vend, double l) { double Vl = 0, Vr = Vmax; for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double res = firstPart + secondPart; if(res < l) Vl = Vm; else Vr = Vm; } this.Vend = Math.min(Vl, Vend); double res = 0.0; { double Vm = Vl; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; //out.println(tSecond); double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double remain = l - firstPart - secondPart; res = tFirst + tSecond + (remain / Vm); } return res; } private double calcSecondPart(double a, double Vmax, double Vstart, double l) { double Vl = Vstart, Vr = Vmax; //out.println(Vstart); for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; if(s < l) Vl = Vm; else Vr = Vm; } double res = 0.0; { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; double remain = l - s; res = t + (remain / Vmax); } return res; } public void solve() throws IOException { double a = nextDouble(), v = nextDouble(), l = nextDouble(), d = nextDouble(), w = nextDouble(); double res = calcFirstSegment(a, v, w, d); res += calcSecondPart(a, v, Vend, l - d); out.println(res); } public static void main(String[] args) { new Solution().run(); } public void run() { try { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader((System.in))); out = new PrintWriter(System.out); st = null; rnd = new Random(); solve(); out.close(); } catch(IOException e) { e.printStackTrace(); } } private String nextToken() throws IOException, NullPointerException { while(st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } private int nextInt() throws IOException { return Integer.parseInt(nextToken()); } private long nextLong() throws IOException { return Long.parseLong(nextToken()); } private double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class Solution implements Runnable { private BufferedReader in; private PrintWriter out; private StringTokenizer st; private Random rnd; double Vend; private double calcFirstSegment(double a, double Vmax, double Vend, double l) { double Vl = 0, Vr = Vmax; for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double res = firstPart + secondPart; if(res < l) Vl = Vm; else Vr = Vm; } this.Vend = Math.min(Vl, Vend); double res = 0.0; { double Vm = Vl; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; //out.println(tSecond); double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double remain = l - firstPart - secondPart; res = tFirst + tSecond + (remain / Vm); } return res; } private double calcSecondPart(double a, double Vmax, double Vstart, double l) { double Vl = Vstart, Vr = Vmax; //out.println(Vstart); for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; if(s < l) Vl = Vm; else Vr = Vm; } double res = 0.0; { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; double remain = l - s; res = t + (remain / Vmax); } return res; } public void solve() throws IOException { double a = nextDouble(), v = nextDouble(), l = nextDouble(), d = nextDouble(), w = nextDouble(); double res = calcFirstSegment(a, v, w, d); res += calcSecondPart(a, v, Vend, l - d); out.println(res); } public static void main(String[] args) { new Solution().run(); } public void run() { try { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader((System.in))); out = new PrintWriter(System.out); st = null; rnd = new Random(); solve(); out.close(); } catch(IOException e) { e.printStackTrace(); } } private String nextToken() throws IOException, NullPointerException { while(st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } private int nextInt() throws IOException { return Integer.parseInt(nextToken()); } private long nextLong() throws IOException { return Long.parseLong(nextToken()); } private double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class Solution implements Runnable { private BufferedReader in; private PrintWriter out; private StringTokenizer st; private Random rnd; double Vend; private double calcFirstSegment(double a, double Vmax, double Vend, double l) { double Vl = 0, Vr = Vmax; for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double res = firstPart + secondPart; if(res < l) Vl = Vm; else Vr = Vm; } this.Vend = Math.min(Vl, Vend); double res = 0.0; { double Vm = Vl; double tFirst = Vm / a; double tSecond = 0; if(Vend < Vm) tSecond = (Vm - Vend) / a; //out.println(tSecond); double firstPart = a * tFirst * tFirst / 2.0; double secondPart = Vm * tSecond - a * tSecond * tSecond / 2.0; double remain = l - firstPart - secondPart; res = tFirst + tSecond + (remain / Vm); } return res; } private double calcSecondPart(double a, double Vmax, double Vstart, double l) { double Vl = Vstart, Vr = Vmax; //out.println(Vstart); for(int it = 0; it < 256; it++) { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; if(s < l) Vl = Vm; else Vr = Vm; } double res = 0.0; { double Vm = (Vl + Vr) / 2.0; double t = (Vm - Vstart) / a; double s = Vstart * t + a * t * t / 2.0; double remain = l - s; res = t + (remain / Vmax); } return res; } public void solve() throws IOException { double a = nextDouble(), v = nextDouble(), l = nextDouble(), d = nextDouble(), w = nextDouble(); double res = calcFirstSegment(a, v, w, d); res += calcSecondPart(a, v, Vend, l - d); out.println(res); } public static void main(String[] args) { new Solution().run(); } public void run() { try { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader((System.in))); out = new PrintWriter(System.out); st = null; rnd = new Random(); solve(); out.close(); } catch(IOException e) { e.printStackTrace(); } } private String nextToken() throws IOException, NullPointerException { while(st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } private int nextInt() throws IOException { return Integer.parseInt(nextToken()); } private long nextLong() throws IOException { return Long.parseLong(nextToken()); } private double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,244
3,119
3,575
import java.util.*; import java.io.*; public class C{ public static void main(String args[]) throws Exception{ Scanner in = new Scanner(new FileReader("input.txt")); PrintWriter out = new PrintWriter(new File("output.txt")); int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[] X = new int[K], Y = new int[K]; for (int i = 0; i < K; i++){ X[i] = in.nextInt(); Y[i] = in.nextInt(); } int d = -1; int a = -1; int b = -1; for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++){ int h = Integer.MAX_VALUE; for (int p = 0; p < K; p++) h = Math.min(h,Math.abs(i-X[p]) + Math.abs(j-Y[p])); if (h > d){ d = h; a = i; b = j; } } out.print(a + " " + b); out.close(); } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; public class C{ public static void main(String args[]) throws Exception{ Scanner in = new Scanner(new FileReader("input.txt")); PrintWriter out = new PrintWriter(new File("output.txt")); int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[] X = new int[K], Y = new int[K]; for (int i = 0; i < K; i++){ X[i] = in.nextInt(); Y[i] = in.nextInt(); } int d = -1; int a = -1; int b = -1; for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++){ int h = Integer.MAX_VALUE; for (int p = 0; p < K; p++) h = Math.min(h,Math.abs(i-X[p]) + Math.abs(j-Y[p])); if (h > d){ d = h; a = i; b = j; } } out.print(a + " " + b); out.close(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; public class C{ public static void main(String args[]) throws Exception{ Scanner in = new Scanner(new FileReader("input.txt")); PrintWriter out = new PrintWriter(new File("output.txt")); int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[] X = new int[K], Y = new int[K]; for (int i = 0; i < K; i++){ X[i] = in.nextInt(); Y[i] = in.nextInt(); } int d = -1; int a = -1; int b = -1; for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++){ int h = Integer.MAX_VALUE; for (int p = 0; p < K; p++) h = Math.min(h,Math.abs(i-X[p]) + Math.abs(j-Y[p])); if (h > d){ d = h; a = i; b = j; } } out.print(a + " " + b); out.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
585
3,567
2,481
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.util.ArrayList; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author beginner1010 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskF2 solver = new TaskF2(); solver.solve(1, in, out); out.close(); } static class TaskF2 { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = in.nextInt(); } HashMap<Long, ArrayList<Interval>> map = new HashMap<>(); for (int i = 0; i < n; i++) { long sum = 0; for (int j = i; j < n; j++) { sum += a[j]; if (map.containsKey(sum) == false) { map.put(sum, new ArrayList<>()); } ArrayList<Interval> arr = map.get(sum); if (arr.isEmpty() || arr.get(arr.size() - 1).r < i) { arr.add(new Interval(i, j)); } else if (arr.get(arr.size() - 1).r >= j) { arr.set(arr.size() - 1, new Interval(i, j)); } } } ArrayList<Interval> best = new ArrayList<>(); for (ArrayList<Interval> arr : map.values()) { if (best.size() < arr.size()) { best = arr; } } out.println(best.size()); for (Interval i : best) { out.println((i.l + 1) + " " + (i.r + 1)); } } class Interval { int l; int r; Interval(int l, int r) { this.l = l; this.r = r; } } } static class InputReader { private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputStream stream; public InputReader(InputStream stream) { this.stream = stream; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = read(); while (isWhitespace(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isWhitespace(c)); return res * sgn; } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.util.ArrayList; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author beginner1010 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskF2 solver = new TaskF2(); solver.solve(1, in, out); out.close(); } static class TaskF2 { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = in.nextInt(); } HashMap<Long, ArrayList<Interval>> map = new HashMap<>(); for (int i = 0; i < n; i++) { long sum = 0; for (int j = i; j < n; j++) { sum += a[j]; if (map.containsKey(sum) == false) { map.put(sum, new ArrayList<>()); } ArrayList<Interval> arr = map.get(sum); if (arr.isEmpty() || arr.get(arr.size() - 1).r < i) { arr.add(new Interval(i, j)); } else if (arr.get(arr.size() - 1).r >= j) { arr.set(arr.size() - 1, new Interval(i, j)); } } } ArrayList<Interval> best = new ArrayList<>(); for (ArrayList<Interval> arr : map.values()) { if (best.size() < arr.size()) { best = arr; } } out.println(best.size()); for (Interval i : best) { out.println((i.l + 1) + " " + (i.r + 1)); } } class Interval { int l; int r; Interval(int l, int r) { this.l = l; this.r = r; } } } static class InputReader { private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputStream stream; public InputReader(InputStream stream) { this.stream = stream; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = read(); while (isWhitespace(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isWhitespace(c)); return res * sgn; } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.util.ArrayList; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author beginner1010 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskF2 solver = new TaskF2(); solver.solve(1, in, out); out.close(); } static class TaskF2 { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = in.nextInt(); } HashMap<Long, ArrayList<Interval>> map = new HashMap<>(); for (int i = 0; i < n; i++) { long sum = 0; for (int j = i; j < n; j++) { sum += a[j]; if (map.containsKey(sum) == false) { map.put(sum, new ArrayList<>()); } ArrayList<Interval> arr = map.get(sum); if (arr.isEmpty() || arr.get(arr.size() - 1).r < i) { arr.add(new Interval(i, j)); } else if (arr.get(arr.size() - 1).r >= j) { arr.set(arr.size() - 1, new Interval(i, j)); } } } ArrayList<Interval> best = new ArrayList<>(); for (ArrayList<Interval> arr : map.values()) { if (best.size() < arr.size()) { best = arr; } } out.println(best.size()); for (Interval i : best) { out.println((i.l + 1) + " " + (i.r + 1)); } } class Interval { int l; int r; Interval(int l, int r) { this.l = l; this.r = r; } } } static class InputReader { private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputStream stream; public InputReader(InputStream stream) { this.stream = stream; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = read(); while (isWhitespace(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isWhitespace(c)); return res * sgn; } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,158
2,476
4,436
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.StringTokenizer; /** * Author - * User: kansal * Date: 9/3/11 * Time: 5:28 PM */ public class CF85E { public static void main(String[] args) { reader = new BufferedReader(new InputStreamReader(System.in)); int height = nextInt(), width = nextInt(); if (width > height) { int t = width; width = height; height = t; } final int INF = height * width + 10; int[][][] dp = new int[height + 1][1 << width][1 << width]; for (int[][] ints : dp) { for (int[] anInt : ints) { Arrays.fill(anInt, INF); } } dp[0][0][0] = 0; for(int r = 0; r < height; ++r) { for(int uncovered = 0; uncovered < (1 << width); ++uncovered) { for(int mask = 0; mask < (1 << width); ++mask) { if (dp[r][uncovered][mask] == INF) { continue; } for(int curMask = uncovered; curMask < (1 << width); curMask = (curMask + 1) | uncovered) { int curUncovered = (1 << width) - 1; for(int i = 0; i < width; ++i) { if (hasBit(mask, i) || hasBit(curMask, i)) { curUncovered &= ~(1 << i); } if (i > 0 && hasBit(curMask, i-1)) { curUncovered &= ~(1 << i); } if (i < width-1 && hasBit(curMask, i+1)) { curUncovered &= ~(1 << i); } } dp[r+1][curUncovered][curMask] = Math.min(dp[r+1][curUncovered][curMask], dp[r][uncovered][mask] + Integer.bitCount(curMask)); } } } } int res = INF; for(int x: dp[height][0]) res = Math.min(res, x); System.out.println(height * width - res); } private static boolean hasBit(int mask, int bit) { return (((mask >> bit) & 1) == 1); } public static BufferedReader reader; public static StringTokenizer tokenizer = null; static String nextToken() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } static public int nextInt() { return Integer.parseInt(nextToken()); } static public long nextLong() { return Long.parseLong(nextToken()); } static public String next() { return nextToken(); } static public String nextLine() { try { return reader.readLine(); } catch (IOException e) { e.printStackTrace(); } return null; } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.StringTokenizer; /** * Author - * User: kansal * Date: 9/3/11 * Time: 5:28 PM */ public class CF85E { public static void main(String[] args) { reader = new BufferedReader(new InputStreamReader(System.in)); int height = nextInt(), width = nextInt(); if (width > height) { int t = width; width = height; height = t; } final int INF = height * width + 10; int[][][] dp = new int[height + 1][1 << width][1 << width]; for (int[][] ints : dp) { for (int[] anInt : ints) { Arrays.fill(anInt, INF); } } dp[0][0][0] = 0; for(int r = 0; r < height; ++r) { for(int uncovered = 0; uncovered < (1 << width); ++uncovered) { for(int mask = 0; mask < (1 << width); ++mask) { if (dp[r][uncovered][mask] == INF) { continue; } for(int curMask = uncovered; curMask < (1 << width); curMask = (curMask + 1) | uncovered) { int curUncovered = (1 << width) - 1; for(int i = 0; i < width; ++i) { if (hasBit(mask, i) || hasBit(curMask, i)) { curUncovered &= ~(1 << i); } if (i > 0 && hasBit(curMask, i-1)) { curUncovered &= ~(1 << i); } if (i < width-1 && hasBit(curMask, i+1)) { curUncovered &= ~(1 << i); } } dp[r+1][curUncovered][curMask] = Math.min(dp[r+1][curUncovered][curMask], dp[r][uncovered][mask] + Integer.bitCount(curMask)); } } } } int res = INF; for(int x: dp[height][0]) res = Math.min(res, x); System.out.println(height * width - res); } private static boolean hasBit(int mask, int bit) { return (((mask >> bit) & 1) == 1); } public static BufferedReader reader; public static StringTokenizer tokenizer = null; static String nextToken() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } static public int nextInt() { return Integer.parseInt(nextToken()); } static public long nextLong() { return Long.parseLong(nextToken()); } static public String next() { return nextToken(); } static public String nextLine() { try { return reader.readLine(); } catch (IOException e) { e.printStackTrace(); } return null; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.StringTokenizer; /** * Author - * User: kansal * Date: 9/3/11 * Time: 5:28 PM */ public class CF85E { public static void main(String[] args) { reader = new BufferedReader(new InputStreamReader(System.in)); int height = nextInt(), width = nextInt(); if (width > height) { int t = width; width = height; height = t; } final int INF = height * width + 10; int[][][] dp = new int[height + 1][1 << width][1 << width]; for (int[][] ints : dp) { for (int[] anInt : ints) { Arrays.fill(anInt, INF); } } dp[0][0][0] = 0; for(int r = 0; r < height; ++r) { for(int uncovered = 0; uncovered < (1 << width); ++uncovered) { for(int mask = 0; mask < (1 << width); ++mask) { if (dp[r][uncovered][mask] == INF) { continue; } for(int curMask = uncovered; curMask < (1 << width); curMask = (curMask + 1) | uncovered) { int curUncovered = (1 << width) - 1; for(int i = 0; i < width; ++i) { if (hasBit(mask, i) || hasBit(curMask, i)) { curUncovered &= ~(1 << i); } if (i > 0 && hasBit(curMask, i-1)) { curUncovered &= ~(1 << i); } if (i < width-1 && hasBit(curMask, i+1)) { curUncovered &= ~(1 << i); } } dp[r+1][curUncovered][curMask] = Math.min(dp[r+1][curUncovered][curMask], dp[r][uncovered][mask] + Integer.bitCount(curMask)); } } } } int res = INF; for(int x: dp[height][0]) res = Math.min(res, x); System.out.println(height * width - res); } private static boolean hasBit(int mask, int bit) { return (((mask >> bit) & 1) == 1); } public static BufferedReader reader; public static StringTokenizer tokenizer = null; static String nextToken() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } static public int nextInt() { return Integer.parseInt(nextToken()); } static public long nextLong() { return Long.parseLong(nextToken()); } static public String next() { return nextToken(); } static public String nextLine() { try { return reader.readLine(); } catch (IOException e) { e.printStackTrace(); } return null; } } </CODE> <EVALUATION_RUBRIC> - O(n): The running time grows linearly with the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,038
4,425
377
import java.util.InputMismatchException; import java.math.BigInteger; import java.io.*; import java.util.*; /** * Generated by Contest helper plug-in * Actual solution is at the bottom */ public class Main { public static void main(String[] args) { InputReader in = new StreamInputReader(System.in); PrintWriter out = new PrintWriter(System.out); run(in, out); } public static void run(InputReader in, PrintWriter out) { Solver solver = new Task(); solver.solve(1, in, out); Exit.exit(in, out); } } abstract class InputReader { private boolean finished = false; public abstract int read(); public long readLong() { return new BigInteger(readString()).longValue(); } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public void setFinished(boolean finished) { this.finished = finished; } public abstract void close(); } class StreamInputReader extends InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar, numChars; public StreamInputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public void close() { try { stream.close(); } catch (IOException ignored) { } } } class Exit { private Exit() { } public static void exit(InputReader in, PrintWriter out) { in.setFinished(true); in.close(); out.close(); } } interface Solver { public void solve(int testNumber, InputReader in, PrintWriter out); } class Task implements Solver { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.readInt(); int a = in.readInt(); int b = in.readInt(); if (a==b) { b = 0; } boolean[] where = new boolean[n]; HashSet<Integer> set = new HashSet<Integer>(); HashMap<Integer,Integer> indexmap = new HashMap<Integer,Integer>(); for (int i = 0; i<n; i++) { int x = in.readInt(); indexmap.put(x, i); set.add(x); } while (set.size() > 0) { int size = set.size(); HashSet<Integer> todo = new HashSet<Integer>(); HashSet<Integer> used = new HashSet<Integer>(); for (int x : set) { if (used.contains(x)) continue; int ax = a-x; int bx = b-x; if ((set.contains(ax) && !used.contains(ax)) && (set.contains(bx) && !used.contains(bx))) { todo.add(x); } else if (set.contains(ax) && !used.contains(ax)) { used.add(x); used.add(ax); todo.remove(ax); //chain bx = b-ax; while (set.contains(bx) && !used.contains(bx)) { x = bx; ax = a-x; if (!set.contains(ax) || used.contains(ax)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(ax); used.add(x); used.add(ax); bx = b-ax; } } else if (set.contains(bx) && !used.contains(bx)) { used.add(x); used.add(bx); todo.remove(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; //chain ax = a-bx; while (set.contains(ax) && !used.contains(ax)) { x = ax; bx = b-x; if (!set.contains(bx) || used.contains(bx)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(bx); used.add(x); used.add(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; ax = a-bx; } } else { System.out.println("NO"); return; } } set = todo; if (set.size() == size) { System.out.println("Set size constant!!"); break; } } System.out.println("YES"); for (int i = 0; i<n; i++) if (where[i]) System.out.print("1 "); else System.out.print("0 "); } } class num { } ///
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.InputMismatchException; import java.math.BigInteger; import java.io.*; import java.util.*; /** * Generated by Contest helper plug-in * Actual solution is at the bottom */ public class Main { public static void main(String[] args) { InputReader in = new StreamInputReader(System.in); PrintWriter out = new PrintWriter(System.out); run(in, out); } public static void run(InputReader in, PrintWriter out) { Solver solver = new Task(); solver.solve(1, in, out); Exit.exit(in, out); } } abstract class InputReader { private boolean finished = false; public abstract int read(); public long readLong() { return new BigInteger(readString()).longValue(); } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public void setFinished(boolean finished) { this.finished = finished; } public abstract void close(); } class StreamInputReader extends InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar, numChars; public StreamInputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public void close() { try { stream.close(); } catch (IOException ignored) { } } } class Exit { private Exit() { } public static void exit(InputReader in, PrintWriter out) { in.setFinished(true); in.close(); out.close(); } } interface Solver { public void solve(int testNumber, InputReader in, PrintWriter out); } class Task implements Solver { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.readInt(); int a = in.readInt(); int b = in.readInt(); if (a==b) { b = 0; } boolean[] where = new boolean[n]; HashSet<Integer> set = new HashSet<Integer>(); HashMap<Integer,Integer> indexmap = new HashMap<Integer,Integer>(); for (int i = 0; i<n; i++) { int x = in.readInt(); indexmap.put(x, i); set.add(x); } while (set.size() > 0) { int size = set.size(); HashSet<Integer> todo = new HashSet<Integer>(); HashSet<Integer> used = new HashSet<Integer>(); for (int x : set) { if (used.contains(x)) continue; int ax = a-x; int bx = b-x; if ((set.contains(ax) && !used.contains(ax)) && (set.contains(bx) && !used.contains(bx))) { todo.add(x); } else if (set.contains(ax) && !used.contains(ax)) { used.add(x); used.add(ax); todo.remove(ax); //chain bx = b-ax; while (set.contains(bx) && !used.contains(bx)) { x = bx; ax = a-x; if (!set.contains(ax) || used.contains(ax)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(ax); used.add(x); used.add(ax); bx = b-ax; } } else if (set.contains(bx) && !used.contains(bx)) { used.add(x); used.add(bx); todo.remove(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; //chain ax = a-bx; while (set.contains(ax) && !used.contains(ax)) { x = ax; bx = b-x; if (!set.contains(bx) || used.contains(bx)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(bx); used.add(x); used.add(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; ax = a-bx; } } else { System.out.println("NO"); return; } } set = todo; if (set.size() == size) { System.out.println("Set size constant!!"); break; } } System.out.println("YES"); for (int i = 0; i<n; i++) if (where[i]) System.out.print("1 "); else System.out.print("0 "); } } class num { } /// </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.InputMismatchException; import java.math.BigInteger; import java.io.*; import java.util.*; /** * Generated by Contest helper plug-in * Actual solution is at the bottom */ public class Main { public static void main(String[] args) { InputReader in = new StreamInputReader(System.in); PrintWriter out = new PrintWriter(System.out); run(in, out); } public static void run(InputReader in, PrintWriter out) { Solver solver = new Task(); solver.solve(1, in, out); Exit.exit(in, out); } } abstract class InputReader { private boolean finished = false; public abstract int read(); public long readLong() { return new BigInteger(readString()).longValue(); } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public void setFinished(boolean finished) { this.finished = finished; } public abstract void close(); } class StreamInputReader extends InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar, numChars; public StreamInputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public void close() { try { stream.close(); } catch (IOException ignored) { } } } class Exit { private Exit() { } public static void exit(InputReader in, PrintWriter out) { in.setFinished(true); in.close(); out.close(); } } interface Solver { public void solve(int testNumber, InputReader in, PrintWriter out); } class Task implements Solver { public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.readInt(); int a = in.readInt(); int b = in.readInt(); if (a==b) { b = 0; } boolean[] where = new boolean[n]; HashSet<Integer> set = new HashSet<Integer>(); HashMap<Integer,Integer> indexmap = new HashMap<Integer,Integer>(); for (int i = 0; i<n; i++) { int x = in.readInt(); indexmap.put(x, i); set.add(x); } while (set.size() > 0) { int size = set.size(); HashSet<Integer> todo = new HashSet<Integer>(); HashSet<Integer> used = new HashSet<Integer>(); for (int x : set) { if (used.contains(x)) continue; int ax = a-x; int bx = b-x; if ((set.contains(ax) && !used.contains(ax)) && (set.contains(bx) && !used.contains(bx))) { todo.add(x); } else if (set.contains(ax) && !used.contains(ax)) { used.add(x); used.add(ax); todo.remove(ax); //chain bx = b-ax; while (set.contains(bx) && !used.contains(bx)) { x = bx; ax = a-x; if (!set.contains(ax) || used.contains(ax)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(ax); used.add(x); used.add(ax); bx = b-ax; } } else if (set.contains(bx) && !used.contains(bx)) { used.add(x); used.add(bx); todo.remove(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; //chain ax = a-bx; while (set.contains(ax) && !used.contains(ax)) { x = ax; bx = b-x; if (!set.contains(bx) || used.contains(bx)) { System.out.println("NO"); return; } todo.remove(x); todo.remove(bx); used.add(x); used.add(bx); where[indexmap.get(bx)] = true; where[indexmap.get(x)] = true; ax = a-bx; } } else { System.out.println("NO"); return; } } set = todo; if (set.size() == size) { System.out.println("Set size constant!!"); break; } } System.out.println("YES"); for (int i = 0; i<n; i++) if (where[i]) System.out.print("1 "); else System.out.print("0 "); } } class num { } /// </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,659
376
2,733
import java.util.*; public class maximus { static long GCD(long a,long b){ if(b==0)return a; return GCD(b,a%b); } public static void main(String [] args){ Scanner in=new Scanner(System.in); long n=in.nextInt(); if(n<=2){ System.out.print(n); return; } if(n%2==1){ System.out.print((n*(n-1)*(n-2))); return; } if(n%2==0 && n<=6){ System.out.print(n*(n-1)*(n-2)/2); return; } long temp=(n*(n-1)*(n-3))/GCD(n,n-3); System.out.print(Math.max((n-1)*(n-2)*(n-3),temp)); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; public class maximus { static long GCD(long a,long b){ if(b==0)return a; return GCD(b,a%b); } public static void main(String [] args){ Scanner in=new Scanner(System.in); long n=in.nextInt(); if(n<=2){ System.out.print(n); return; } if(n%2==1){ System.out.print((n*(n-1)*(n-2))); return; } if(n%2==0 && n<=6){ System.out.print(n*(n-1)*(n-2)/2); return; } long temp=(n*(n-1)*(n-3))/GCD(n,n-3); System.out.print(Math.max((n-1)*(n-2)*(n-3),temp)); } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.*; public class maximus { static long GCD(long a,long b){ if(b==0)return a; return GCD(b,a%b); } public static void main(String [] args){ Scanner in=new Scanner(System.in); long n=in.nextInt(); if(n<=2){ System.out.print(n); return; } if(n%2==1){ System.out.print((n*(n-1)*(n-2))); return; } if(n%2==0 && n<=6){ System.out.print(n*(n-1)*(n-2)/2); return; } long temp=(n*(n-1)*(n-3))/GCD(n,n-3); System.out.print(Math.max((n-1)*(n-2)*(n-3),temp)); } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
534
2,727
4,072
import java.io.*; import java.util.*; public class Main { BufferedReader in = null; PrintWriter out = null; int dist(int x1, int y1, int x2, int y2) { int dx = Math.abs(x1 - x2); int dy = Math.abs(y1 - y2); return dx * dx + dy * dy; } boolean testBit(int use, int p) { return ((use >> p) & 1) == 1; } int rec(int use, int a[][], int dp[], int next[]) { if (dp[use] != -1) { return dp[use]; } if (use == 0) { return dp[use] = 0; } int n = a.length; int ix = -1; for (int i = 0; i < n; ++i) { if (testBit(use, i)) { if (ix == -1) { ix = i; break; } /* int t = rec(use ^ (1 << i), a, dp); t += a[i][i]; r = Math.min(r, t);*/ } } int r = rec(use ^ (1 << ix), a, dp, next) + a[ix][ix]; next[use] = use ^ (1 << ix); for (int i = ix + 1; i < n; ++i) { if (!testBit(use, i)) { continue; } int t = rec(use ^ (1 << ix) ^ (1 << i), a, dp, next); t += a[ix][i]; if (t < r) { r = t; next[use] = use ^ (1 << ix) ^ (1 << i); } } return dp[use] = r; } void print(int use1, int use2, int n) { for (int i = 0; i < n; ++i) { if (testBit(use1, i) ^ testBit(use2, i)) { int x = i + 1; out.print(x + " "); } } } void solve() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); //in = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter(System.out); StringTokenizer st; st = new StringTokenizer(in.readLine()); int sx = Integer.valueOf(st.nextToken()); int sy = Integer.valueOf(st.nextToken()); st = new StringTokenizer(in.readLine()); int n = Integer.valueOf(st.nextToken()); int x[] = new int[n]; int y[] = new int[n]; for (int i = 0; i < n; ++i) { st = new StringTokenizer(in.readLine()); x[i] = Integer.valueOf(st.nextToken()); y[i] = Integer.valueOf(st.nextToken()); } int a[][] = new int[n][n]; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { a[i][j] = dist(x[i], y[i], sx, sy) + dist(x[j], y[j], sx, sy) + dist(x[i], y[i], x[j], y[j]); } } int dp[] = new int[1 << n]; Arrays.fill(dp, -1); int next[] = new int[1 << n]; int ans = rec((1 << n) - 1, a, dp, next); out.println(ans); int use = (1 << n) - 1; while (true) { int nuse = next[use]; out.print("0 "); print(use, nuse, n); if (nuse == 0) break; use = nuse; } out.println("0"); in.close(); out.close(); } public static void main(String[] args) throws IOException { new Main().solve(); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { BufferedReader in = null; PrintWriter out = null; int dist(int x1, int y1, int x2, int y2) { int dx = Math.abs(x1 - x2); int dy = Math.abs(y1 - y2); return dx * dx + dy * dy; } boolean testBit(int use, int p) { return ((use >> p) & 1) == 1; } int rec(int use, int a[][], int dp[], int next[]) { if (dp[use] != -1) { return dp[use]; } if (use == 0) { return dp[use] = 0; } int n = a.length; int ix = -1; for (int i = 0; i < n; ++i) { if (testBit(use, i)) { if (ix == -1) { ix = i; break; } /* int t = rec(use ^ (1 << i), a, dp); t += a[i][i]; r = Math.min(r, t);*/ } } int r = rec(use ^ (1 << ix), a, dp, next) + a[ix][ix]; next[use] = use ^ (1 << ix); for (int i = ix + 1; i < n; ++i) { if (!testBit(use, i)) { continue; } int t = rec(use ^ (1 << ix) ^ (1 << i), a, dp, next); t += a[ix][i]; if (t < r) { r = t; next[use] = use ^ (1 << ix) ^ (1 << i); } } return dp[use] = r; } void print(int use1, int use2, int n) { for (int i = 0; i < n; ++i) { if (testBit(use1, i) ^ testBit(use2, i)) { int x = i + 1; out.print(x + " "); } } } void solve() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); //in = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter(System.out); StringTokenizer st; st = new StringTokenizer(in.readLine()); int sx = Integer.valueOf(st.nextToken()); int sy = Integer.valueOf(st.nextToken()); st = new StringTokenizer(in.readLine()); int n = Integer.valueOf(st.nextToken()); int x[] = new int[n]; int y[] = new int[n]; for (int i = 0; i < n; ++i) { st = new StringTokenizer(in.readLine()); x[i] = Integer.valueOf(st.nextToken()); y[i] = Integer.valueOf(st.nextToken()); } int a[][] = new int[n][n]; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { a[i][j] = dist(x[i], y[i], sx, sy) + dist(x[j], y[j], sx, sy) + dist(x[i], y[i], x[j], y[j]); } } int dp[] = new int[1 << n]; Arrays.fill(dp, -1); int next[] = new int[1 << n]; int ans = rec((1 << n) - 1, a, dp, next); out.println(ans); int use = (1 << n) - 1; while (true) { int nuse = next[use]; out.print("0 "); print(use, nuse, n); if (nuse == 0) break; use = nuse; } out.println("0"); in.close(); out.close(); } public static void main(String[] args) throws IOException { new Main().solve(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { BufferedReader in = null; PrintWriter out = null; int dist(int x1, int y1, int x2, int y2) { int dx = Math.abs(x1 - x2); int dy = Math.abs(y1 - y2); return dx * dx + dy * dy; } boolean testBit(int use, int p) { return ((use >> p) & 1) == 1; } int rec(int use, int a[][], int dp[], int next[]) { if (dp[use] != -1) { return dp[use]; } if (use == 0) { return dp[use] = 0; } int n = a.length; int ix = -1; for (int i = 0; i < n; ++i) { if (testBit(use, i)) { if (ix == -1) { ix = i; break; } /* int t = rec(use ^ (1 << i), a, dp); t += a[i][i]; r = Math.min(r, t);*/ } } int r = rec(use ^ (1 << ix), a, dp, next) + a[ix][ix]; next[use] = use ^ (1 << ix); for (int i = ix + 1; i < n; ++i) { if (!testBit(use, i)) { continue; } int t = rec(use ^ (1 << ix) ^ (1 << i), a, dp, next); t += a[ix][i]; if (t < r) { r = t; next[use] = use ^ (1 << ix) ^ (1 << i); } } return dp[use] = r; } void print(int use1, int use2, int n) { for (int i = 0; i < n; ++i) { if (testBit(use1, i) ^ testBit(use2, i)) { int x = i + 1; out.print(x + " "); } } } void solve() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); //in = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter(System.out); StringTokenizer st; st = new StringTokenizer(in.readLine()); int sx = Integer.valueOf(st.nextToken()); int sy = Integer.valueOf(st.nextToken()); st = new StringTokenizer(in.readLine()); int n = Integer.valueOf(st.nextToken()); int x[] = new int[n]; int y[] = new int[n]; for (int i = 0; i < n; ++i) { st = new StringTokenizer(in.readLine()); x[i] = Integer.valueOf(st.nextToken()); y[i] = Integer.valueOf(st.nextToken()); } int a[][] = new int[n][n]; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { a[i][j] = dist(x[i], y[i], sx, sy) + dist(x[j], y[j], sx, sy) + dist(x[i], y[i], x[j], y[j]); } } int dp[] = new int[1 << n]; Arrays.fill(dp, -1); int next[] = new int[1 << n]; int ans = rec((1 << n) - 1, a, dp, next); out.println(ans); int use = (1 << n) - 1; while (true) { int nuse = next[use]; out.print("0 "); print(use, nuse, n); if (nuse == 0) break; use = nuse; } out.println("0"); in.close(); out.close(); } public static void main(String[] args) throws IOException { new Main().solve(); } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,220
4,061
2,142
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.io.OutputStream; import java.io.IOException; import java.io.InputStreamReader; import java.util.TreeSet; import java.io.BufferedOutputStream; import java.util.StringTokenizer; import java.io.BufferedReader; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author Aeroui */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; Kattio in = new Kattio(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, Kattio in, PrintWriter out) { int n = in.nextInt(); int r = in.nextInt(); double[] xs = new double[n]; double[] ys = new double[n]; TreeSet<TaskC.Pair> set = new TreeSet<>(); for (int i = 0; i < n; ++i) { xs[i] = in.nextDouble(); ys[i] = (double) Integer.MIN_VALUE; if (i == 0) { // the first one out.printf("%f", (double) r); ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); } else { for (TaskC.Pair p : set) { double maximum = p.x; double diffX = (xs[i] - maximum) * (xs[i] - maximum); if (diffX <= r * r * 4.0) { ys[i] = Math.max(ys[i], p.y + Math.sqrt(r * r * 4.0 - diffX)); continue; } } if (ys[i] < 0) ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); out.printf(" %f", ys[i]); } } } private static class Pair implements Comparable<TaskC.Pair> { double x; double y; public Pair(double x, double y) { this.x = x; this.y = y; } public int compareTo(TaskC.Pair p) { if (this.y - p.y < 0) return 1; return -1; } } } static class Kattio extends PrintWriter { private BufferedReader r; private String line; private StringTokenizer st; private String token; public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public int nextInt() { return Integer.parseInt(nextToken()); } public double nextDouble() { return Double.parseDouble(nextToken()); } private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.io.OutputStream; import java.io.IOException; import java.io.InputStreamReader; import java.util.TreeSet; import java.io.BufferedOutputStream; import java.util.StringTokenizer; import java.io.BufferedReader; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author Aeroui */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; Kattio in = new Kattio(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, Kattio in, PrintWriter out) { int n = in.nextInt(); int r = in.nextInt(); double[] xs = new double[n]; double[] ys = new double[n]; TreeSet<TaskC.Pair> set = new TreeSet<>(); for (int i = 0; i < n; ++i) { xs[i] = in.nextDouble(); ys[i] = (double) Integer.MIN_VALUE; if (i == 0) { // the first one out.printf("%f", (double) r); ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); } else { for (TaskC.Pair p : set) { double maximum = p.x; double diffX = (xs[i] - maximum) * (xs[i] - maximum); if (diffX <= r * r * 4.0) { ys[i] = Math.max(ys[i], p.y + Math.sqrt(r * r * 4.0 - diffX)); continue; } } if (ys[i] < 0) ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); out.printf(" %f", ys[i]); } } } private static class Pair implements Comparable<TaskC.Pair> { double x; double y; public Pair(double x, double y) { this.x = x; this.y = y; } public int compareTo(TaskC.Pair p) { if (this.y - p.y < 0) return 1; return -1; } } } static class Kattio extends PrintWriter { private BufferedReader r; private String line; private StringTokenizer st; private String token; public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public int nextInt() { return Integer.parseInt(nextToken()); } public double nextDouble() { return Double.parseDouble(nextToken()); } private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.io.OutputStream; import java.io.IOException; import java.io.InputStreamReader; import java.util.TreeSet; import java.io.BufferedOutputStream; import java.util.StringTokenizer; import java.io.BufferedReader; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author Aeroui */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; Kattio in = new Kattio(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, Kattio in, PrintWriter out) { int n = in.nextInt(); int r = in.nextInt(); double[] xs = new double[n]; double[] ys = new double[n]; TreeSet<TaskC.Pair> set = new TreeSet<>(); for (int i = 0; i < n; ++i) { xs[i] = in.nextDouble(); ys[i] = (double) Integer.MIN_VALUE; if (i == 0) { // the first one out.printf("%f", (double) r); ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); } else { for (TaskC.Pair p : set) { double maximum = p.x; double diffX = (xs[i] - maximum) * (xs[i] - maximum); if (diffX <= r * r * 4.0) { ys[i] = Math.max(ys[i], p.y + Math.sqrt(r * r * 4.0 - diffX)); continue; } } if (ys[i] < 0) ys[i] = (double) r; set.add(new TaskC.Pair(xs[i], ys[i])); out.printf(" %f", ys[i]); } } } private static class Pair implements Comparable<TaskC.Pair> { double x; double y; public Pair(double x, double y) { this.x = x; this.y = y; } public int compareTo(TaskC.Pair p) { if (this.y - p.y < 0) return 1; return -1; } } } static class Kattio extends PrintWriter { private BufferedReader r; private String line; private StringTokenizer st; private String token; public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public int nextInt() { return Integer.parseInt(nextToken()); } public double nextDouble() { return Double.parseDouble(nextToken()); } private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,118
2,138
761
//package baobab; import java.io.*; import java.util.*; public class C { public static void main(String[] args) { CIO io = new CIO(); try { Csolver solver = new Csolver(io); solver.solve(); } finally { io.close(); } } } class Csolver { CIO io; public Csolver(CIO io) { this.io = io; } public void solve() { int n = io.nextInt(); String input = io.next(); int[] count = new int[1000]; int pokemonsInRange = 0; int answer = Integer.MAX_VALUE; int a = 0; int b = 0; for (; b < n ; b++) { char end = input.charAt(b); if (count[end] == 0) { pokemonsInRange++; answer = Integer.MAX_VALUE; } count[end]++; while (count[input.charAt(a)] > 1) { count[input.charAt(a)]--; a++; } answer = Math.min(answer, b-a+1); } io.println(answer); } private static class Pair implements Comparable<Pair> { int id; long val; public Pair(long val, int id) { this.val = val; this.id = id; } @Override public int compareTo(Pair o) { if (this.val < o.val) return -1; if (this.val > o.val) return 1; return this.id - o.id; } } private List<Integer>[] toGraph(CIO io, int n) { List<Integer>[] g = new ArrayList[n+1]; for (int i=1; i<=n; i++) g[i] = new ArrayList<>(); for (int i=1; i<=n-1; i++) { int a = io.nextInt(); int b = io.nextInt(); g[a].add(b); g[b].add(a); } return g; } } class CIO extends PrintWriter { private InputStreamReader r; private static final int BUFSIZE = 1 << 15; private char[] buf; private int bufc; private int bufi; private StringBuilder sb; public CIO() { super(new BufferedOutputStream(System.out)); r = new InputStreamReader(System.in); buf = new char[BUFSIZE]; bufc = 0; bufi = 0; sb = new StringBuilder(); } private void fillBuf() throws IOException { bufi = 0; bufc = 0; while(bufc == 0) { bufc = r.read(buf, 0, BUFSIZE); if(bufc == -1) { bufc = 0; return; } } } private boolean pumpBuf() throws IOException { if(bufi == bufc) { fillBuf(); } return bufc != 0; } private boolean isDelimiter(char c) { return c == ' ' || c == '\t' || c == '\n' || c == '\r' || c == '\f'; } private void eatDelimiters() throws IOException { while(true) { if(bufi == bufc) { fillBuf(); if(bufc == 0) throw new RuntimeException("IO: Out of input."); } if(!isDelimiter(buf[bufi])) break; ++bufi; } } public String next() { try { sb.setLength(0); eatDelimiters(); int start = bufi; while(true) { if(bufi == bufc) { sb.append(buf, start, bufi - start); fillBuf(); start = 0; if(bufc == 0) break; } if(isDelimiter(buf[bufi])) break; ++bufi; } sb.append(buf, start, bufi - start); return sb.toString(); } catch(IOException e) { throw new RuntimeException("IO.next: Caught IOException."); } } public int nextInt() { try { int ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextInt: Invalid int."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextInt: Invalid int."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -214748364) throw new RuntimeException("IO.nextInt: Invalid int."); ret *= 10; ret -= (int)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextInt: Invalid int."); } else { throw new RuntimeException("IO.nextInt: Invalid int."); } ++bufi; } if(positive) { if(ret == -2147483648) throw new RuntimeException("IO.nextInt: Invalid int."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextInt: Caught IOException."); } } public long nextLong() { try { long ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextLong: Invalid long."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextLong: Invalid long."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -922337203685477580L) throw new RuntimeException("IO.nextLong: Invalid long."); ret *= 10; ret -= (long)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextLong: Invalid long."); } else { throw new RuntimeException("IO.nextLong: Invalid long."); } ++bufi; } if(positive) { if(ret == -9223372036854775808L) throw new RuntimeException("IO.nextLong: Invalid long."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextLong: Caught IOException."); } } public double nextDouble() { return Double.parseDouble(next()); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package baobab; import java.io.*; import java.util.*; public class C { public static void main(String[] args) { CIO io = new CIO(); try { Csolver solver = new Csolver(io); solver.solve(); } finally { io.close(); } } } class Csolver { CIO io; public Csolver(CIO io) { this.io = io; } public void solve() { int n = io.nextInt(); String input = io.next(); int[] count = new int[1000]; int pokemonsInRange = 0; int answer = Integer.MAX_VALUE; int a = 0; int b = 0; for (; b < n ; b++) { char end = input.charAt(b); if (count[end] == 0) { pokemonsInRange++; answer = Integer.MAX_VALUE; } count[end]++; while (count[input.charAt(a)] > 1) { count[input.charAt(a)]--; a++; } answer = Math.min(answer, b-a+1); } io.println(answer); } private static class Pair implements Comparable<Pair> { int id; long val; public Pair(long val, int id) { this.val = val; this.id = id; } @Override public int compareTo(Pair o) { if (this.val < o.val) return -1; if (this.val > o.val) return 1; return this.id - o.id; } } private List<Integer>[] toGraph(CIO io, int n) { List<Integer>[] g = new ArrayList[n+1]; for (int i=1; i<=n; i++) g[i] = new ArrayList<>(); for (int i=1; i<=n-1; i++) { int a = io.nextInt(); int b = io.nextInt(); g[a].add(b); g[b].add(a); } return g; } } class CIO extends PrintWriter { private InputStreamReader r; private static final int BUFSIZE = 1 << 15; private char[] buf; private int bufc; private int bufi; private StringBuilder sb; public CIO() { super(new BufferedOutputStream(System.out)); r = new InputStreamReader(System.in); buf = new char[BUFSIZE]; bufc = 0; bufi = 0; sb = new StringBuilder(); } private void fillBuf() throws IOException { bufi = 0; bufc = 0; while(bufc == 0) { bufc = r.read(buf, 0, BUFSIZE); if(bufc == -1) { bufc = 0; return; } } } private boolean pumpBuf() throws IOException { if(bufi == bufc) { fillBuf(); } return bufc != 0; } private boolean isDelimiter(char c) { return c == ' ' || c == '\t' || c == '\n' || c == '\r' || c == '\f'; } private void eatDelimiters() throws IOException { while(true) { if(bufi == bufc) { fillBuf(); if(bufc == 0) throw new RuntimeException("IO: Out of input."); } if(!isDelimiter(buf[bufi])) break; ++bufi; } } public String next() { try { sb.setLength(0); eatDelimiters(); int start = bufi; while(true) { if(bufi == bufc) { sb.append(buf, start, bufi - start); fillBuf(); start = 0; if(bufc == 0) break; } if(isDelimiter(buf[bufi])) break; ++bufi; } sb.append(buf, start, bufi - start); return sb.toString(); } catch(IOException e) { throw new RuntimeException("IO.next: Caught IOException."); } } public int nextInt() { try { int ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextInt: Invalid int."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextInt: Invalid int."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -214748364) throw new RuntimeException("IO.nextInt: Invalid int."); ret *= 10; ret -= (int)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextInt: Invalid int."); } else { throw new RuntimeException("IO.nextInt: Invalid int."); } ++bufi; } if(positive) { if(ret == -2147483648) throw new RuntimeException("IO.nextInt: Invalid int."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextInt: Caught IOException."); } } public long nextLong() { try { long ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextLong: Invalid long."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextLong: Invalid long."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -922337203685477580L) throw new RuntimeException("IO.nextLong: Invalid long."); ret *= 10; ret -= (long)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextLong: Invalid long."); } else { throw new RuntimeException("IO.nextLong: Invalid long."); } ++bufi; } if(positive) { if(ret == -9223372036854775808L) throw new RuntimeException("IO.nextLong: Invalid long."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextLong: Caught IOException."); } } public double nextDouble() { return Double.parseDouble(next()); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> //package baobab; import java.io.*; import java.util.*; public class C { public static void main(String[] args) { CIO io = new CIO(); try { Csolver solver = new Csolver(io); solver.solve(); } finally { io.close(); } } } class Csolver { CIO io; public Csolver(CIO io) { this.io = io; } public void solve() { int n = io.nextInt(); String input = io.next(); int[] count = new int[1000]; int pokemonsInRange = 0; int answer = Integer.MAX_VALUE; int a = 0; int b = 0; for (; b < n ; b++) { char end = input.charAt(b); if (count[end] == 0) { pokemonsInRange++; answer = Integer.MAX_VALUE; } count[end]++; while (count[input.charAt(a)] > 1) { count[input.charAt(a)]--; a++; } answer = Math.min(answer, b-a+1); } io.println(answer); } private static class Pair implements Comparable<Pair> { int id; long val; public Pair(long val, int id) { this.val = val; this.id = id; } @Override public int compareTo(Pair o) { if (this.val < o.val) return -1; if (this.val > o.val) return 1; return this.id - o.id; } } private List<Integer>[] toGraph(CIO io, int n) { List<Integer>[] g = new ArrayList[n+1]; for (int i=1; i<=n; i++) g[i] = new ArrayList<>(); for (int i=1; i<=n-1; i++) { int a = io.nextInt(); int b = io.nextInt(); g[a].add(b); g[b].add(a); } return g; } } class CIO extends PrintWriter { private InputStreamReader r; private static final int BUFSIZE = 1 << 15; private char[] buf; private int bufc; private int bufi; private StringBuilder sb; public CIO() { super(new BufferedOutputStream(System.out)); r = new InputStreamReader(System.in); buf = new char[BUFSIZE]; bufc = 0; bufi = 0; sb = new StringBuilder(); } private void fillBuf() throws IOException { bufi = 0; bufc = 0; while(bufc == 0) { bufc = r.read(buf, 0, BUFSIZE); if(bufc == -1) { bufc = 0; return; } } } private boolean pumpBuf() throws IOException { if(bufi == bufc) { fillBuf(); } return bufc != 0; } private boolean isDelimiter(char c) { return c == ' ' || c == '\t' || c == '\n' || c == '\r' || c == '\f'; } private void eatDelimiters() throws IOException { while(true) { if(bufi == bufc) { fillBuf(); if(bufc == 0) throw new RuntimeException("IO: Out of input."); } if(!isDelimiter(buf[bufi])) break; ++bufi; } } public String next() { try { sb.setLength(0); eatDelimiters(); int start = bufi; while(true) { if(bufi == bufc) { sb.append(buf, start, bufi - start); fillBuf(); start = 0; if(bufc == 0) break; } if(isDelimiter(buf[bufi])) break; ++bufi; } sb.append(buf, start, bufi - start); return sb.toString(); } catch(IOException e) { throw new RuntimeException("IO.next: Caught IOException."); } } public int nextInt() { try { int ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextInt: Invalid int."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextInt: Invalid int."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -214748364) throw new RuntimeException("IO.nextInt: Invalid int."); ret *= 10; ret -= (int)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextInt: Invalid int."); } else { throw new RuntimeException("IO.nextInt: Invalid int."); } ++bufi; } if(positive) { if(ret == -2147483648) throw new RuntimeException("IO.nextInt: Invalid int."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextInt: Caught IOException."); } } public long nextLong() { try { long ret = 0; eatDelimiters(); boolean positive = true; if(buf[bufi] == '-') { ++bufi; if(!pumpBuf()) throw new RuntimeException("IO.nextLong: Invalid long."); positive = false; } boolean first = true; while(true) { if(!pumpBuf()) break; if(isDelimiter(buf[bufi])) { if(first) throw new RuntimeException("IO.nextLong: Invalid long."); break; } first = false; if(buf[bufi] >= '0' && buf[bufi] <= '9') { if(ret < -922337203685477580L) throw new RuntimeException("IO.nextLong: Invalid long."); ret *= 10; ret -= (long)(buf[bufi] - '0'); if(ret > 0) throw new RuntimeException("IO.nextLong: Invalid long."); } else { throw new RuntimeException("IO.nextLong: Invalid long."); } ++bufi; } if(positive) { if(ret == -9223372036854775808L) throw new RuntimeException("IO.nextLong: Invalid long."); ret = -ret; } return ret; } catch(IOException e) { throw new RuntimeException("IO.nextLong: Caught IOException."); } } public double nextDouble() { return Double.parseDouble(next()); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,864
760
561
import java.util.*; import java.lang.*; import java.io.*; public class Main { PrintWriter out = new PrintWriter(System.out); BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer tok = new StringTokenizer(""); String next() throws IOException { if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); } return tok.nextToken(); } int ni() throws IOException { return Integer.parseInt(next()); } long nl() throws IOException { return Long.parseLong(next()); } long mod=1000000007; void solve() throws IOException { for (int tc=ni();tc>0;tc--) { long n=ni(); long q=1; long p=1; boolean f=false; while (true) { if (p*2==n || p*4==n) { f=true; break; } q++; p=q*q; if (p>n) break; } if (f) out.println("YES"); else out.println("NO"); } out.flush(); } int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); } long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); } long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; } public static void main(String[] args) throws IOException { new Main().solve(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.io.*; public class Main { PrintWriter out = new PrintWriter(System.out); BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer tok = new StringTokenizer(""); String next() throws IOException { if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); } return tok.nextToken(); } int ni() throws IOException { return Integer.parseInt(next()); } long nl() throws IOException { return Long.parseLong(next()); } long mod=1000000007; void solve() throws IOException { for (int tc=ni();tc>0;tc--) { long n=ni(); long q=1; long p=1; boolean f=false; while (true) { if (p*2==n || p*4==n) { f=true; break; } q++; p=q*q; if (p>n) break; } if (f) out.println("YES"); else out.println("NO"); } out.flush(); } int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); } long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); } long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; } public static void main(String[] args) throws IOException { new Main().solve(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.io.*; public class Main { PrintWriter out = new PrintWriter(System.out); BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer tok = new StringTokenizer(""); String next() throws IOException { if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); } return tok.nextToken(); } int ni() throws IOException { return Integer.parseInt(next()); } long nl() throws IOException { return Long.parseLong(next()); } long mod=1000000007; void solve() throws IOException { for (int tc=ni();tc>0;tc--) { long n=ni(); long q=1; long p=1; boolean f=false; while (true) { if (p*2==n || p*4==n) { f=true; break; } q++; p=q*q; if (p>n) break; } if (f) out.println("YES"); else out.println("NO"); } out.flush(); } int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); } long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); } long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; } public static void main(String[] args) throws IOException { new Main().solve(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
680
560
2,432
import java.util.*; import java.io.*; import java.util.Queue; import java.util.LinkedList; import java.math.*; public class Sample implements Runnable { public static void solve() { int n=i(); int[] a=new int[n]; for(int i=0;i<n;i++)a[i]=i(); int temp=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { if(a[j]<a[i])temp++; } } boolean even=(temp%2==0)?true:false; int m=i(); while(m-->0) { int l=i(); int r=i(); long tt=(long)(Math.floor(r-l+1)/2); if(tt%2==1) { if(even) { out.println("odd"); even=false; } else { out.println("even"); even=true; } }else { if(even) { out.println("even"); } else { out.println("odd"); } } } } public void run() { solve(); out.close(); } public static void main(String[] args) throws IOException { new Thread(null, new Sample(), "whatever", 1<<26).start(); } abstract static class Pair implements Comparable<Pair> { long a; int b; Pair(){} Pair(long a,int b) { this.a=a; this.b=b; } public int compareTo(Pair x) { return Long.compare(x.a,this.a); } } ////////////////////////////////////////////////////// Merge Sort //////////////////////////////////////////////////////////////////////// static class Merge { public static void sort(long inputArr[]) { int length = inputArr.length; doMergeSort(inputArr,0, length - 1); } private static void doMergeSort(long[] arr,int lowerIndex, int higherIndex) { if (lowerIndex < higherIndex) { int middle = lowerIndex + (higherIndex - lowerIndex) / 2; doMergeSort(arr,lowerIndex, middle); doMergeSort(arr,middle + 1, higherIndex); mergeParts(arr,lowerIndex, middle, higherIndex); } } private static void mergeParts(long[]array,int lowerIndex, int middle, int higherIndex) { long[] temp=new long[higherIndex-lowerIndex+1]; for (int i = lowerIndex; i <= higherIndex; i++) { temp[i-lowerIndex] = array[i]; } int i = lowerIndex; int j = middle + 1; int k = lowerIndex; while (i <= middle && j <= higherIndex) { if (temp[i-lowerIndex] < temp[j-lowerIndex]) { array[k] = temp[i-lowerIndex]; i++; } else { array[k] = temp[j-lowerIndex]; j++; } k++; } while (i <= middle) { array[k] = temp[i-lowerIndex]; k++; i++; } while(j<=higherIndex) { array[k]=temp[j-lowerIndex]; k++; j++; } } } /////////////////////////////////////////////////////////// Methods //////////////////////////////////////////////////////////////////////// static boolean isPal(String s) { for(int i=0, j=s.length()-1;i<=j;i++,j--) { if(s.charAt(i)!=s.charAt(j)) return false; } return true; } static String rev(String s) { StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static int gcd(int a,int b){return (a==0)?b:gcd(b%a,a);} static long gcdExtended(long a,long b,long[] x) { if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } boolean findSum(int set[], int n, long sum) { if (sum == 0) return true; if (n == 0 && sum != 0) return false; if (set[n-1] > sum) return findSum(set, n-1, sum); return findSum(set, n-1, sum) ||findSum(set, n-1, sum-set[n-1]); } public static long modPow(long base, long exp, long mod) { base = base % mod; long result = 1; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % mod; } base = (base * base) % mod; exp = exp >> 1; } return result; } static long[] fac; static long[] inv; static long mod=(long)1e9+7; public static void cal() { fac = new long[1000005]; inv = new long[1000005]; fac[0] = 1; inv[0] = 1; for (int i = 1; i <= 1000000; i++) { fac[i] = (fac[i - 1] * i) % mod; inv[i] = (inv[i - 1] * modPow(i, mod - 2, mod)) % mod; } } public static long ncr(int n, int r) { return (((fac[n] * inv[r]) % mod) * inv[n - r]) % mod; } ////////////////////////////////////////// Input Reader //////////////////////////////////////////////////////////////////////////////////////////////////// static InputReader sc = new InputReader(System.in); static PrintWriter out= new PrintWriter(System.out); static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static int i() { return sc.nextInt(); } static long l(){ return sc.nextLong(); } static int[] iarr(int n) { return sc.nextIntArray(n); } static long[] larr(int n) { return sc.nextLongArray(n); } static String s(){ return sc.nextLine(); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.io.*; import java.util.Queue; import java.util.LinkedList; import java.math.*; public class Sample implements Runnable { public static void solve() { int n=i(); int[] a=new int[n]; for(int i=0;i<n;i++)a[i]=i(); int temp=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { if(a[j]<a[i])temp++; } } boolean even=(temp%2==0)?true:false; int m=i(); while(m-->0) { int l=i(); int r=i(); long tt=(long)(Math.floor(r-l+1)/2); if(tt%2==1) { if(even) { out.println("odd"); even=false; } else { out.println("even"); even=true; } }else { if(even) { out.println("even"); } else { out.println("odd"); } } } } public void run() { solve(); out.close(); } public static void main(String[] args) throws IOException { new Thread(null, new Sample(), "whatever", 1<<26).start(); } abstract static class Pair implements Comparable<Pair> { long a; int b; Pair(){} Pair(long a,int b) { this.a=a; this.b=b; } public int compareTo(Pair x) { return Long.compare(x.a,this.a); } } ////////////////////////////////////////////////////// Merge Sort //////////////////////////////////////////////////////////////////////// static class Merge { public static void sort(long inputArr[]) { int length = inputArr.length; doMergeSort(inputArr,0, length - 1); } private static void doMergeSort(long[] arr,int lowerIndex, int higherIndex) { if (lowerIndex < higherIndex) { int middle = lowerIndex + (higherIndex - lowerIndex) / 2; doMergeSort(arr,lowerIndex, middle); doMergeSort(arr,middle + 1, higherIndex); mergeParts(arr,lowerIndex, middle, higherIndex); } } private static void mergeParts(long[]array,int lowerIndex, int middle, int higherIndex) { long[] temp=new long[higherIndex-lowerIndex+1]; for (int i = lowerIndex; i <= higherIndex; i++) { temp[i-lowerIndex] = array[i]; } int i = lowerIndex; int j = middle + 1; int k = lowerIndex; while (i <= middle && j <= higherIndex) { if (temp[i-lowerIndex] < temp[j-lowerIndex]) { array[k] = temp[i-lowerIndex]; i++; } else { array[k] = temp[j-lowerIndex]; j++; } k++; } while (i <= middle) { array[k] = temp[i-lowerIndex]; k++; i++; } while(j<=higherIndex) { array[k]=temp[j-lowerIndex]; k++; j++; } } } /////////////////////////////////////////////////////////// Methods //////////////////////////////////////////////////////////////////////// static boolean isPal(String s) { for(int i=0, j=s.length()-1;i<=j;i++,j--) { if(s.charAt(i)!=s.charAt(j)) return false; } return true; } static String rev(String s) { StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static int gcd(int a,int b){return (a==0)?b:gcd(b%a,a);} static long gcdExtended(long a,long b,long[] x) { if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } boolean findSum(int set[], int n, long sum) { if (sum == 0) return true; if (n == 0 && sum != 0) return false; if (set[n-1] > sum) return findSum(set, n-1, sum); return findSum(set, n-1, sum) ||findSum(set, n-1, sum-set[n-1]); } public static long modPow(long base, long exp, long mod) { base = base % mod; long result = 1; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % mod; } base = (base * base) % mod; exp = exp >> 1; } return result; } static long[] fac; static long[] inv; static long mod=(long)1e9+7; public static void cal() { fac = new long[1000005]; inv = new long[1000005]; fac[0] = 1; inv[0] = 1; for (int i = 1; i <= 1000000; i++) { fac[i] = (fac[i - 1] * i) % mod; inv[i] = (inv[i - 1] * modPow(i, mod - 2, mod)) % mod; } } public static long ncr(int n, int r) { return (((fac[n] * inv[r]) % mod) * inv[n - r]) % mod; } ////////////////////////////////////////// Input Reader //////////////////////////////////////////////////////////////////////////////////////////////////// static InputReader sc = new InputReader(System.in); static PrintWriter out= new PrintWriter(System.out); static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static int i() { return sc.nextInt(); } static long l(){ return sc.nextLong(); } static int[] iarr(int n) { return sc.nextIntArray(n); } static long[] larr(int n) { return sc.nextLongArray(n); } static String s(){ return sc.nextLine(); } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.io.*; import java.util.Queue; import java.util.LinkedList; import java.math.*; public class Sample implements Runnable { public static void solve() { int n=i(); int[] a=new int[n]; for(int i=0;i<n;i++)a[i]=i(); int temp=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { if(a[j]<a[i])temp++; } } boolean even=(temp%2==0)?true:false; int m=i(); while(m-->0) { int l=i(); int r=i(); long tt=(long)(Math.floor(r-l+1)/2); if(tt%2==1) { if(even) { out.println("odd"); even=false; } else { out.println("even"); even=true; } }else { if(even) { out.println("even"); } else { out.println("odd"); } } } } public void run() { solve(); out.close(); } public static void main(String[] args) throws IOException { new Thread(null, new Sample(), "whatever", 1<<26).start(); } abstract static class Pair implements Comparable<Pair> { long a; int b; Pair(){} Pair(long a,int b) { this.a=a; this.b=b; } public int compareTo(Pair x) { return Long.compare(x.a,this.a); } } ////////////////////////////////////////////////////// Merge Sort //////////////////////////////////////////////////////////////////////// static class Merge { public static void sort(long inputArr[]) { int length = inputArr.length; doMergeSort(inputArr,0, length - 1); } private static void doMergeSort(long[] arr,int lowerIndex, int higherIndex) { if (lowerIndex < higherIndex) { int middle = lowerIndex + (higherIndex - lowerIndex) / 2; doMergeSort(arr,lowerIndex, middle); doMergeSort(arr,middle + 1, higherIndex); mergeParts(arr,lowerIndex, middle, higherIndex); } } private static void mergeParts(long[]array,int lowerIndex, int middle, int higherIndex) { long[] temp=new long[higherIndex-lowerIndex+1]; for (int i = lowerIndex; i <= higherIndex; i++) { temp[i-lowerIndex] = array[i]; } int i = lowerIndex; int j = middle + 1; int k = lowerIndex; while (i <= middle && j <= higherIndex) { if (temp[i-lowerIndex] < temp[j-lowerIndex]) { array[k] = temp[i-lowerIndex]; i++; } else { array[k] = temp[j-lowerIndex]; j++; } k++; } while (i <= middle) { array[k] = temp[i-lowerIndex]; k++; i++; } while(j<=higherIndex) { array[k]=temp[j-lowerIndex]; k++; j++; } } } /////////////////////////////////////////////////////////// Methods //////////////////////////////////////////////////////////////////////// static boolean isPal(String s) { for(int i=0, j=s.length()-1;i<=j;i++,j--) { if(s.charAt(i)!=s.charAt(j)) return false; } return true; } static String rev(String s) { StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static int gcd(int a,int b){return (a==0)?b:gcd(b%a,a);} static long gcdExtended(long a,long b,long[] x) { if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } boolean findSum(int set[], int n, long sum) { if (sum == 0) return true; if (n == 0 && sum != 0) return false; if (set[n-1] > sum) return findSum(set, n-1, sum); return findSum(set, n-1, sum) ||findSum(set, n-1, sum-set[n-1]); } public static long modPow(long base, long exp, long mod) { base = base % mod; long result = 1; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % mod; } base = (base * base) % mod; exp = exp >> 1; } return result; } static long[] fac; static long[] inv; static long mod=(long)1e9+7; public static void cal() { fac = new long[1000005]; inv = new long[1000005]; fac[0] = 1; inv[0] = 1; for (int i = 1; i <= 1000000; i++) { fac[i] = (fac[i - 1] * i) % mod; inv[i] = (inv[i - 1] * modPow(i, mod - 2, mod)) % mod; } } public static long ncr(int n, int r) { return (((fac[n] * inv[r]) % mod) * inv[n - r]) % mod; } ////////////////////////////////////////// Input Reader //////////////////////////////////////////////////////////////////////////////////////////////////// static InputReader sc = new InputReader(System.in); static PrintWriter out= new PrintWriter(System.out); static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static int i() { return sc.nextInt(); } static long l(){ return sc.nextLong(); } static int[] iarr(int n) { return sc.nextIntArray(n); } static long[] larr(int n) { return sc.nextLongArray(n); } static String s(){ return sc.nextLine(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
2,559
2,427
586
import java.util.*; import java.io.*; public class A6 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int d = in.nextInt(); int ans=2; int[] a = new int[n]; for(int i=0;i<n;i++) a[i] = in.nextInt(); for(int i=1;i<n;i++) { if(a[i]-a[i-1]>2*d) { ans += 2; } else if(a[i]-a[i-1]==2*d) ans += 1; } System.out.println(ans); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; public class A6 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int d = in.nextInt(); int ans=2; int[] a = new int[n]; for(int i=0;i<n;i++) a[i] = in.nextInt(); for(int i=1;i<n;i++) { if(a[i]-a[i-1]>2*d) { ans += 2; } else if(a[i]-a[i-1]==2*d) ans += 1; } System.out.println(ans); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; public class A6 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int d = in.nextInt(); int ans=2; int[] a = new int[n]; for(int i=0;i<n;i++) a[i] = in.nextInt(); for(int i=1;i<n;i++) { if(a[i]-a[i-1]>2*d) { ans += 2; } else if(a[i]-a[i-1]==2*d) ans += 1; } System.out.println(ans); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
487
585
2,822
import java.util.Scanner; public class subtractions { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); while(n-->0){ int a=sc.nextInt(); int b=sc.nextInt(); int c=0; while(a!=0 && b!=0){ if(a>b){ int t=a; a=b; b=t; } c+=b/a; b=b%a; } System.out.println(c); } } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Scanner; public class subtractions { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); while(n-->0){ int a=sc.nextInt(); int b=sc.nextInt(); int c=0; while(a!=0 && b!=0){ if(a>b){ int t=a; a=b; b=t; } c+=b/a; b=b%a; } System.out.println(c); } } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Scanner; public class subtractions { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); while(n-->0){ int a=sc.nextInt(); int b=sc.nextInt(); int c=0; while(a!=0 && b!=0){ if(a>b){ int t=a; a=b; b=t; } c+=b/a; b=b%a; } System.out.println(c); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
467
2,816
631
import java.util.Scanner; public class R025A { int n; int[] nums; public R025A() { Scanner scanner = new Scanner(System.in); n = scanner.nextInt(); nums = new int[n]; for(int i=0; i<n; i++) { nums[i] = scanner.nextInt(); } } private void process() { int[] c = new int[2]; int[] r = new int[2]; for(int i=0; i<n; i++) { c[nums[i] % 2]++; if(r[nums[i] %2] == 0) { r[nums[i] % 2] = i+1; } } System.out.println(r[c[0]==1 ? 0 : 1]); } public static void main(String[] args) { new R025A().process(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Scanner; public class R025A { int n; int[] nums; public R025A() { Scanner scanner = new Scanner(System.in); n = scanner.nextInt(); nums = new int[n]; for(int i=0; i<n; i++) { nums[i] = scanner.nextInt(); } } private void process() { int[] c = new int[2]; int[] r = new int[2]; for(int i=0; i<n; i++) { c[nums[i] % 2]++; if(r[nums[i] %2] == 0) { r[nums[i] % 2] = i+1; } } System.out.println(r[c[0]==1 ? 0 : 1]); } public static void main(String[] args) { new R025A().process(); } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class R025A { int n; int[] nums; public R025A() { Scanner scanner = new Scanner(System.in); n = scanner.nextInt(); nums = new int[n]; for(int i=0; i<n; i++) { nums[i] = scanner.nextInt(); } } private void process() { int[] c = new int[2]; int[] r = new int[2]; for(int i=0; i<n; i++) { c[nums[i] % 2]++; if(r[nums[i] %2] == 0) { r[nums[i] % 2] = i+1; } } System.out.println(r[c[0]==1 ? 0 : 1]); } public static void main(String[] args) { new R025A().process(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
554
630
592
import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; public class A { static StringTokenizer st; static BufferedReader br; static PrintWriter pw; public static void main(String[] args) throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int n = nextInt(); int d = nextInt(); int[]x = new int[n]; for (int i = 0; i < n; i++) { x[i] = nextInt(); } int ans = 2; for (int i = 1; i < n; i++) { if (x[i]-x[i-1]==2*d) ans++; else if (x[i]-x[i-1] > 2*d) ans += 2; } System.out.println(ans); pw.close(); } private static int nextInt() throws IOException { return Integer.parseInt(next()); } private static long nextLong() throws IOException { return Long.parseLong(next()); } private static double nextDouble() throws IOException { return Double.parseDouble(next()); } private static String next() throws IOException { while (st==null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; public class A { static StringTokenizer st; static BufferedReader br; static PrintWriter pw; public static void main(String[] args) throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int n = nextInt(); int d = nextInt(); int[]x = new int[n]; for (int i = 0; i < n; i++) { x[i] = nextInt(); } int ans = 2; for (int i = 1; i < n; i++) { if (x[i]-x[i-1]==2*d) ans++; else if (x[i]-x[i-1] > 2*d) ans += 2; } System.out.println(ans); pw.close(); } private static int nextInt() throws IOException { return Integer.parseInt(next()); } private static long nextLong() throws IOException { return Long.parseLong(next()); } private static double nextDouble() throws IOException { return Double.parseDouble(next()); } private static String next() throws IOException { while (st==null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; public class A { static StringTokenizer st; static BufferedReader br; static PrintWriter pw; public static void main(String[] args) throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int n = nextInt(); int d = nextInt(); int[]x = new int[n]; for (int i = 0; i < n; i++) { x[i] = nextInt(); } int ans = 2; for (int i = 1; i < n; i++) { if (x[i]-x[i-1]==2*d) ans++; else if (x[i]-x[i-1] > 2*d) ans += 2; } System.out.println(ans); pw.close(); } private static int nextInt() throws IOException { return Integer.parseInt(next()); } private static long nextLong() throws IOException { return Long.parseLong(next()); } private static double nextDouble() throws IOException { return Double.parseDouble(next()); } private static String next() throws IOException { while (st==null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
654
591
3,552
import java.util.LinkedList; import java.io.FileInputStream; import java.util.Arrays; import java.util.ArrayList; import java.io.InputStream; import java.io.InputStreamReader; import java.io.BufferedReader; import java.util.Collection; import java.io.OutputStream; import java.io.PrintWriter; import java.util.Queue; import java.io.IOException; import java.io.FileOutputStream; import java.util.StringTokenizer; /** * Built using CHelper plug-in * Actual solution is at the top * @author nasko */ public class Main { public static void main(String[] args) { InputStream inputStream; try { inputStream = new FileInputStream("input.txt"); } catch (IOException e) { throw new RuntimeException(e); } OutputStream outputStream; try { outputStream = new FileOutputStream("output.txt"); } catch (IOException e) { throw new RuntimeException(e); } InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } } class TaskC { public void solve(int testNumber, InputReader in, PrintWriter out) { int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[][] dist = new int[N+1][M+1]; for(int[] ini : dist) Arrays.fill(ini,1 << 30); int best = 0; ArrayList<Integer> result = new ArrayList<Integer>(); Queue<Integer> q = new LinkedList<Integer>(); for(int k = 0; k < K; ++k) { int x = in.nextInt(); int y = in.nextInt(); dist[x][y] = 0; q.offer(x); q.offer(y); } int[] dx = new int[] { 1,-1,0,0 }; int[] dy = new int[] {0,0,1,-1}; while(!q.isEmpty()) { int a = q.poll(); int b = q.poll(); for(int r = 0; r < 4; ++r) { int x = a + dx[r]; int y = b + dy[r]; if(x >= 1 && x <= N && y >=1 && y <= M && dist[x][y] > dist[a][b] + 1) { dist[x][y] = dist[a][b] + 1; q.offer(x); q.offer(y); } } } for(int i = 1; i <= N; ++i) for(int j = 1; j <= M; ++j) best = Math.max(best,dist[i][j]); for(int a = 1; a <= N; ++a) for(int b = 1; b <= M; ++b) if(dist[a][b] == best) { result.add(a); result.add(b); } if(result.size() > 0) { out.print(result.get(0) + " " + result.get(1)); } out.println(); } } class InputReader { public BufferedReader reader; public StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.LinkedList; import java.io.FileInputStream; import java.util.Arrays; import java.util.ArrayList; import java.io.InputStream; import java.io.InputStreamReader; import java.io.BufferedReader; import java.util.Collection; import java.io.OutputStream; import java.io.PrintWriter; import java.util.Queue; import java.io.IOException; import java.io.FileOutputStream; import java.util.StringTokenizer; /** * Built using CHelper plug-in * Actual solution is at the top * @author nasko */ public class Main { public static void main(String[] args) { InputStream inputStream; try { inputStream = new FileInputStream("input.txt"); } catch (IOException e) { throw new RuntimeException(e); } OutputStream outputStream; try { outputStream = new FileOutputStream("output.txt"); } catch (IOException e) { throw new RuntimeException(e); } InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } } class TaskC { public void solve(int testNumber, InputReader in, PrintWriter out) { int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[][] dist = new int[N+1][M+1]; for(int[] ini : dist) Arrays.fill(ini,1 << 30); int best = 0; ArrayList<Integer> result = new ArrayList<Integer>(); Queue<Integer> q = new LinkedList<Integer>(); for(int k = 0; k < K; ++k) { int x = in.nextInt(); int y = in.nextInt(); dist[x][y] = 0; q.offer(x); q.offer(y); } int[] dx = new int[] { 1,-1,0,0 }; int[] dy = new int[] {0,0,1,-1}; while(!q.isEmpty()) { int a = q.poll(); int b = q.poll(); for(int r = 0; r < 4; ++r) { int x = a + dx[r]; int y = b + dy[r]; if(x >= 1 && x <= N && y >=1 && y <= M && dist[x][y] > dist[a][b] + 1) { dist[x][y] = dist[a][b] + 1; q.offer(x); q.offer(y); } } } for(int i = 1; i <= N; ++i) for(int j = 1; j <= M; ++j) best = Math.max(best,dist[i][j]); for(int a = 1; a <= N; ++a) for(int b = 1; b <= M; ++b) if(dist[a][b] == best) { result.add(a); result.add(b); } if(result.size() > 0) { out.print(result.get(0) + " " + result.get(1)); } out.println(); } } class InputReader { public BufferedReader reader; public StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.LinkedList; import java.io.FileInputStream; import java.util.Arrays; import java.util.ArrayList; import java.io.InputStream; import java.io.InputStreamReader; import java.io.BufferedReader; import java.util.Collection; import java.io.OutputStream; import java.io.PrintWriter; import java.util.Queue; import java.io.IOException; import java.io.FileOutputStream; import java.util.StringTokenizer; /** * Built using CHelper plug-in * Actual solution is at the top * @author nasko */ public class Main { public static void main(String[] args) { InputStream inputStream; try { inputStream = new FileInputStream("input.txt"); } catch (IOException e) { throw new RuntimeException(e); } OutputStream outputStream; try { outputStream = new FileOutputStream("output.txt"); } catch (IOException e) { throw new RuntimeException(e); } InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } } class TaskC { public void solve(int testNumber, InputReader in, PrintWriter out) { int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); int[][] dist = new int[N+1][M+1]; for(int[] ini : dist) Arrays.fill(ini,1 << 30); int best = 0; ArrayList<Integer> result = new ArrayList<Integer>(); Queue<Integer> q = new LinkedList<Integer>(); for(int k = 0; k < K; ++k) { int x = in.nextInt(); int y = in.nextInt(); dist[x][y] = 0; q.offer(x); q.offer(y); } int[] dx = new int[] { 1,-1,0,0 }; int[] dy = new int[] {0,0,1,-1}; while(!q.isEmpty()) { int a = q.poll(); int b = q.poll(); for(int r = 0; r < 4; ++r) { int x = a + dx[r]; int y = b + dy[r]; if(x >= 1 && x <= N && y >=1 && y <= M && dist[x][y] > dist[a][b] + 1) { dist[x][y] = dist[a][b] + 1; q.offer(x); q.offer(y); } } } for(int i = 1; i <= N; ++i) for(int j = 1; j <= M; ++j) best = Math.max(best,dist[i][j]); for(int a = 1; a <= N; ++a) for(int b = 1; b <= M; ++b) if(dist[a][b] == best) { result.add(a); result.add(b); } if(result.size() > 0) { out.print(result.get(0) + " " + result.get(1)); } out.println(); } } class InputReader { public BufferedReader reader; public StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } } </CODE> <EVALUATION_RUBRIC> - O(n): The running time grows linearly with the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,128
3,545
1,109
import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; public class Div1_503B { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int N = Integer.parseInt(reader.readLine()); printer.println("? 1"); printer.flush(); int v1 = Integer.parseInt(reader.readLine()); printer.println("? " + (1 + N / 2)); printer.flush(); int v2 = Integer.parseInt(reader.readLine()); if ((v1 + v2) % 2 != 0) { printer.println("! -1"); printer.close(); return; } if (v1 == v2) { printer.println("! 1"); printer.close(); return; } boolean less = v1 < v2; int low = 1; int high = (1 + N / 2); while (low != high) { int mid = (low + high) >> 1; printer.println("? " + mid); printer.flush(); int r1 = Integer.parseInt(reader.readLine()); int q2 = (mid + N / 2); if (q2 > N) { q2 -= N; } printer.println("? " + q2); printer.flush(); int r2 = Integer.parseInt(reader.readLine()); if (r1 == r2) { printer.println("! " + mid); printer.close(); return; } if (r1 < r2 == less) { low = mid + 1; } else { high = mid - 1; } } printer.println("! " + low); printer.close(); return; } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; public class Div1_503B { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int N = Integer.parseInt(reader.readLine()); printer.println("? 1"); printer.flush(); int v1 = Integer.parseInt(reader.readLine()); printer.println("? " + (1 + N / 2)); printer.flush(); int v2 = Integer.parseInt(reader.readLine()); if ((v1 + v2) % 2 != 0) { printer.println("! -1"); printer.close(); return; } if (v1 == v2) { printer.println("! 1"); printer.close(); return; } boolean less = v1 < v2; int low = 1; int high = (1 + N / 2); while (low != high) { int mid = (low + high) >> 1; printer.println("? " + mid); printer.flush(); int r1 = Integer.parseInt(reader.readLine()); int q2 = (mid + N / 2); if (q2 > N) { q2 -= N; } printer.println("? " + q2); printer.flush(); int r2 = Integer.parseInt(reader.readLine()); if (r1 == r2) { printer.println("! " + mid); printer.close(); return; } if (r1 < r2 == less) { low = mid + 1; } else { high = mid - 1; } } printer.println("! " + low); printer.close(); return; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; public class Div1_503B { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int N = Integer.parseInt(reader.readLine()); printer.println("? 1"); printer.flush(); int v1 = Integer.parseInt(reader.readLine()); printer.println("? " + (1 + N / 2)); printer.flush(); int v2 = Integer.parseInt(reader.readLine()); if ((v1 + v2) % 2 != 0) { printer.println("! -1"); printer.close(); return; } if (v1 == v2) { printer.println("! 1"); printer.close(); return; } boolean less = v1 < v2; int low = 1; int high = (1 + N / 2); while (low != high) { int mid = (low + high) >> 1; printer.println("? " + mid); printer.flush(); int r1 = Integer.parseInt(reader.readLine()); int q2 = (mid + N / 2); if (q2 > N) { q2 -= N; } printer.println("? " + q2); printer.flush(); int r2 = Integer.parseInt(reader.readLine()); if (r1 == r2) { printer.println("! " + mid); printer.close(); return; } if (r1 < r2 == less) { low = mid + 1; } else { high = mid - 1; } } printer.println("! " + low); printer.close(); return; } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
759
1,108
2,406
/* * Author Ayub Subhaniya * Institute DA-IICT */ import java.io.*; import java.math.*; import java.util.*; public class A { InputStream in; PrintWriter out; void solve() { int n=ni(); int a[]=na(n); int INV=0; for (int i=0;i<n;i++) for (int j=i+1;j<n;j++) if (a[i]>a[j]) INV++; boolean even=INV%2==0; int q=ni(); while (q-->0) { int l=ni(); int r=ni(); int len=r-l+1; len=(len-1)*(len)/2; if (len%2==1) even=!even; if (even) out.println("even"); else out.println("odd"); } } int MAX = (int)1e5; long factorial[]; void findfactorial() { factorial = new long[MAX + 1]; factorial[0] = 1; for (int i = 1; i < MAX + 1; i++) { factorial[i] = mul(i,factorial[i - 1]); } } long mod=(long)1e9+7; long add(long a,long b) { long x=(a+b); while(x>=mod) x-=mod; return x; } long sub(long a,long b) { long x=(a-b); while(x<0) x+=mod; return x; } long mul(long a,long b) { a%=mod; b%=mod; long x=(a*b); return x%mod; } int max(int a,int b) { if(a>b) return a; else return b; } int min(int a,int b) { if(a>b) return b; else return a; } long max(long a,long b) { if(a>b) return a; else return b; } long min(long a,long b) { if(a>b) return b; else return a; } void run() throws Exception { String INPUT = "C:/Users/ayubs/Desktop/input.txt"; in = oj ? System.in : new FileInputStream(INPUT); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new A().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = in.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean inSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && inSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(inSpaceChar(b))) { // when nextLine, (inSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(inSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* * Author Ayub Subhaniya * Institute DA-IICT */ import java.io.*; import java.math.*; import java.util.*; public class A { InputStream in; PrintWriter out; void solve() { int n=ni(); int a[]=na(n); int INV=0; for (int i=0;i<n;i++) for (int j=i+1;j<n;j++) if (a[i]>a[j]) INV++; boolean even=INV%2==0; int q=ni(); while (q-->0) { int l=ni(); int r=ni(); int len=r-l+1; len=(len-1)*(len)/2; if (len%2==1) even=!even; if (even) out.println("even"); else out.println("odd"); } } int MAX = (int)1e5; long factorial[]; void findfactorial() { factorial = new long[MAX + 1]; factorial[0] = 1; for (int i = 1; i < MAX + 1; i++) { factorial[i] = mul(i,factorial[i - 1]); } } long mod=(long)1e9+7; long add(long a,long b) { long x=(a+b); while(x>=mod) x-=mod; return x; } long sub(long a,long b) { long x=(a-b); while(x<0) x+=mod; return x; } long mul(long a,long b) { a%=mod; b%=mod; long x=(a*b); return x%mod; } int max(int a,int b) { if(a>b) return a; else return b; } int min(int a,int b) { if(a>b) return b; else return a; } long max(long a,long b) { if(a>b) return a; else return b; } long min(long a,long b) { if(a>b) return b; else return a; } void run() throws Exception { String INPUT = "C:/Users/ayubs/Desktop/input.txt"; in = oj ? System.in : new FileInputStream(INPUT); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new A().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = in.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean inSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && inSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(inSpaceChar(b))) { // when nextLine, (inSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(inSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* * Author Ayub Subhaniya * Institute DA-IICT */ import java.io.*; import java.math.*; import java.util.*; public class A { InputStream in; PrintWriter out; void solve() { int n=ni(); int a[]=na(n); int INV=0; for (int i=0;i<n;i++) for (int j=i+1;j<n;j++) if (a[i]>a[j]) INV++; boolean even=INV%2==0; int q=ni(); while (q-->0) { int l=ni(); int r=ni(); int len=r-l+1; len=(len-1)*(len)/2; if (len%2==1) even=!even; if (even) out.println("even"); else out.println("odd"); } } int MAX = (int)1e5; long factorial[]; void findfactorial() { factorial = new long[MAX + 1]; factorial[0] = 1; for (int i = 1; i < MAX + 1; i++) { factorial[i] = mul(i,factorial[i - 1]); } } long mod=(long)1e9+7; long add(long a,long b) { long x=(a+b); while(x>=mod) x-=mod; return x; } long sub(long a,long b) { long x=(a-b); while(x<0) x+=mod; return x; } long mul(long a,long b) { a%=mod; b%=mod; long x=(a*b); return x%mod; } int max(int a,int b) { if(a>b) return a; else return b; } int min(int a,int b) { if(a>b) return b; else return a; } long max(long a,long b) { if(a>b) return a; else return b; } long min(long a,long b) { if(a>b) return b; else return a; } void run() throws Exception { String INPUT = "C:/Users/ayubs/Desktop/input.txt"; in = oj ? System.in : new FileInputStream(INPUT); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new A().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = in.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean inSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && inSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(inSpaceChar(b))) { // when nextLine, (inSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(inSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The time complexity does not fit to any of the given categories or is ambiguous. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,820
2,401
3,785
import java.io.*; import java.util.*; public class Main { public static void main(String args[]) { FastReader input=new FastReader(); PrintWriter out=new PrintWriter(System.out); int T=1; while(T-->0) { int n=input.nextInt(); int m=input.nextInt(); int k=input.nextInt(); int arr1[][]=new int[n+1][m]; for(int i=1;i<=n;i++) { for(int j=1;j<m;j++) { arr1[i][j]=input.nextInt(); } } int arr2[][]=new int[n][m+1]; for(int i=1;i<n;i++) { for(int j=1;j<=m;j++) { arr2[i][j]=input.nextInt(); } } if(k%2==0) { int dp[][][]=new int[n+1][m+1][k+1]; for(int l=2;l<=k;l+=2) { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { int min=Integer.MAX_VALUE; if(j+1<=m) { min=Math.min(min,dp[i][j+1][l-2]+2*arr1[i][j]); } if(i+1<=n) { min=Math.min(min,dp[i+1][j][l-2]+2*arr2[i][j]); } if(j-1>=1) { min=Math.min(min,dp[i][j-1][l-2]+2*arr1[i][j-1]); } if(i-1>=1) { min=Math.min(min,dp[i-1][j][l-2]+2*arr2[i-1][j]); } dp[i][j][l]=min; } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(dp[i][j][k]+" "); } out.println(); } } else { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(-1+" "); } out.println(); } } } out.close(); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str=""; try { str=br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { public static void main(String args[]) { FastReader input=new FastReader(); PrintWriter out=new PrintWriter(System.out); int T=1; while(T-->0) { int n=input.nextInt(); int m=input.nextInt(); int k=input.nextInt(); int arr1[][]=new int[n+1][m]; for(int i=1;i<=n;i++) { for(int j=1;j<m;j++) { arr1[i][j]=input.nextInt(); } } int arr2[][]=new int[n][m+1]; for(int i=1;i<n;i++) { for(int j=1;j<=m;j++) { arr2[i][j]=input.nextInt(); } } if(k%2==0) { int dp[][][]=new int[n+1][m+1][k+1]; for(int l=2;l<=k;l+=2) { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { int min=Integer.MAX_VALUE; if(j+1<=m) { min=Math.min(min,dp[i][j+1][l-2]+2*arr1[i][j]); } if(i+1<=n) { min=Math.min(min,dp[i+1][j][l-2]+2*arr2[i][j]); } if(j-1>=1) { min=Math.min(min,dp[i][j-1][l-2]+2*arr1[i][j-1]); } if(i-1>=1) { min=Math.min(min,dp[i-1][j][l-2]+2*arr2[i-1][j]); } dp[i][j][l]=min; } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(dp[i][j][k]+" "); } out.println(); } } else { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(-1+" "); } out.println(); } } } out.close(); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str=""; try { str=br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { public static void main(String args[]) { FastReader input=new FastReader(); PrintWriter out=new PrintWriter(System.out); int T=1; while(T-->0) { int n=input.nextInt(); int m=input.nextInt(); int k=input.nextInt(); int arr1[][]=new int[n+1][m]; for(int i=1;i<=n;i++) { for(int j=1;j<m;j++) { arr1[i][j]=input.nextInt(); } } int arr2[][]=new int[n][m+1]; for(int i=1;i<n;i++) { for(int j=1;j<=m;j++) { arr2[i][j]=input.nextInt(); } } if(k%2==0) { int dp[][][]=new int[n+1][m+1][k+1]; for(int l=2;l<=k;l+=2) { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { int min=Integer.MAX_VALUE; if(j+1<=m) { min=Math.min(min,dp[i][j+1][l-2]+2*arr1[i][j]); } if(i+1<=n) { min=Math.min(min,dp[i+1][j][l-2]+2*arr2[i][j]); } if(j-1>=1) { min=Math.min(min,dp[i][j-1][l-2]+2*arr1[i][j-1]); } if(i-1>=1) { min=Math.min(min,dp[i-1][j][l-2]+2*arr2[i-1][j]); } dp[i][j][l]=min; } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(dp[i][j][k]+" "); } out.println(); } } else { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { out.print(-1+" "); } out.println(); } } } out.close(); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str=""; try { str=br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,038
3,777
398
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.*; public class B { BufferedReader reader; StringTokenizer tokenizer; PrintWriter out; public void solve() throws IOException { int N = nextInt(); int A = nextInt(); int B = nextInt(); int[] nsA = new int[N]; int[] nsB = new int[N]; char[] ans = new char[N]; Arrays.fill(nsA, -1); Arrays.fill(nsB, -1); Arrays.fill(ans, 'C'); HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); int[] P = new int[N]; for (int i = 0; i < N; i++) { P[i] = nextInt(); map.put(P[i], i); } // if A == B if (A == B) { for (int i = 0; i < N; i++) { if (!map.containsKey(A - P[i])) { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(0 + " "); } out.println(); return; } for (int i = 0; i < N; i++) { int oppA = A - P[i]; int oppB = B - P[i]; if (map.containsKey(oppA)) { nsA[i] = map.get(oppA); } if (map.containsKey(oppB)) { nsB[i] = map.get(oppB); } } for (int i = 0; i < N; i++) { if (nsA[i] == -1 && nsB[i] == -1) { out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == -1) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else if (nsB[i] == -1) { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == i || nsB[i] == i) { if (nsA[i] == i) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] == 'C') { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(ans[i] == 'A'? 0: 1); out.print(" "); } out.println(); } public boolean go(int cur, char link, char[] ans, int[] nsA, int[] nsB) { while (ans[cur] == 'C') { int next = link == 'A'? nsA[cur]: nsB[cur]; if (next == -1) return false; if (ans[next] != 'C') return false; ans[cur] = link; ans[next] = link; int nextNext = link == 'A'? nsB[next]: nsA[next]; cur = nextNext; if (cur == -1) return true; } return true; } /** * @param args */ public static void main(String[] args) { new B().run(); } public void run() { try { reader = new BufferedReader(new InputStreamReader(System.in)); tokenizer = null; out = new PrintWriter(System.out); solve(); reader.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.*; public class B { BufferedReader reader; StringTokenizer tokenizer; PrintWriter out; public void solve() throws IOException { int N = nextInt(); int A = nextInt(); int B = nextInt(); int[] nsA = new int[N]; int[] nsB = new int[N]; char[] ans = new char[N]; Arrays.fill(nsA, -1); Arrays.fill(nsB, -1); Arrays.fill(ans, 'C'); HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); int[] P = new int[N]; for (int i = 0; i < N; i++) { P[i] = nextInt(); map.put(P[i], i); } // if A == B if (A == B) { for (int i = 0; i < N; i++) { if (!map.containsKey(A - P[i])) { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(0 + " "); } out.println(); return; } for (int i = 0; i < N; i++) { int oppA = A - P[i]; int oppB = B - P[i]; if (map.containsKey(oppA)) { nsA[i] = map.get(oppA); } if (map.containsKey(oppB)) { nsB[i] = map.get(oppB); } } for (int i = 0; i < N; i++) { if (nsA[i] == -1 && nsB[i] == -1) { out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == -1) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else if (nsB[i] == -1) { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == i || nsB[i] == i) { if (nsA[i] == i) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] == 'C') { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(ans[i] == 'A'? 0: 1); out.print(" "); } out.println(); } public boolean go(int cur, char link, char[] ans, int[] nsA, int[] nsB) { while (ans[cur] == 'C') { int next = link == 'A'? nsA[cur]: nsB[cur]; if (next == -1) return false; if (ans[next] != 'C') return false; ans[cur] = link; ans[next] = link; int nextNext = link == 'A'? nsB[next]: nsA[next]; cur = nextNext; if (cur == -1) return true; } return true; } /** * @param args */ public static void main(String[] args) { new B().run(); } public void run() { try { reader = new BufferedReader(new InputStreamReader(System.in)); tokenizer = null; out = new PrintWriter(System.out); solve(); reader.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.*; public class B { BufferedReader reader; StringTokenizer tokenizer; PrintWriter out; public void solve() throws IOException { int N = nextInt(); int A = nextInt(); int B = nextInt(); int[] nsA = new int[N]; int[] nsB = new int[N]; char[] ans = new char[N]; Arrays.fill(nsA, -1); Arrays.fill(nsB, -1); Arrays.fill(ans, 'C'); HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); int[] P = new int[N]; for (int i = 0; i < N; i++) { P[i] = nextInt(); map.put(P[i], i); } // if A == B if (A == B) { for (int i = 0; i < N; i++) { if (!map.containsKey(A - P[i])) { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(0 + " "); } out.println(); return; } for (int i = 0; i < N; i++) { int oppA = A - P[i]; int oppB = B - P[i]; if (map.containsKey(oppA)) { nsA[i] = map.get(oppA); } if (map.containsKey(oppB)) { nsB[i] = map.get(oppB); } } for (int i = 0; i < N; i++) { if (nsA[i] == -1 && nsB[i] == -1) { out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == -1) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else if (nsB[i] == -1) { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (nsA[i] == i || nsB[i] == i) { if (nsA[i] == i) { if (!go(i, 'B', ans, nsA, nsB) ){ out.println("NO"); return; } } else { if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } } } for (int i = 0; i < N; i++) { if (ans[i] != 'C') continue; if (!go(i, 'A', ans, nsA, nsB) ){ out.println("NO"); return; } } for (int i = 0; i < N; i++) { if (ans[i] == 'C') { out.println("NO"); return; } } out.println("YES"); for (int i = 0; i < N; i++) { out.print(ans[i] == 'A'? 0: 1); out.print(" "); } out.println(); } public boolean go(int cur, char link, char[] ans, int[] nsA, int[] nsB) { while (ans[cur] == 'C') { int next = link == 'A'? nsA[cur]: nsB[cur]; if (next == -1) return false; if (ans[next] != 'C') return false; ans[cur] = link; ans[next] = link; int nextNext = link == 'A'? nsB[next]: nsA[next]; cur = nextNext; if (cur == -1) return true; } return true; } /** * @param args */ public static void main(String[] args) { new B().run(); } public void run() { try { reader = new BufferedReader(new InputStreamReader(System.in)); tokenizer = null; out = new PrintWriter(System.out); solve(); reader.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,445
397
1,665
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.HashSet; import java.util.Set; public class ProblemA { public static void main(String[] args) throws IOException { BufferedReader s = new BufferedReader(new InputStreamReader(System.in)); String[] data = s.readLine().split(" "); int n = Integer.valueOf(data[0]); int a = Integer.valueOf(data[1]); int b = Integer.valueOf(data[2]); long[] h = new long[n]; String[] line = s.readLine().split(" "); for (int i = 0 ; i < n ; i++) { h[i] = Integer.valueOf(line[i]); } Arrays.sort(h); System.out.println(h[b] - h[b-1]); } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.HashSet; import java.util.Set; public class ProblemA { public static void main(String[] args) throws IOException { BufferedReader s = new BufferedReader(new InputStreamReader(System.in)); String[] data = s.readLine().split(" "); int n = Integer.valueOf(data[0]); int a = Integer.valueOf(data[1]); int b = Integer.valueOf(data[2]); long[] h = new long[n]; String[] line = s.readLine().split(" "); for (int i = 0 ; i < n ; i++) { h[i] = Integer.valueOf(line[i]); } Arrays.sort(h); System.out.println(h[b] - h[b-1]); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.HashSet; import java.util.Set; public class ProblemA { public static void main(String[] args) throws IOException { BufferedReader s = new BufferedReader(new InputStreamReader(System.in)); String[] data = s.readLine().split(" "); int n = Integer.valueOf(data[0]); int a = Integer.valueOf(data[1]); int b = Integer.valueOf(data[2]); long[] h = new long[n]; String[] line = s.readLine().split(" "); for (int i = 0 ; i < n ; i++) { h[i] = Integer.valueOf(line[i]); } Arrays.sort(h); System.out.println(h[b] - h[b-1]); } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
523
1,662
4,238
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.io.FilterInputStream; import java.io.BufferedInputStream; import java.util.Random; import java.util.Comparator; import java.io.InputStream; /** * @author khokharnikunj8 */ public class Main { public static void main(String[] args) { new Thread(null, new Runnable() { public void run() { new Main().solve(); } }, "1", 1 << 26).start(); } void solve() { InputStream inputStream = System.in; OutputStream outputStream = System.out; ScanReader in = new ScanReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = in.scanInt(); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { int[][] dp; int[] cur; public void solve(int testNumber, ScanReader in, PrintWriter out) { int n = in.scanInt(); int m = in.scanInt(); int[][] ar = new int[n][m]; int[][] max = new int[m][2]; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) ar[i][j] = in.scanInt(); } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { max[i][0] = Math.max(max[i][0], ar[j][i]); } max[i][1] = i; } CodeHash.shuffle(max); Arrays.sort(max, new Comparator<int[]>() { public int compare(int[] o1, int[] o2) { return -o1[0] + o2[0]; } }); dp = new int[2][1 << n]; cur = new int[1 << n]; for (int i = 0; i < Math.min(m, n); i++) { Arrays.fill(cur, 0); Arrays.fill(dp[i & 1], 0); for (int j = 0; j < 1 << n; j++) { for (int k = 0; k < n; k++) { int sum = 0; for (int l = 0; l < n; l++) { if ((j & (1 << l)) != 0) { sum += (ar[(k + l) % n][max[i][1]]); } } cur[j] = Math.max(cur[j], sum); } } for (int j = 0; j < (1 << n); j++) { for (int k = j; ; k = (k - 1) & j) { dp[i & 1][j] = Math.max(dp[i & 1][j], dp[(i - 1) & 1][k] + cur[j ^ k]); if (k == 0) break; } } } out.println(dp[Math.min(n, m) & 1 ^ 1][(1 << n) - 1]); } } static class CodeHash { public static void shuffle(int[][] ar) { Random rd = new Random(new Random().nextInt()); for (int i = 0; i < ar.length; i++) { int index = rd.nextInt(ar.length); int[] temp = ar[i]; ar[i] = ar[index]; ar[index] = temp; } } } static class ScanReader { private byte[] buf = new byte[4 * 1024]; private int index; private BufferedInputStream in; private int total; public ScanReader(InputStream inputStream) { in = new BufferedInputStream(inputStream); } private int scan() { if (index >= total) { index = 0; try { total = in.read(buf); } catch (Exception e) { e.printStackTrace(); } if (total <= 0) return -1; } return buf[index++]; } public int scanInt() { int integer = 0; int n = scan(); while (isWhiteSpace(n)) n = scan(); int neg = 1; if (n == '-') { neg = -1; n = scan(); } while (!isWhiteSpace(n)) { if (n >= '0' && n <= '9') { integer *= 10; integer += n - '0'; n = scan(); } } return neg * integer; } private boolean isWhiteSpace(int n) { if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true; else return false; } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.io.FilterInputStream; import java.io.BufferedInputStream; import java.util.Random; import java.util.Comparator; import java.io.InputStream; /** * @author khokharnikunj8 */ public class Main { public static void main(String[] args) { new Thread(null, new Runnable() { public void run() { new Main().solve(); } }, "1", 1 << 26).start(); } void solve() { InputStream inputStream = System.in; OutputStream outputStream = System.out; ScanReader in = new ScanReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = in.scanInt(); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { int[][] dp; int[] cur; public void solve(int testNumber, ScanReader in, PrintWriter out) { int n = in.scanInt(); int m = in.scanInt(); int[][] ar = new int[n][m]; int[][] max = new int[m][2]; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) ar[i][j] = in.scanInt(); } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { max[i][0] = Math.max(max[i][0], ar[j][i]); } max[i][1] = i; } CodeHash.shuffle(max); Arrays.sort(max, new Comparator<int[]>() { public int compare(int[] o1, int[] o2) { return -o1[0] + o2[0]; } }); dp = new int[2][1 << n]; cur = new int[1 << n]; for (int i = 0; i < Math.min(m, n); i++) { Arrays.fill(cur, 0); Arrays.fill(dp[i & 1], 0); for (int j = 0; j < 1 << n; j++) { for (int k = 0; k < n; k++) { int sum = 0; for (int l = 0; l < n; l++) { if ((j & (1 << l)) != 0) { sum += (ar[(k + l) % n][max[i][1]]); } } cur[j] = Math.max(cur[j], sum); } } for (int j = 0; j < (1 << n); j++) { for (int k = j; ; k = (k - 1) & j) { dp[i & 1][j] = Math.max(dp[i & 1][j], dp[(i - 1) & 1][k] + cur[j ^ k]); if (k == 0) break; } } } out.println(dp[Math.min(n, m) & 1 ^ 1][(1 << n) - 1]); } } static class CodeHash { public static void shuffle(int[][] ar) { Random rd = new Random(new Random().nextInt()); for (int i = 0; i < ar.length; i++) { int index = rd.nextInt(ar.length); int[] temp = ar[i]; ar[i] = ar[index]; ar[index] = temp; } } } static class ScanReader { private byte[] buf = new byte[4 * 1024]; private int index; private BufferedInputStream in; private int total; public ScanReader(InputStream inputStream) { in = new BufferedInputStream(inputStream); } private int scan() { if (index >= total) { index = 0; try { total = in.read(buf); } catch (Exception e) { e.printStackTrace(); } if (total <= 0) return -1; } return buf[index++]; } public int scanInt() { int integer = 0; int n = scan(); while (isWhiteSpace(n)) n = scan(); int neg = 1; if (n == '-') { neg = -1; n = scan(); } while (!isWhiteSpace(n)) { if (n >= '0' && n <= '9') { integer *= 10; integer += n - '0'; n = scan(); } } return neg * integer; } private boolean isWhiteSpace(int n) { if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true; else return false; } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.io.FilterInputStream; import java.io.BufferedInputStream; import java.util.Random; import java.util.Comparator; import java.io.InputStream; /** * @author khokharnikunj8 */ public class Main { public static void main(String[] args) { new Thread(null, new Runnable() { public void run() { new Main().solve(); } }, "1", 1 << 26).start(); } void solve() { InputStream inputStream = System.in; OutputStream outputStream = System.out; ScanReader in = new ScanReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = in.scanInt(); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { int[][] dp; int[] cur; public void solve(int testNumber, ScanReader in, PrintWriter out) { int n = in.scanInt(); int m = in.scanInt(); int[][] ar = new int[n][m]; int[][] max = new int[m][2]; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) ar[i][j] = in.scanInt(); } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { max[i][0] = Math.max(max[i][0], ar[j][i]); } max[i][1] = i; } CodeHash.shuffle(max); Arrays.sort(max, new Comparator<int[]>() { public int compare(int[] o1, int[] o2) { return -o1[0] + o2[0]; } }); dp = new int[2][1 << n]; cur = new int[1 << n]; for (int i = 0; i < Math.min(m, n); i++) { Arrays.fill(cur, 0); Arrays.fill(dp[i & 1], 0); for (int j = 0; j < 1 << n; j++) { for (int k = 0; k < n; k++) { int sum = 0; for (int l = 0; l < n; l++) { if ((j & (1 << l)) != 0) { sum += (ar[(k + l) % n][max[i][1]]); } } cur[j] = Math.max(cur[j], sum); } } for (int j = 0; j < (1 << n); j++) { for (int k = j; ; k = (k - 1) & j) { dp[i & 1][j] = Math.max(dp[i & 1][j], dp[(i - 1) & 1][k] + cur[j ^ k]); if (k == 0) break; } } } out.println(dp[Math.min(n, m) & 1 ^ 1][(1 << n) - 1]); } } static class CodeHash { public static void shuffle(int[][] ar) { Random rd = new Random(new Random().nextInt()); for (int i = 0; i < ar.length; i++) { int index = rd.nextInt(ar.length); int[] temp = ar[i]; ar[i] = ar[index]; ar[index] = temp; } } } static class ScanReader { private byte[] buf = new byte[4 * 1024]; private int index; private BufferedInputStream in; private int total; public ScanReader(InputStream inputStream) { in = new BufferedInputStream(inputStream); } private int scan() { if (index >= total) { index = 0; try { total = in.read(buf); } catch (Exception e) { e.printStackTrace(); } if (total <= 0) return -1; } return buf[index++]; } public int scanInt() { int integer = 0; int n = scan(); while (isWhiteSpace(n)) n = scan(); int neg = 1; if (n == '-') { neg = -1; n = scan(); } while (!isWhiteSpace(n)) { if (n >= '0' && n <= '9') { integer *= 10; integer += n - '0'; n = scan(); } } return neg * integer; } private boolean isWhiteSpace(int n) { if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true; else return false; } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,462
4,227
2,836
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.io.StreamTokenizer; import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.LinkedList; import java.util.Scanner; import java.util.TreeSet; /** * * @author таня */ public class KF { /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { B jk = new B(); } } class B { PrintWriter out = new PrintWriter(System.out); Scanner in = new Scanner(System.in); long l = in.nextLong(); long r=in.nextLong(); B(){ if(Math.abs(r-l)>=2){ if(l%2==0){ out.print(l+" "+(l+1)+" "+(l+2)); } else{ if(Math.abs(r-l)==2){ out.print("-1"); }else{ out.print((l+1)+" "+(l+2)+" "+(l+3)); } } }else{ out.print("-1"); } out.flush(); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.io.StreamTokenizer; import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.LinkedList; import java.util.Scanner; import java.util.TreeSet; /** * * @author таня */ public class KF { /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { B jk = new B(); } } class B { PrintWriter out = new PrintWriter(System.out); Scanner in = new Scanner(System.in); long l = in.nextLong(); long r=in.nextLong(); B(){ if(Math.abs(r-l)>=2){ if(l%2==0){ out.print(l+" "+(l+1)+" "+(l+2)); } else{ if(Math.abs(r-l)==2){ out.print("-1"); }else{ out.print((l+1)+" "+(l+2)+" "+(l+3)); } } }else{ out.print("-1"); } out.flush(); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.io.StreamTokenizer; import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.LinkedList; import java.util.Scanner; import java.util.TreeSet; /** * * @author таня */ public class KF { /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { B jk = new B(); } } class B { PrintWriter out = new PrintWriter(System.out); Scanner in = new Scanner(System.in); long l = in.nextLong(); long r=in.nextLong(); B(){ if(Math.abs(r-l)>=2){ if(l%2==0){ out.print(l+" "+(l+1)+" "+(l+2)); } else{ if(Math.abs(r-l)==2){ out.print("-1"); }else{ out.print((l+1)+" "+(l+2)+" "+(l+3)); } } }else{ out.print("-1"); } out.flush(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
578
2,830
3,968
import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; // Petya and Spiders // 2012/08/16 public class P111C{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int h, w; void run(){ h=sc.nextInt(); w=sc.nextInt(); solve(); } void solve(){ n=w*h; g=new long[n]; int[] dx={0, 0, -1, 1}; int[] dy={-1, 1, 0, 0}; for(int y=0; y<h; y++){ for(int x=0; x<w; x++){ for(int k=0; k<4; k++){ int x2=x+dx[k]; int y2=y+dy[k]; if(x2>=0&&x2<w&&y2>=0&&y2<h){ g[y*w+x]|=1L<<(y2*w+x2); } } } } mds=(1L<<n)-1; mds(0, 0, 0); println((n-Long.bitCount(mds))+""); } int n; long[] g; long mds; void mds(long choosed, long removed, long covered){ if(Long.bitCount(choosed)>=Long.bitCount(mds)) return; if(covered==((1L<<n)-1)){ mds=choosed; return; } long s=covered; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); s|=(1L<<i)|g[i]; } if(s!=((1L<<n)-1)) return; int k=-1; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); if((covered>>>i&1)==1){ if(Long.bitCount(g[i]&~covered)==0){ mds(choosed, removed|(1L<<i), covered); return; }else if(Long.bitCount(g[i]&~covered)==1 &&(g[i]&~covered&~removed)!=0){ mds(choosed, removed|(1L<<i), covered); return; } }else{ if(Long.bitCount(g[i]&~removed)==0){ mds(choosed|(1L<<i), removed|(1L<<i), covered|(1L<<i)|g[i]); return; }else if(Long.bitCount(g[i]&~removed)==1 &&((g[i]&~removed)|(g[i]&~covered))==(g[i]&~removed)){ int j=Long.numberOfTrailingZeros(g[i]&~removed); mds(choosed|(1L<<j), removed|(1L<<i)|(1L<<j), covered |(1L<<j)|g[j]); return; } } if(k==-1||Long.bitCount(g[i]&~covered)>Long.bitCount(g[k]&~covered)) k=i; } if(k==-1) return; mds(choosed|(1L<<k), removed|(1L<<k), covered|(1L<<k)|g[k]); mds(choosed, removed|(1L<<k), covered); } void println(String s){ System.out.println(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new P111C().run(); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; // Petya and Spiders // 2012/08/16 public class P111C{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int h, w; void run(){ h=sc.nextInt(); w=sc.nextInt(); solve(); } void solve(){ n=w*h; g=new long[n]; int[] dx={0, 0, -1, 1}; int[] dy={-1, 1, 0, 0}; for(int y=0; y<h; y++){ for(int x=0; x<w; x++){ for(int k=0; k<4; k++){ int x2=x+dx[k]; int y2=y+dy[k]; if(x2>=0&&x2<w&&y2>=0&&y2<h){ g[y*w+x]|=1L<<(y2*w+x2); } } } } mds=(1L<<n)-1; mds(0, 0, 0); println((n-Long.bitCount(mds))+""); } int n; long[] g; long mds; void mds(long choosed, long removed, long covered){ if(Long.bitCount(choosed)>=Long.bitCount(mds)) return; if(covered==((1L<<n)-1)){ mds=choosed; return; } long s=covered; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); s|=(1L<<i)|g[i]; } if(s!=((1L<<n)-1)) return; int k=-1; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); if((covered>>>i&1)==1){ if(Long.bitCount(g[i]&~covered)==0){ mds(choosed, removed|(1L<<i), covered); return; }else if(Long.bitCount(g[i]&~covered)==1 &&(g[i]&~covered&~removed)!=0){ mds(choosed, removed|(1L<<i), covered); return; } }else{ if(Long.bitCount(g[i]&~removed)==0){ mds(choosed|(1L<<i), removed|(1L<<i), covered|(1L<<i)|g[i]); return; }else if(Long.bitCount(g[i]&~removed)==1 &&((g[i]&~removed)|(g[i]&~covered))==(g[i]&~removed)){ int j=Long.numberOfTrailingZeros(g[i]&~removed); mds(choosed|(1L<<j), removed|(1L<<i)|(1L<<j), covered |(1L<<j)|g[j]); return; } } if(k==-1||Long.bitCount(g[i]&~covered)>Long.bitCount(g[k]&~covered)) k=i; } if(k==-1) return; mds(choosed|(1L<<k), removed|(1L<<k), covered|(1L<<k)|g[k]); mds(choosed, removed|(1L<<k), covered); } void println(String s){ System.out.println(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new P111C().run(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; // Petya and Spiders // 2012/08/16 public class P111C{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int h, w; void run(){ h=sc.nextInt(); w=sc.nextInt(); solve(); } void solve(){ n=w*h; g=new long[n]; int[] dx={0, 0, -1, 1}; int[] dy={-1, 1, 0, 0}; for(int y=0; y<h; y++){ for(int x=0; x<w; x++){ for(int k=0; k<4; k++){ int x2=x+dx[k]; int y2=y+dy[k]; if(x2>=0&&x2<w&&y2>=0&&y2<h){ g[y*w+x]|=1L<<(y2*w+x2); } } } } mds=(1L<<n)-1; mds(0, 0, 0); println((n-Long.bitCount(mds))+""); } int n; long[] g; long mds; void mds(long choosed, long removed, long covered){ if(Long.bitCount(choosed)>=Long.bitCount(mds)) return; if(covered==((1L<<n)-1)){ mds=choosed; return; } long s=covered; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); s|=(1L<<i)|g[i]; } if(s!=((1L<<n)-1)) return; int k=-1; for(long remained=~removed&((1L<<n)-1); remained!=0; remained&=remained-1){ int i=Long.numberOfTrailingZeros(remained); if((covered>>>i&1)==1){ if(Long.bitCount(g[i]&~covered)==0){ mds(choosed, removed|(1L<<i), covered); return; }else if(Long.bitCount(g[i]&~covered)==1 &&(g[i]&~covered&~removed)!=0){ mds(choosed, removed|(1L<<i), covered); return; } }else{ if(Long.bitCount(g[i]&~removed)==0){ mds(choosed|(1L<<i), removed|(1L<<i), covered|(1L<<i)|g[i]); return; }else if(Long.bitCount(g[i]&~removed)==1 &&((g[i]&~removed)|(g[i]&~covered))==(g[i]&~removed)){ int j=Long.numberOfTrailingZeros(g[i]&~removed); mds(choosed|(1L<<j), removed|(1L<<i)|(1L<<j), covered |(1L<<j)|g[j]); return; } } if(k==-1||Long.bitCount(g[i]&~covered)>Long.bitCount(g[k]&~covered)) k=i; } if(k==-1) return; mds(choosed|(1L<<k), removed|(1L<<k), covered|(1L<<k)|g[k]); mds(choosed, removed|(1L<<k), covered); } void println(String s){ System.out.println(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new P111C().run(); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,207
3,957
1,440
import java.io.*; import java.util.*; public class Example { BufferedReader in; PrintWriter out; StringTokenizer st; String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(in.readLine()); } catch (Exception e) { } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } public void run() throws Exception { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.flush(); out.close(); in.close(); } long n, x, y, c; boolean check(long z) { long res = 1 + 2 * z * (z + 1); long bx1 = z - x + 1; long by1 = z - y + 1; long bxn = z - n + x; long byn = z - n + y; if (bx1 > 0) { res -= bx1 * bx1; } if (bxn > 0) { res -= bxn * bxn; } if (by1 > 0) { res -= by1 * by1; } if (byn > 0) { res -= byn * byn; } if (z > x + y - 1) { long m = z - x - y + 1; res += m * (m + 1) / 2; } if (z > x + n - y) { long m = z - x - n + y; res += m * (m + 1) / 2; } if (z > n - x + y) { long m = z - n - y + x; res += m * (m + 1) / 2; } if (z > n - x + n - y + 1) { long m = z - 2 * n + x + y - 1; res += m * (m + 1) / 2; } return res >= c; } public void solve() throws Exception { n = nextLong(); x = nextLong(); y = nextLong(); c = nextLong(); long l = 0; long r = 2 * n; while (r > l) { long mid = (r + l) / 2; if (check(mid)) { r = mid; } else { l = mid + 1; } } out.println(r); } public static void main(String[] args) throws Exception { new Example().run(); } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Example { BufferedReader in; PrintWriter out; StringTokenizer st; String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(in.readLine()); } catch (Exception e) { } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } public void run() throws Exception { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.flush(); out.close(); in.close(); } long n, x, y, c; boolean check(long z) { long res = 1 + 2 * z * (z + 1); long bx1 = z - x + 1; long by1 = z - y + 1; long bxn = z - n + x; long byn = z - n + y; if (bx1 > 0) { res -= bx1 * bx1; } if (bxn > 0) { res -= bxn * bxn; } if (by1 > 0) { res -= by1 * by1; } if (byn > 0) { res -= byn * byn; } if (z > x + y - 1) { long m = z - x - y + 1; res += m * (m + 1) / 2; } if (z > x + n - y) { long m = z - x - n + y; res += m * (m + 1) / 2; } if (z > n - x + y) { long m = z - n - y + x; res += m * (m + 1) / 2; } if (z > n - x + n - y + 1) { long m = z - 2 * n + x + y - 1; res += m * (m + 1) / 2; } return res >= c; } public void solve() throws Exception { n = nextLong(); x = nextLong(); y = nextLong(); c = nextLong(); long l = 0; long r = 2 * n; while (r > l) { long mid = (r + l) / 2; if (check(mid)) { r = mid; } else { l = mid + 1; } } out.println(r); } public static void main(String[] args) throws Exception { new Example().run(); } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Example { BufferedReader in; PrintWriter out; StringTokenizer st; String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(in.readLine()); } catch (Exception e) { } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } public void run() throws Exception { //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.flush(); out.close(); in.close(); } long n, x, y, c; boolean check(long z) { long res = 1 + 2 * z * (z + 1); long bx1 = z - x + 1; long by1 = z - y + 1; long bxn = z - n + x; long byn = z - n + y; if (bx1 > 0) { res -= bx1 * bx1; } if (bxn > 0) { res -= bxn * bxn; } if (by1 > 0) { res -= by1 * by1; } if (byn > 0) { res -= byn * byn; } if (z > x + y - 1) { long m = z - x - y + 1; res += m * (m + 1) / 2; } if (z > x + n - y) { long m = z - x - n + y; res += m * (m + 1) / 2; } if (z > n - x + y) { long m = z - n - y + x; res += m * (m + 1) / 2; } if (z > n - x + n - y + 1) { long m = z - 2 * n + x + y - 1; res += m * (m + 1) / 2; } return res >= c; } public void solve() throws Exception { n = nextLong(); x = nextLong(); y = nextLong(); c = nextLong(); long l = 0; long r = 2 * n; while (r > l) { long mid = (r + l) / 2; if (check(mid)) { r = mid; } else { l = mid + 1; } } out.println(r); } public static void main(String[] args) throws Exception { new Example().run(); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,007
1,438
1,822
import java.io.*; import java.util.*; public class ProblemA { InputReader in; PrintWriter out; void solve() { int n = in.nextInt(); int m = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); Arrays.sort(a); int d = k; int cur = n - 1; int ans = 0; while (d < m && cur >= 0) { d += a[cur] - 1; cur--; ans++; } if (d >= m) out.println(ans); else out.println("-1"); } ProblemA(){ boolean oj = System.getProperty("ONLINE_JUDGE") != null; try { if (oj) { in = new InputReader(System.in); out = new PrintWriter(System.out); } else { Writer w = new FileWriter("output.txt"); in = new InputReader(new FileReader("input.txt")); out = new PrintWriter(w); } } catch(Exception e) { throw new RuntimeException(e); } solve(); out.close(); } public static void main(String[] args){ new ProblemA(); } } class InputReader { private BufferedReader reader; private StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public InputReader(FileReader fr) { reader = new BufferedReader(fr); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public double nextDouble() { return Double.parseDouble(next()); } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class ProblemA { InputReader in; PrintWriter out; void solve() { int n = in.nextInt(); int m = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); Arrays.sort(a); int d = k; int cur = n - 1; int ans = 0; while (d < m && cur >= 0) { d += a[cur] - 1; cur--; ans++; } if (d >= m) out.println(ans); else out.println("-1"); } ProblemA(){ boolean oj = System.getProperty("ONLINE_JUDGE") != null; try { if (oj) { in = new InputReader(System.in); out = new PrintWriter(System.out); } else { Writer w = new FileWriter("output.txt"); in = new InputReader(new FileReader("input.txt")); out = new PrintWriter(w); } } catch(Exception e) { throw new RuntimeException(e); } solve(); out.close(); } public static void main(String[] args){ new ProblemA(); } } class InputReader { private BufferedReader reader; private StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public InputReader(FileReader fr) { reader = new BufferedReader(fr); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public double nextDouble() { return Double.parseDouble(next()); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class ProblemA { InputReader in; PrintWriter out; void solve() { int n = in.nextInt(); int m = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); Arrays.sort(a); int d = k; int cur = n - 1; int ans = 0; while (d < m && cur >= 0) { d += a[cur] - 1; cur--; ans++; } if (d >= m) out.println(ans); else out.println("-1"); } ProblemA(){ boolean oj = System.getProperty("ONLINE_JUDGE") != null; try { if (oj) { in = new InputReader(System.in); out = new PrintWriter(System.out); } else { Writer w = new FileWriter("output.txt"); in = new InputReader(new FileReader("input.txt")); out = new PrintWriter(w); } } catch(Exception e) { throw new RuntimeException(e); } solve(); out.close(); } public static void main(String[] args){ new ProblemA(); } } class InputReader { private BufferedReader reader; private StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream)); tokenizer = null; } public InputReader(FileReader fr) { reader = new BufferedReader(fr); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public double nextDouble() { return Double.parseDouble(next()); } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
779
1,818
2,175
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.Reader; import java.util.*; public class C { public static void main(String[] args) { FastScanner sc = new FastScanner(); int n = sc.nextInt(); long r = sc.nextInt(); double d = 2 * r; long[] xs = sc.readLongArray(n); P[] points = new P[n]; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { if (i > 0) sb.append(' '); double y = r; for (int j = 0; j < i; j++) { long diff = Math.abs(xs[i] - points[j].x); if (diff <= 2 * r) { double dy = Math.sqrt(d * d - diff * diff); double testY = points[j].y + dy; y = Math.max(y, testY); } } sb.append(y); points[i] = new P(xs[i], y); } System.out.println(sb); } static class P { final long x; final double y; public P(long x, double y) { this.x = x; this.y = y; } } static void shuffle(int[] arr) { Random rng = new Random(); int length = arr.length; for (int idx = 0; idx < arr.length; idx++) { int toSwap = idx + rng.nextInt(length - idx); int tmp = arr[idx]; arr[idx] = arr[toSwap]; arr[toSwap] = tmp; } } public static class FastScanner { BufferedReader br; StringTokenizer st; public FastScanner(Reader in) { br = new BufferedReader(in); } public FastScanner() { this(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String readNextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } int[] readIntArray(int n) { int[] a = new int[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextInt(); } return a; } long[] readLongArray(int n) { long[] a = new long[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextLong(); } return a; } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.Reader; import java.util.*; public class C { public static void main(String[] args) { FastScanner sc = new FastScanner(); int n = sc.nextInt(); long r = sc.nextInt(); double d = 2 * r; long[] xs = sc.readLongArray(n); P[] points = new P[n]; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { if (i > 0) sb.append(' '); double y = r; for (int j = 0; j < i; j++) { long diff = Math.abs(xs[i] - points[j].x); if (diff <= 2 * r) { double dy = Math.sqrt(d * d - diff * diff); double testY = points[j].y + dy; y = Math.max(y, testY); } } sb.append(y); points[i] = new P(xs[i], y); } System.out.println(sb); } static class P { final long x; final double y; public P(long x, double y) { this.x = x; this.y = y; } } static void shuffle(int[] arr) { Random rng = new Random(); int length = arr.length; for (int idx = 0; idx < arr.length; idx++) { int toSwap = idx + rng.nextInt(length - idx); int tmp = arr[idx]; arr[idx] = arr[toSwap]; arr[toSwap] = tmp; } } public static class FastScanner { BufferedReader br; StringTokenizer st; public FastScanner(Reader in) { br = new BufferedReader(in); } public FastScanner() { this(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String readNextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } int[] readIntArray(int n) { int[] a = new int[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextInt(); } return a; } long[] readLongArray(int n) { long[] a = new long[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextLong(); } return a; } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.Reader; import java.util.*; public class C { public static void main(String[] args) { FastScanner sc = new FastScanner(); int n = sc.nextInt(); long r = sc.nextInt(); double d = 2 * r; long[] xs = sc.readLongArray(n); P[] points = new P[n]; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { if (i > 0) sb.append(' '); double y = r; for (int j = 0; j < i; j++) { long diff = Math.abs(xs[i] - points[j].x); if (diff <= 2 * r) { double dy = Math.sqrt(d * d - diff * diff); double testY = points[j].y + dy; y = Math.max(y, testY); } } sb.append(y); points[i] = new P(xs[i], y); } System.out.println(sb); } static class P { final long x; final double y; public P(long x, double y) { this.x = x; this.y = y; } } static void shuffle(int[] arr) { Random rng = new Random(); int length = arr.length; for (int idx = 0; idx < arr.length; idx++) { int toSwap = idx + rng.nextInt(length - idx); int tmp = arr[idx]; arr[idx] = arr[toSwap]; arr[toSwap] = tmp; } } public static class FastScanner { BufferedReader br; StringTokenizer st; public FastScanner(Reader in) { br = new BufferedReader(in); } public FastScanner() { this(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String readNextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } int[] readIntArray(int n) { int[] a = new int[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextInt(); } return a; } long[] readLongArray(int n) { long[] a = new long[n]; for (int idx = 0; idx < n; idx++) { a[idx] = nextLong(); } return a; } } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
978
2,171
314
import java.io.*; import java.util.*; public class Main{ public static void main(String[] args){ try { new Main().solve(); } catch (Exception e) { e.printStackTrace(); } } ArrayList<Edge>[]edge; int n,m,cnt=0; int ord; int[]order;int[]vis; Edge[] e; private void solve() throws Exception{ InputReader in = new InputReader(System.in); PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); n=in.nextInt();m=in.nextInt(); edge=new ArrayList[n+1]; e=new Edge[m]; vis=new int[n+1]; order=new int[n+1]; for(int i=1;i<=n;i++){ edge[i]=new ArrayList<>(); } for(int i=1;i<=m;i++){ int s=in.nextInt(),t=in.nextInt(),c=in.nextInt(); edge[s].add(new Edge(s,t,c,i)); } int l=0,r=1000000000; while (l<r){ int mid=(l+r)>>>1; if(judge(mid,false))r=mid; else l=mid+1; } out.print(l+" "); judge(l,true); Arrays.sort(e,0,cnt,Comparator.comparingInt(x->x.id)); int ans=0; int[]a=new int[m]; for(int i=0;i<cnt;i++){ if(order[e[i].s]<order[e[i].t])a[ans++]=e[i].id; } out.println(ans); for(int i=0;i<ans;i++){ out.print(a[i]+" "); } out.println(); out.flush(); } boolean judge(int min,boolean mod){ Arrays.fill(vis,0); cycle=false; for(int i=1;i<=n;i++){ if(vis[i]==0){ dfs(i,min,mod); if(cycle)return false; } } return true; } boolean cycle=false; void dfs(int cur,int min,boolean mod){ if(cycle)return; vis[cur]=1; for(Edge e:edge[cur]){ if(e.c<=min){ if(mod)this.e[cnt++]=e; continue; } if(vis[e.t]==1){ cycle=true;return; } else if(vis[e.t]==0)dfs(e.t,min,mod); } vis[cur]=2; if(mod)order[cur]=ord++; } } class Edge{ int s,t,c,id; Edge(int a,int b,int c,int d){ s=a;t=b;this.c=c;id=d; } } class InputReader{ StreamTokenizer tokenizer; public InputReader(InputStream stream){ tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream))); tokenizer.ordinaryChars(33,126); tokenizer.wordChars(33,126); } public String next() throws IOException { tokenizer.nextToken(); return tokenizer.sval; } public int nextInt() throws IOException { return Integer.parseInt(next()); } public boolean hasNext() throws IOException { int res=tokenizer.nextToken(); tokenizer.pushBack(); return res!=tokenizer.TT_EOF; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main{ public static void main(String[] args){ try { new Main().solve(); } catch (Exception e) { e.printStackTrace(); } } ArrayList<Edge>[]edge; int n,m,cnt=0; int ord; int[]order;int[]vis; Edge[] e; private void solve() throws Exception{ InputReader in = new InputReader(System.in); PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); n=in.nextInt();m=in.nextInt(); edge=new ArrayList[n+1]; e=new Edge[m]; vis=new int[n+1]; order=new int[n+1]; for(int i=1;i<=n;i++){ edge[i]=new ArrayList<>(); } for(int i=1;i<=m;i++){ int s=in.nextInt(),t=in.nextInt(),c=in.nextInt(); edge[s].add(new Edge(s,t,c,i)); } int l=0,r=1000000000; while (l<r){ int mid=(l+r)>>>1; if(judge(mid,false))r=mid; else l=mid+1; } out.print(l+" "); judge(l,true); Arrays.sort(e,0,cnt,Comparator.comparingInt(x->x.id)); int ans=0; int[]a=new int[m]; for(int i=0;i<cnt;i++){ if(order[e[i].s]<order[e[i].t])a[ans++]=e[i].id; } out.println(ans); for(int i=0;i<ans;i++){ out.print(a[i]+" "); } out.println(); out.flush(); } boolean judge(int min,boolean mod){ Arrays.fill(vis,0); cycle=false; for(int i=1;i<=n;i++){ if(vis[i]==0){ dfs(i,min,mod); if(cycle)return false; } } return true; } boolean cycle=false; void dfs(int cur,int min,boolean mod){ if(cycle)return; vis[cur]=1; for(Edge e:edge[cur]){ if(e.c<=min){ if(mod)this.e[cnt++]=e; continue; } if(vis[e.t]==1){ cycle=true;return; } else if(vis[e.t]==0)dfs(e.t,min,mod); } vis[cur]=2; if(mod)order[cur]=ord++; } } class Edge{ int s,t,c,id; Edge(int a,int b,int c,int d){ s=a;t=b;this.c=c;id=d; } } class InputReader{ StreamTokenizer tokenizer; public InputReader(InputStream stream){ tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream))); tokenizer.ordinaryChars(33,126); tokenizer.wordChars(33,126); } public String next() throws IOException { tokenizer.nextToken(); return tokenizer.sval; } public int nextInt() throws IOException { return Integer.parseInt(next()); } public boolean hasNext() throws IOException { int res=tokenizer.nextToken(); tokenizer.pushBack(); return res!=tokenizer.TT_EOF; } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main{ public static void main(String[] args){ try { new Main().solve(); } catch (Exception e) { e.printStackTrace(); } } ArrayList<Edge>[]edge; int n,m,cnt=0; int ord; int[]order;int[]vis; Edge[] e; private void solve() throws Exception{ InputReader in = new InputReader(System.in); PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); n=in.nextInt();m=in.nextInt(); edge=new ArrayList[n+1]; e=new Edge[m]; vis=new int[n+1]; order=new int[n+1]; for(int i=1;i<=n;i++){ edge[i]=new ArrayList<>(); } for(int i=1;i<=m;i++){ int s=in.nextInt(),t=in.nextInt(),c=in.nextInt(); edge[s].add(new Edge(s,t,c,i)); } int l=0,r=1000000000; while (l<r){ int mid=(l+r)>>>1; if(judge(mid,false))r=mid; else l=mid+1; } out.print(l+" "); judge(l,true); Arrays.sort(e,0,cnt,Comparator.comparingInt(x->x.id)); int ans=0; int[]a=new int[m]; for(int i=0;i<cnt;i++){ if(order[e[i].s]<order[e[i].t])a[ans++]=e[i].id; } out.println(ans); for(int i=0;i<ans;i++){ out.print(a[i]+" "); } out.println(); out.flush(); } boolean judge(int min,boolean mod){ Arrays.fill(vis,0); cycle=false; for(int i=1;i<=n;i++){ if(vis[i]==0){ dfs(i,min,mod); if(cycle)return false; } } return true; } boolean cycle=false; void dfs(int cur,int min,boolean mod){ if(cycle)return; vis[cur]=1; for(Edge e:edge[cur]){ if(e.c<=min){ if(mod)this.e[cnt++]=e; continue; } if(vis[e.t]==1){ cycle=true;return; } else if(vis[e.t]==0)dfs(e.t,min,mod); } vis[cur]=2; if(mod)order[cur]=ord++; } } class Edge{ int s,t,c,id; Edge(int a,int b,int c,int d){ s=a;t=b;this.c=c;id=d; } } class InputReader{ StreamTokenizer tokenizer; public InputReader(InputStream stream){ tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream))); tokenizer.ordinaryChars(33,126); tokenizer.wordChars(33,126); } public String next() throws IOException { tokenizer.nextToken(); return tokenizer.sval; } public int nextInt() throws IOException { return Integer.parseInt(next()); } public boolean hasNext() throws IOException { int res=tokenizer.nextToken(); tokenizer.pushBack(); return res!=tokenizer.TT_EOF; } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,043
314
3,018
//package arbuz; import java.util.Scanner; public class Arbuz { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); if (n % 4 == 0 || n % 7 == 0 || n % 47 == 0 || n % 74 == 0 || n % 444 == 0 || n % 447 == 0 || n % 474 == 0 || n % 477 == 0) { System.out.println("YES"); } else { System.out.println("NO"); } } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> //package arbuz; import java.util.Scanner; public class Arbuz { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); if (n % 4 == 0 || n % 7 == 0 || n % 47 == 0 || n % 74 == 0 || n % 444 == 0 || n % 447 == 0 || n % 474 == 0 || n % 477 == 0) { System.out.println("YES"); } else { System.out.println("NO"); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package arbuz; import java.util.Scanner; public class Arbuz { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); if (n % 4 == 0 || n % 7 == 0 || n % 47 == 0 || n % 74 == 0 || n % 444 == 0 || n % 447 == 0 || n % 474 == 0 || n % 477 == 0) { System.out.println("YES"); } else { System.out.println("NO"); } } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
471
3,012
1,882
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.math.BigInteger; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastScanner in = new FastScanner(inputStream); PrintWriter out = new PrintWriter(outputStream); D903 solver = new D903(); solver.solve(1, in, out); out.close(); } static class D903 { int N; long ripple; BigInteger tot; long[] nums; BigInteger[] cs; public void solve(int testNumber, FastScanner s, PrintWriter out) { N = s.nextInt(); nums = s.nextLongArray(N); tot = new BigInteger("0"); cs = new BigInteger[N + 1]; cs[0] = new BigInteger("0"); ripple = 0; for (int i = 1; i <= N; i++) cs[i] = cs[i - 1].add(new BigInteger("" + nums[i - 1])); for (int i = 1; i <= N; i++) { long cur = nums[i - 1]; // out.printf("%d: cs %d, minus %d%n", i, (cs[N] - cs[i]), cur * (N - i)); tot = tot.add(cs[N].subtract(cs[i])).subtract(new BigInteger("" + (cur * (N - i)))); } HashMap<Long, Integer> seen = new HashMap<>(); for (long i : nums) { Integer lo = seen.get(i - 1); Integer hi = seen.get(i + 1); if (lo != null) tot = tot.subtract(new BigInteger("" + lo)); if (hi != null) tot = tot.add(new BigInteger("" + hi)); if (!seen.containsKey(i)) seen.put(i, 0); seen.put(i, seen.get(i) + 1); } out.println(tot); } } static class FastScanner { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastScanner(InputStream stream) { this.stream = stream; } int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public long[] nextLongArray(int N) { long[] ret = new long[N]; for (int i = 0; i < N; i++) ret[i] = this.nextLong(); return ret; } } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.math.BigInteger; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastScanner in = new FastScanner(inputStream); PrintWriter out = new PrintWriter(outputStream); D903 solver = new D903(); solver.solve(1, in, out); out.close(); } static class D903 { int N; long ripple; BigInteger tot; long[] nums; BigInteger[] cs; public void solve(int testNumber, FastScanner s, PrintWriter out) { N = s.nextInt(); nums = s.nextLongArray(N); tot = new BigInteger("0"); cs = new BigInteger[N + 1]; cs[0] = new BigInteger("0"); ripple = 0; for (int i = 1; i <= N; i++) cs[i] = cs[i - 1].add(new BigInteger("" + nums[i - 1])); for (int i = 1; i <= N; i++) { long cur = nums[i - 1]; // out.printf("%d: cs %d, minus %d%n", i, (cs[N] - cs[i]), cur * (N - i)); tot = tot.add(cs[N].subtract(cs[i])).subtract(new BigInteger("" + (cur * (N - i)))); } HashMap<Long, Integer> seen = new HashMap<>(); for (long i : nums) { Integer lo = seen.get(i - 1); Integer hi = seen.get(i + 1); if (lo != null) tot = tot.subtract(new BigInteger("" + lo)); if (hi != null) tot = tot.add(new BigInteger("" + hi)); if (!seen.containsKey(i)) seen.put(i, 0); seen.put(i, seen.get(i) + 1); } out.println(tot); } } static class FastScanner { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastScanner(InputStream stream) { this.stream = stream; } int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public long[] nextLongArray(int N) { long[] ret = new long[N]; for (int i = 0; i < N; i++) ret[i] = this.nextLong(); return ret; } } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.math.BigInteger; import java.util.HashMap; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastScanner in = new FastScanner(inputStream); PrintWriter out = new PrintWriter(outputStream); D903 solver = new D903(); solver.solve(1, in, out); out.close(); } static class D903 { int N; long ripple; BigInteger tot; long[] nums; BigInteger[] cs; public void solve(int testNumber, FastScanner s, PrintWriter out) { N = s.nextInt(); nums = s.nextLongArray(N); tot = new BigInteger("0"); cs = new BigInteger[N + 1]; cs[0] = new BigInteger("0"); ripple = 0; for (int i = 1; i <= N; i++) cs[i] = cs[i - 1].add(new BigInteger("" + nums[i - 1])); for (int i = 1; i <= N; i++) { long cur = nums[i - 1]; // out.printf("%d: cs %d, minus %d%n", i, (cs[N] - cs[i]), cur * (N - i)); tot = tot.add(cs[N].subtract(cs[i])).subtract(new BigInteger("" + (cur * (N - i)))); } HashMap<Long, Integer> seen = new HashMap<>(); for (long i : nums) { Integer lo = seen.get(i - 1); Integer hi = seen.get(i + 1); if (lo != null) tot = tot.subtract(new BigInteger("" + lo)); if (hi != null) tot = tot.add(new BigInteger("" + hi)); if (!seen.containsKey(i)) seen.put(i, 0); seen.put(i, seen.get(i) + 1); } out.println(tot); } } static class FastScanner { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastScanner(InputStream stream) { this.stream = stream; } int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public long[] nextLongArray(int N) { long[] ret = new long[N]; for (int i = 0; i < N; i++) ret[i] = this.nextLong(); return ret; } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,137
1,878
2,586
import java.util.ArrayList; import java.util.Scanner; public class PaintTheNumber { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); ArrayList<Integer> l=new ArrayList<Integer>(); for(int i=0; i<n; i++) { l.add(sc.nextInt()); } boolean c=false; for(int i=0; i<l.size(); i++) { if(l.get(i)==-1) continue; for(int j=0; j<l.size(); j++) { if(i==j || l.get(j)==-1) continue; else { if(l.get(j)%l.get(i)==0) { l.set(j, -1); } } } } int nbr=0; for(int i=0; i<l.size(); i++) if(l.get(i)!=-1) nbr++; System.out.println(nbr); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.ArrayList; import java.util.Scanner; public class PaintTheNumber { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); ArrayList<Integer> l=new ArrayList<Integer>(); for(int i=0; i<n; i++) { l.add(sc.nextInt()); } boolean c=false; for(int i=0; i<l.size(); i++) { if(l.get(i)==-1) continue; for(int j=0; j<l.size(); j++) { if(i==j || l.get(j)==-1) continue; else { if(l.get(j)%l.get(i)==0) { l.set(j, -1); } } } } int nbr=0; for(int i=0; i<l.size(); i++) if(l.get(i)!=-1) nbr++; System.out.println(nbr); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.ArrayList; import java.util.Scanner; public class PaintTheNumber { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); ArrayList<Integer> l=new ArrayList<Integer>(); for(int i=0; i<n; i++) { l.add(sc.nextInt()); } boolean c=false; for(int i=0; i<l.size(); i++) { if(l.get(i)==-1) continue; for(int j=0; j<l.size(); j++) { if(i==j || l.get(j)==-1) continue; else { if(l.get(j)%l.get(i)==0) { l.set(j, -1); } } } } int nbr=0; for(int i=0; i<l.size(); i++) if(l.get(i)!=-1) nbr++; System.out.println(nbr); } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
545
2,580
4,441
/*(c) gorlum0 [at] gmail.com*/ import java.io.*; import java.util.*; import java.math.*; public class E { int bitcount(int x) { int c = 0; for ( ; x != 0; c++) x &= x-1; return c; } boolean bit(int x, int i) { if (i < 0) return false; return (x>>i & 1) == 1 ? true : false; } int solve(int n, int m) { if (m > n) { int x = m; m = n; n = x; } int maxmask = 1<<m; int[][][] dp = new int[n+1][maxmask][maxmask]; for (int i = 0; i <= n; i++) for (int j = 0; j < maxmask; j++) for (int k = 0; k < maxmask; k++) dp[i][j][k] = inf; for (int i = 0; i < maxmask; i++) dp[0][0][i] = bitcount(i); for (int i = 1; i <= n; i++) for (int b = 0; b < maxmask; b++) for (int c = 0; c < maxmask; c++) for (int a = 0; a < maxmask; a++) { boolean nospider = false; for (int j = 0; j < m; j++) if (not(bit(a,j) || bit(c,j) || bit(b,j-1) || bit(b,j) || bit(b,j+1))) { nospider = true; break; } if (nospider) continue; dp[i][b][c] = Math.min(dp[i][b][c], dp[i-1][a][b] + bitcount(c)); } int res = inf; for (int b = 0; b < maxmask; b++) res = Math.min(res, dp[n][b][0]); return n*m - res; } void main() throws IOException { int n; while ((n = nextInt()) != EOF) { int m = nextInt(); out.println(solve(n, m)); } } public static void main(String[] args) { new E().run(); } // ====================================================================== int inf = (int) 1e9; final int EOF = -1; boolean not(boolean p) { return !p; } int sqr(int x) { return x*x; } long sqr(long x) { return x*x; } double sqr(double x) { return x*x; } BufferedReader fin; StringTokenizer st; PrintWriter out; public void run() { try { fin = new BufferedReader(new InputStreamReader(System.in)); st = null; out = new PrintWriter(System.out); main(); fin.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String line = fin.readLine(); if (line == null) return "-1"; else st = new StringTokenizer(line); } return st.nextToken(); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> /*(c) gorlum0 [at] gmail.com*/ import java.io.*; import java.util.*; import java.math.*; public class E { int bitcount(int x) { int c = 0; for ( ; x != 0; c++) x &= x-1; return c; } boolean bit(int x, int i) { if (i < 0) return false; return (x>>i & 1) == 1 ? true : false; } int solve(int n, int m) { if (m > n) { int x = m; m = n; n = x; } int maxmask = 1<<m; int[][][] dp = new int[n+1][maxmask][maxmask]; for (int i = 0; i <= n; i++) for (int j = 0; j < maxmask; j++) for (int k = 0; k < maxmask; k++) dp[i][j][k] = inf; for (int i = 0; i < maxmask; i++) dp[0][0][i] = bitcount(i); for (int i = 1; i <= n; i++) for (int b = 0; b < maxmask; b++) for (int c = 0; c < maxmask; c++) for (int a = 0; a < maxmask; a++) { boolean nospider = false; for (int j = 0; j < m; j++) if (not(bit(a,j) || bit(c,j) || bit(b,j-1) || bit(b,j) || bit(b,j+1))) { nospider = true; break; } if (nospider) continue; dp[i][b][c] = Math.min(dp[i][b][c], dp[i-1][a][b] + bitcount(c)); } int res = inf; for (int b = 0; b < maxmask; b++) res = Math.min(res, dp[n][b][0]); return n*m - res; } void main() throws IOException { int n; while ((n = nextInt()) != EOF) { int m = nextInt(); out.println(solve(n, m)); } } public static void main(String[] args) { new E().run(); } // ====================================================================== int inf = (int) 1e9; final int EOF = -1; boolean not(boolean p) { return !p; } int sqr(int x) { return x*x; } long sqr(long x) { return x*x; } double sqr(double x) { return x*x; } BufferedReader fin; StringTokenizer st; PrintWriter out; public void run() { try { fin = new BufferedReader(new InputStreamReader(System.in)); st = null; out = new PrintWriter(System.out); main(); fin.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String line = fin.readLine(); if (line == null) return "-1"; else st = new StringTokenizer(line); } return st.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> /*(c) gorlum0 [at] gmail.com*/ import java.io.*; import java.util.*; import java.math.*; public class E { int bitcount(int x) { int c = 0; for ( ; x != 0; c++) x &= x-1; return c; } boolean bit(int x, int i) { if (i < 0) return false; return (x>>i & 1) == 1 ? true : false; } int solve(int n, int m) { if (m > n) { int x = m; m = n; n = x; } int maxmask = 1<<m; int[][][] dp = new int[n+1][maxmask][maxmask]; for (int i = 0; i <= n; i++) for (int j = 0; j < maxmask; j++) for (int k = 0; k < maxmask; k++) dp[i][j][k] = inf; for (int i = 0; i < maxmask; i++) dp[0][0][i] = bitcount(i); for (int i = 1; i <= n; i++) for (int b = 0; b < maxmask; b++) for (int c = 0; c < maxmask; c++) for (int a = 0; a < maxmask; a++) { boolean nospider = false; for (int j = 0; j < m; j++) if (not(bit(a,j) || bit(c,j) || bit(b,j-1) || bit(b,j) || bit(b,j+1))) { nospider = true; break; } if (nospider) continue; dp[i][b][c] = Math.min(dp[i][b][c], dp[i-1][a][b] + bitcount(c)); } int res = inf; for (int b = 0; b < maxmask; b++) res = Math.min(res, dp[n][b][0]); return n*m - res; } void main() throws IOException { int n; while ((n = nextInt()) != EOF) { int m = nextInt(); out.println(solve(n, m)); } } public static void main(String[] args) { new E().run(); } // ====================================================================== int inf = (int) 1e9; final int EOF = -1; boolean not(boolean p) { return !p; } int sqr(int x) { return x*x; } long sqr(long x) { return x*x; } double sqr(double x) { return x*x; } BufferedReader fin; StringTokenizer st; PrintWriter out; public void run() { try { fin = new BufferedReader(new InputStreamReader(System.in)); st = null; out = new PrintWriter(System.out); main(); fin.close(); out.close(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String line = fin.readLine(); if (line == null) return "-1"; else st = new StringTokenizer(line); } return st.nextToken(); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,129
4,430
3,076
import java.util.*; import java.io.*; public class A { public static void main(String[] args) { Kattio io = new Kattio(System.in); int n = io.getInt(); int ans = 0; int V = n; int A = n; A -= n/2; ans += n/2; V -= A; ans += A; A -= n/2; ans += n/2; io.println(ans); io.flush(); } } class Kattio extends PrintWriter { public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public boolean hasMoreTokens() { return peekToken() != null; } public int getInt() { return Integer.parseInt(nextToken()); } public double getDouble() { return Double.parseDouble(nextToken()); } public long getLong() { return Long.parseLong(nextToken()); } public String getWord() { return nextToken(); } private BufferedReader r; private String line; private StringTokenizer st; private String token; private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.io.*; public class A { public static void main(String[] args) { Kattio io = new Kattio(System.in); int n = io.getInt(); int ans = 0; int V = n; int A = n; A -= n/2; ans += n/2; V -= A; ans += A; A -= n/2; ans += n/2; io.println(ans); io.flush(); } } class Kattio extends PrintWriter { public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public boolean hasMoreTokens() { return peekToken() != null; } public int getInt() { return Integer.parseInt(nextToken()); } public double getDouble() { return Double.parseDouble(nextToken()); } public long getLong() { return Long.parseLong(nextToken()); } public String getWord() { return nextToken(); } private BufferedReader r; private String line; private StringTokenizer st; private String token; private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; public class A { public static void main(String[] args) { Kattio io = new Kattio(System.in); int n = io.getInt(); int ans = 0; int V = n; int A = n; A -= n/2; ans += n/2; V -= A; ans += A; A -= n/2; ans += n/2; io.println(ans); io.flush(); } } class Kattio extends PrintWriter { public Kattio(InputStream i) { super(new BufferedOutputStream(System.out)); r = new BufferedReader(new InputStreamReader(i)); } public Kattio(InputStream i, OutputStream o) { super(new BufferedOutputStream(o)); r = new BufferedReader(new InputStreamReader(i)); } public boolean hasMoreTokens() { return peekToken() != null; } public int getInt() { return Integer.parseInt(nextToken()); } public double getDouble() { return Double.parseDouble(nextToken()); } public long getLong() { return Long.parseLong(nextToken()); } public String getWord() { return nextToken(); } private BufferedReader r; private String line; private StringTokenizer st; private String token; private String peekToken() { if (token == null) try { while (st == null || !st.hasMoreTokens()) { line = r.readLine(); if (line == null) return null; st = new StringTokenizer(line); } token = st.nextToken(); } catch (IOException e) { } return token; } private String nextToken() { String ans = peekToken(); token = null; return ans; } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
719
3,070
2,391
import java.util.Scanner; public class d { public static void main(String[] args) { Scanner in = new Scanner(System.in); int size = in.nextInt(); int[] vals = new int[size]; long[] cum = new long[size]; for(int i=0; i<size; i++){ vals[i] = in.nextInt(); int c = 0; for(int j=0; j<i; j++) if(vals[j] > vals[i]) c++; if(i != 0) cum[i] = cum[i-1]+c; else cum[i] = c; } long tot = cum[size-1]; int q = in.nextInt(); int[] nv = new int[size]; for(int i=0; i<q; i++) { int l = in.nextInt()-1; int r = in.nextInt()-1; int n = (r-l); long add = (n*(n+1))/2 - (cum[r] - cum[l]); tot = tot - (cum[r] - cum[l]) + add; if(tot%2 == 0) System.out.println("even"); else System.out.println("odd"); // for(int j=0; j<=r-l; j++) // nv[l+j] = vals[r-j]; // // for(int j=0; j<=r-l; j++) // vals[l+j] = nv[l+j]; } } } /* 3 1 2 3 2 1 2 2 3 4 1 2 4 3 4 1 1 1 4 1 4 2 3 */
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class d { public static void main(String[] args) { Scanner in = new Scanner(System.in); int size = in.nextInt(); int[] vals = new int[size]; long[] cum = new long[size]; for(int i=0; i<size; i++){ vals[i] = in.nextInt(); int c = 0; for(int j=0; j<i; j++) if(vals[j] > vals[i]) c++; if(i != 0) cum[i] = cum[i-1]+c; else cum[i] = c; } long tot = cum[size-1]; int q = in.nextInt(); int[] nv = new int[size]; for(int i=0; i<q; i++) { int l = in.nextInt()-1; int r = in.nextInt()-1; int n = (r-l); long add = (n*(n+1))/2 - (cum[r] - cum[l]); tot = tot - (cum[r] - cum[l]) + add; if(tot%2 == 0) System.out.println("even"); else System.out.println("odd"); // for(int j=0; j<=r-l; j++) // nv[l+j] = vals[r-j]; // // for(int j=0; j<=r-l; j++) // vals[l+j] = nv[l+j]; } } } /* 3 1 2 3 2 1 2 2 3 4 1 2 4 3 4 1 1 1 4 1 4 2 3 */ </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class d { public static void main(String[] args) { Scanner in = new Scanner(System.in); int size = in.nextInt(); int[] vals = new int[size]; long[] cum = new long[size]; for(int i=0; i<size; i++){ vals[i] = in.nextInt(); int c = 0; for(int j=0; j<i; j++) if(vals[j] > vals[i]) c++; if(i != 0) cum[i] = cum[i-1]+c; else cum[i] = c; } long tot = cum[size-1]; int q = in.nextInt(); int[] nv = new int[size]; for(int i=0; i<q; i++) { int l = in.nextInt()-1; int r = in.nextInt()-1; int n = (r-l); long add = (n*(n+1))/2 - (cum[r] - cum[l]); tot = tot - (cum[r] - cum[l]) + add; if(tot%2 == 0) System.out.println("even"); else System.out.println("odd"); // for(int j=0; j<=r-l; j++) // nv[l+j] = vals[r-j]; // // for(int j=0; j<=r-l; j++) // vals[l+j] = nv[l+j]; } } } /* 3 1 2 3 2 1 2 2 3 4 1 2 4 3 4 1 1 1 4 1 4 2 3 */ </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
712
2,386
2,467
// Author: aman_robotics //package math_codet; import java.io.*; import java.util.*; public class lets_do { FastReader in; PrintWriter out; Helper_class h; final long mod = 1000000009; final int MAXN = 1000005; final int lgN = 20; final long INF = (long)1e18; final long MAX_Ai = (long)1e12; public static void main(String[] args) throws java.lang.Exception{ new lets_do().run(); } void run() throws Exception{ in=new FastReader(System.in); out = new PrintWriter(System.out); h = new Helper_class(); int t = 1; while(t--> 0) solve(); out.flush(); out.close(); } void solve(){ int n = h.ni(); long[] arr = new long[n]; int i = 0, j = 0; for(i = 0; i < n; i++) arr[i] = h.nl(); HashMap<Long, Integer> hmap = new HashMap<Long, Integer>(); int cnt = 0; for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; Integer x = hmap.get(sum); if(x == null) hmap.put(sum, cnt++); } } TreeSet<Pair>[] tset = new TreeSet[cnt]; for(i = 0; i < cnt; i++) tset[i] = new TreeSet<Pair>(com); for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; tset[hmap.get(sum)].add(new Pair(i, j)); } } int max = 0; int ind = -1; int max_x = 0, max_y = 0; for(i = 0; i < cnt; i++){ int curr_y = tset[i].first().y; int cnt1 = 1; for(Pair yo : tset[i]){ if(yo.x > curr_y) { cnt1++; curr_y = yo.y; } } if(max < cnt1) { max = cnt1; ind = i; } } h.pn(max); Pair hola_yee = new Pair(tset[ind].first().x, tset[ind].first().y); h.pn((tset[ind].first().x + 1) +" "+(tset[ind].first().y + 1)); int curr_y = tset[ind].first().y; for(Pair yo : tset[ind]){ if(yo.x > curr_y) { curr_y = yo.y; h.pn((yo.x + 1) +" "+(yo.y + 1)); } } } static final Comparator<Pair> com=new Comparator<Pair>(){ public int compare(Pair a, Pair b){ if(Integer.compare(a.y, b.y) != 0) return Integer.compare(a.y, b.y); else return Integer.compare(a.x, b.x); } }; class Pair{ int x; int y; Pair(int p, int q){ x = p; y = q; } } class Edge{ int u , v; long wt; Edge(int a, int b, long w){ u = a; v = b; wt = w; } int other(int x) { return u ^ v ^ x; } } class Helper_class{ long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);} int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);} int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));} void p(Object o){out.print(o);} void pn(Object o){out.println(o);} void pni(Object o){out.println(o);out.flush();} String n(){return in.next();} String nln(){return in.nextLine();} int ni(){return Integer.parseInt(in.next());} long nl(){return Long.parseLong(in.next());} double nd(){return Double.parseDouble(in.next());} long mul(long a,long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; a*=b; if(a>=mod)a%=mod; return a; } long modPow(long a, long p){ long o = 1; while(p>0){ if((p&1)==1)o = mul(o,a); a = mul(a,a); p>>=1; } return o; } long add(long a, long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; if(b<0)b+=mod; a+=b; if(a>=mod)a-=mod; return a; } } class FastReader{ private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new UnknownError(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new UnknownError(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int peek() { if (numChars == -1) return -1; if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { return -1; } if (numChars <= 0) return -1; } return buf[curChar]; } public void skip(int x) { while (x-- > 0) read(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String nextString() { return next(); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { StringBuffer buf = new StringBuffer(); int c = read(); while (c != '\n' && c != -1) { if (c != '\r') buf.appendCodePoint(c); c = read(); } return buf.toString(); } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public boolean hasNext() { int value; while (isSpaceChar(value = peek()) && value != -1) read(); return value != -1; } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> // Author: aman_robotics //package math_codet; import java.io.*; import java.util.*; public class lets_do { FastReader in; PrintWriter out; Helper_class h; final long mod = 1000000009; final int MAXN = 1000005; final int lgN = 20; final long INF = (long)1e18; final long MAX_Ai = (long)1e12; public static void main(String[] args) throws java.lang.Exception{ new lets_do().run(); } void run() throws Exception{ in=new FastReader(System.in); out = new PrintWriter(System.out); h = new Helper_class(); int t = 1; while(t--> 0) solve(); out.flush(); out.close(); } void solve(){ int n = h.ni(); long[] arr = new long[n]; int i = 0, j = 0; for(i = 0; i < n; i++) arr[i] = h.nl(); HashMap<Long, Integer> hmap = new HashMap<Long, Integer>(); int cnt = 0; for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; Integer x = hmap.get(sum); if(x == null) hmap.put(sum, cnt++); } } TreeSet<Pair>[] tset = new TreeSet[cnt]; for(i = 0; i < cnt; i++) tset[i] = new TreeSet<Pair>(com); for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; tset[hmap.get(sum)].add(new Pair(i, j)); } } int max = 0; int ind = -1; int max_x = 0, max_y = 0; for(i = 0; i < cnt; i++){ int curr_y = tset[i].first().y; int cnt1 = 1; for(Pair yo : tset[i]){ if(yo.x > curr_y) { cnt1++; curr_y = yo.y; } } if(max < cnt1) { max = cnt1; ind = i; } } h.pn(max); Pair hola_yee = new Pair(tset[ind].first().x, tset[ind].first().y); h.pn((tset[ind].first().x + 1) +" "+(tset[ind].first().y + 1)); int curr_y = tset[ind].first().y; for(Pair yo : tset[ind]){ if(yo.x > curr_y) { curr_y = yo.y; h.pn((yo.x + 1) +" "+(yo.y + 1)); } } } static final Comparator<Pair> com=new Comparator<Pair>(){ public int compare(Pair a, Pair b){ if(Integer.compare(a.y, b.y) != 0) return Integer.compare(a.y, b.y); else return Integer.compare(a.x, b.x); } }; class Pair{ int x; int y; Pair(int p, int q){ x = p; y = q; } } class Edge{ int u , v; long wt; Edge(int a, int b, long w){ u = a; v = b; wt = w; } int other(int x) { return u ^ v ^ x; } } class Helper_class{ long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);} int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);} int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));} void p(Object o){out.print(o);} void pn(Object o){out.println(o);} void pni(Object o){out.println(o);out.flush();} String n(){return in.next();} String nln(){return in.nextLine();} int ni(){return Integer.parseInt(in.next());} long nl(){return Long.parseLong(in.next());} double nd(){return Double.parseDouble(in.next());} long mul(long a,long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; a*=b; if(a>=mod)a%=mod; return a; } long modPow(long a, long p){ long o = 1; while(p>0){ if((p&1)==1)o = mul(o,a); a = mul(a,a); p>>=1; } return o; } long add(long a, long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; if(b<0)b+=mod; a+=b; if(a>=mod)a-=mod; return a; } } class FastReader{ private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new UnknownError(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new UnknownError(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int peek() { if (numChars == -1) return -1; if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { return -1; } if (numChars <= 0) return -1; } return buf[curChar]; } public void skip(int x) { while (x-- > 0) read(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String nextString() { return next(); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { StringBuffer buf = new StringBuffer(); int c = read(); while (c != '\n' && c != -1) { if (c != '\r') buf.appendCodePoint(c); c = read(); } return buf.toString(); } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public boolean hasNext() { int value; while (isSpaceChar(value = peek()) && value != -1) read(); return value != -1; } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> // Author: aman_robotics //package math_codet; import java.io.*; import java.util.*; public class lets_do { FastReader in; PrintWriter out; Helper_class h; final long mod = 1000000009; final int MAXN = 1000005; final int lgN = 20; final long INF = (long)1e18; final long MAX_Ai = (long)1e12; public static void main(String[] args) throws java.lang.Exception{ new lets_do().run(); } void run() throws Exception{ in=new FastReader(System.in); out = new PrintWriter(System.out); h = new Helper_class(); int t = 1; while(t--> 0) solve(); out.flush(); out.close(); } void solve(){ int n = h.ni(); long[] arr = new long[n]; int i = 0, j = 0; for(i = 0; i < n; i++) arr[i] = h.nl(); HashMap<Long, Integer> hmap = new HashMap<Long, Integer>(); int cnt = 0; for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; Integer x = hmap.get(sum); if(x == null) hmap.put(sum, cnt++); } } TreeSet<Pair>[] tset = new TreeSet[cnt]; for(i = 0; i < cnt; i++) tset[i] = new TreeSet<Pair>(com); for(i = 0; i < n; i++){ long sum = 0; for(j = i; j < n; j++){ sum += arr[j]; tset[hmap.get(sum)].add(new Pair(i, j)); } } int max = 0; int ind = -1; int max_x = 0, max_y = 0; for(i = 0; i < cnt; i++){ int curr_y = tset[i].first().y; int cnt1 = 1; for(Pair yo : tset[i]){ if(yo.x > curr_y) { cnt1++; curr_y = yo.y; } } if(max < cnt1) { max = cnt1; ind = i; } } h.pn(max); Pair hola_yee = new Pair(tset[ind].first().x, tset[ind].first().y); h.pn((tset[ind].first().x + 1) +" "+(tset[ind].first().y + 1)); int curr_y = tset[ind].first().y; for(Pair yo : tset[ind]){ if(yo.x > curr_y) { curr_y = yo.y; h.pn((yo.x + 1) +" "+(yo.y + 1)); } } } static final Comparator<Pair> com=new Comparator<Pair>(){ public int compare(Pair a, Pair b){ if(Integer.compare(a.y, b.y) != 0) return Integer.compare(a.y, b.y); else return Integer.compare(a.x, b.x); } }; class Pair{ int x; int y; Pair(int p, int q){ x = p; y = q; } } class Edge{ int u , v; long wt; Edge(int a, int b, long w){ u = a; v = b; wt = w; } int other(int x) { return u ^ v ^ x; } } class Helper_class{ long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);} int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);} int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));} void p(Object o){out.print(o);} void pn(Object o){out.println(o);} void pni(Object o){out.println(o);out.flush();} String n(){return in.next();} String nln(){return in.nextLine();} int ni(){return Integer.parseInt(in.next());} long nl(){return Long.parseLong(in.next());} double nd(){return Double.parseDouble(in.next());} long mul(long a,long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; a*=b; if(a>=mod)a%=mod; return a; } long modPow(long a, long p){ long o = 1; while(p>0){ if((p&1)==1)o = mul(o,a); a = mul(a,a); p>>=1; } return o; } long add(long a, long b){ if(a>=mod)a%=mod; if(b>=mod)b%=mod; if(b<0)b+=mod; a+=b; if(a>=mod)a-=mod; return a; } } class FastReader{ private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public FastReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new UnknownError(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new UnknownError(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int peek() { if (numChars == -1) return -1; if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { return -1; } if (numChars <= 0) return -1; } return buf[curChar]; } public void skip(int x) { while (x-- > 0) read(); } public int nextInt() { return Integer.parseInt(next()); } public long nextLong() { return Long.parseLong(next()); } public String nextString() { return next(); } public String next() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { StringBuffer buf = new StringBuffer(); int c = read(); while (c != '\n' && c != -1) { if (c != '\r') buf.appendCodePoint(c); c = read(); } return buf.toString(); } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public boolean hasNext() { int value; while (isSpaceChar(value = peek()) && value != -1) read(); return value != -1; } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
2,246
2,462
494
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author prakhar897 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskA solver = new TaskA(); solver.solve(1, in, out); out.close(); } static class TaskA { public void solve(int testNumber, InputReader in, OutputWriter out) { int n = in.nextInt(); String arr1[] = new String[n]; String arr2[] = new String[n]; int i, j, count = 0; for (i = 0; i < n; i++) { arr1[i] = in.nextString(); } for (i = 0; i < n; i++) { arr2[i] = in.nextString(); for (j = 0; j < n; j++) { if (arr2[i].equals(arr1[j])) { arr1[j] = ""; count++; break; } } } out.println(n - count); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = read(); while (isSpaceChar(c)) { c = read(); } StringBuilder res = new StringBuilder(); do { if (Character.isValidCodePoint(c)) { res.appendCodePoint(c); } c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void close() { writer.close(); } public void println(int i) { writer.println(i); } } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author prakhar897 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskA solver = new TaskA(); solver.solve(1, in, out); out.close(); } static class TaskA { public void solve(int testNumber, InputReader in, OutputWriter out) { int n = in.nextInt(); String arr1[] = new String[n]; String arr2[] = new String[n]; int i, j, count = 0; for (i = 0; i < n; i++) { arr1[i] = in.nextString(); } for (i = 0; i < n; i++) { arr2[i] = in.nextString(); for (j = 0; j < n; j++) { if (arr2[i].equals(arr1[j])) { arr1[j] = ""; count++; break; } } } out.println(n - count); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = read(); while (isSpaceChar(c)) { c = read(); } StringBuilder res = new StringBuilder(); do { if (Character.isValidCodePoint(c)) { res.appendCodePoint(c); } c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void close() { writer.close(); } public void println(int i) { writer.println(i); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top * * @author prakhar897 */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskA solver = new TaskA(); solver.solve(1, in, out); out.close(); } static class TaskA { public void solve(int testNumber, InputReader in, OutputWriter out) { int n = in.nextInt(); String arr1[] = new String[n]; String arr2[] = new String[n]; int i, j, count = 0; for (i = 0; i < n; i++) { arr1[i] = in.nextString(); } for (i = 0; i < n; i++) { arr2[i] = in.nextString(); for (j = 0; j < n; j++) { if (arr2[i].equals(arr1[j])) { arr1[j] = ""; count++; break; } } } out.println(n - count); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = read(); while (isSpaceChar(c)) { c = read(); } StringBuilder res = new StringBuilder(); do { if (Character.isValidCodePoint(c)) { res.appendCodePoint(c); } c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void close() { writer.close(); } public void println(int i) { writer.println(i); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,235
493
4,310
import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.Arrays; import java.util.Locale; import java.util.StringTokenizer; import java.util.TreeMap; import java.util.Map.Entry; public class Solution implements Runnable { public static void main(String[] args) { (new Thread(new Solution())).start(); } BufferedReader in; PrintWriter out; StringTokenizer st; String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String r = in.readLine(); if (r == null) return null; st = new StringTokenizer(r); } return st.nextToken(); } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } int n; int[] b, l; double ans; int[] sw; int a; void sol() { for (int i = 0; i < n; i++) l[i] += sw[i] * 10; double yes = 0; for (int q = 0; q < (1 << n); q++) { double p = 1; int bb = 0; int cnt = 0; for (int i = 0; i < n; i++) { if ((q & (1 << i)) == 0) { p *= (1.0 - (double)l[i] / 100); bb += b[i]; } else { p *= 1.0 * (double)l[i] / 100; cnt++; } } if (cnt > n / 2) { yes += p; } else { yes += p * (double)a / (double)(a + bb); } } if (ans < yes) ans = yes; for (int i = 0; i < n; i++) l[i] -= sw[i] * 10; } void rek(int i, int k) { if (i == n) sol(); else { for (int q = 0; q <= k && l[i] + q * 10 <= 100; q++) { sw[i] = q; rek(i + 1, k - q); } } } void solve() throws Exception { n = nextInt(); int k = nextInt(); a = nextInt(); b = new int[n]; l = new int[n]; sw = new int[n]; for (int i = 0; i < n; i++) { b[i] = nextInt(); l[i] = nextInt(); } rek(0, k); out.printf("%.10f", ans); } public void run() { Locale.setDefault(Locale.UK); try { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); // in = new BufferedReader(new FileReader("knights.in")); // out = new PrintWriter("knights.out"); solve(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } finally { out.flush(); } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.Arrays; import java.util.Locale; import java.util.StringTokenizer; import java.util.TreeMap; import java.util.Map.Entry; public class Solution implements Runnable { public static void main(String[] args) { (new Thread(new Solution())).start(); } BufferedReader in; PrintWriter out; StringTokenizer st; String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String r = in.readLine(); if (r == null) return null; st = new StringTokenizer(r); } return st.nextToken(); } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } int n; int[] b, l; double ans; int[] sw; int a; void sol() { for (int i = 0; i < n; i++) l[i] += sw[i] * 10; double yes = 0; for (int q = 0; q < (1 << n); q++) { double p = 1; int bb = 0; int cnt = 0; for (int i = 0; i < n; i++) { if ((q & (1 << i)) == 0) { p *= (1.0 - (double)l[i] / 100); bb += b[i]; } else { p *= 1.0 * (double)l[i] / 100; cnt++; } } if (cnt > n / 2) { yes += p; } else { yes += p * (double)a / (double)(a + bb); } } if (ans < yes) ans = yes; for (int i = 0; i < n; i++) l[i] -= sw[i] * 10; } void rek(int i, int k) { if (i == n) sol(); else { for (int q = 0; q <= k && l[i] + q * 10 <= 100; q++) { sw[i] = q; rek(i + 1, k - q); } } } void solve() throws Exception { n = nextInt(); int k = nextInt(); a = nextInt(); b = new int[n]; l = new int[n]; sw = new int[n]; for (int i = 0; i < n; i++) { b[i] = nextInt(); l[i] = nextInt(); } rek(0, k); out.printf("%.10f", ans); } public void run() { Locale.setDefault(Locale.UK); try { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); // in = new BufferedReader(new FileReader("knights.in")); // out = new PrintWriter("knights.out"); solve(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } finally { out.flush(); } } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.Arrays; import java.util.Locale; import java.util.StringTokenizer; import java.util.TreeMap; import java.util.Map.Entry; public class Solution implements Runnable { public static void main(String[] args) { (new Thread(new Solution())).start(); } BufferedReader in; PrintWriter out; StringTokenizer st; String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { String r = in.readLine(); if (r == null) return null; st = new StringTokenizer(r); } return st.nextToken(); } int nextInt() throws IOException { return Integer.parseInt(nextToken()); } long nextLong() throws IOException { return Long.parseLong(nextToken()); } double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } int n; int[] b, l; double ans; int[] sw; int a; void sol() { for (int i = 0; i < n; i++) l[i] += sw[i] * 10; double yes = 0; for (int q = 0; q < (1 << n); q++) { double p = 1; int bb = 0; int cnt = 0; for (int i = 0; i < n; i++) { if ((q & (1 << i)) == 0) { p *= (1.0 - (double)l[i] / 100); bb += b[i]; } else { p *= 1.0 * (double)l[i] / 100; cnt++; } } if (cnt > n / 2) { yes += p; } else { yes += p * (double)a / (double)(a + bb); } } if (ans < yes) ans = yes; for (int i = 0; i < n; i++) l[i] -= sw[i] * 10; } void rek(int i, int k) { if (i == n) sol(); else { for (int q = 0; q <= k && l[i] + q * 10 <= 100; q++) { sw[i] = q; rek(i + 1, k - q); } } } void solve() throws Exception { n = nextInt(); int k = nextInt(); a = nextInt(); b = new int[n]; l = new int[n]; sw = new int[n]; for (int i = 0; i < n; i++) { b[i] = nextInt(); l[i] = nextInt(); } rek(0, k); out.printf("%.10f", ans); } public void run() { Locale.setDefault(Locale.UK); try { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); // in = new BufferedReader(new FileReader("knights.in")); // out = new PrintWriter("knights.out"); solve(); } catch (Exception e) { e.printStackTrace(); System.exit(1); } finally { out.flush(); } } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,080
4,299
1,554
import java.io.*; import java.math.*; import java.util.*; import java.lang.*; public class Main { static Input in; static Output out; public static void main(String[] args) throws IOException { in = new Input(System.in); out = new Output(System.out); run(); out.close(); System.exit(0); } private static void run() throws IOException { int n = in.nextInt(), k = in.nextInt(); int[] A = new int[n]; for (int i = 0; i < n; i++) { A[i] = in.nextInt(); } Arrays.sort(A); int count = n; boolean[] hash = new boolean[n]; for (int i = n-1; i > 0; i--) { if(!hash[i]) { int a = A[i]; if(a % k == 0) { int p = a / k; int j = Arrays.binarySearch(A, p); if(j >= 0 && j < i) { hash[j] = true; count--; } } } } out.print(count); } } class Input { final int SIZE = 8192; private InputStream in; private byte[] buf = new byte[SIZE]; private int last, current, total; public Input(InputStream stream) throws IOException { in = stream; last = read(); } private int read() throws IOException { if (total == -1) return -1; if (current >= total) { current = 0; total = in.read(buf); if (total <= 0) return -1; } return buf[current++]; } private void advance() throws IOException { while (true) { if (last == -1) return; if (!isValidChar(last)) last = read(); else break; } } private boolean isValidChar(int c) { return c > 32 && c < 127; } public boolean isEOF() throws IOException { advance(); return last == -1; } public String nextString() throws IOException { advance(); if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (!isValidChar(last)) break; } return s.toString(); } public String nextLine() throws IOException { if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (last == '\n' || last == -1) break; } return s.toString(); } public String nextLine(boolean ignoreIfEmpty) throws IOException { if (!ignoreIfEmpty) return nextLine(); String s = nextLine(); while (s.trim().length() == 0) s = nextLine(); return s; } public int nextInt() throws IOException { advance(); if (last == -1) throw new EOFException(); int n = 0, s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public long nextLong() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } long n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public BigInteger nextBigInt() throws IOException { return new BigInteger(nextString()); } public char nextChar() throws IOException { advance(); return (char) last; } public double nextDouble() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } double n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last) || last == '.') break; } if (last == '.') { last = read(); if (last == -1) throw new EOFException(); double m = 1; while (true) { m = m / 10; n = n + (last - '0') * m; last = read(); if (!isValidChar(last)) break; } } return n * s; } public BigDecimal nextBigDecimal() throws IOException { return new BigDecimal(nextString()); } public void close() throws IOException { in.close(); in = null; } } class Output { final int SIZE = 8192; private Writer out; private char cb[] = new char[SIZE]; private int nChars = SIZE, nextChar = 0; private char lineSeparator = '\n'; public Output(OutputStream stream) { out = new OutputStreamWriter(stream); } void flushBuffer() throws IOException { if (nextChar == 0) return; out.write(cb, 0, nextChar); nextChar = 0; } void write(int c) throws IOException { if (nextChar >= nChars) flushBuffer(); cb[nextChar++] = (char) c; } void write(String s, int off, int len) throws IOException { int b = off, t = off + len; while (b < t) { int a = nChars - nextChar, a1 = t - b; int d = a < a1 ? a : a1; s.getChars(b, b + d, cb, nextChar); b += d; nextChar += d; if (nextChar >= nChars) flushBuffer(); } } void write(String s) throws IOException { write(s, 0, s.length()); } public void print(Object obj) throws IOException { write(String.valueOf(obj)); } public void println(Object obj) throws IOException { write(String.valueOf(obj)); write(lineSeparator); } public void printf(String format, Object... obj) throws IOException { write(String.format(format, obj)); } public void close() throws IOException { flushBuffer(); out.close(); out = null; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.math.*; import java.util.*; import java.lang.*; public class Main { static Input in; static Output out; public static void main(String[] args) throws IOException { in = new Input(System.in); out = new Output(System.out); run(); out.close(); System.exit(0); } private static void run() throws IOException { int n = in.nextInt(), k = in.nextInt(); int[] A = new int[n]; for (int i = 0; i < n; i++) { A[i] = in.nextInt(); } Arrays.sort(A); int count = n; boolean[] hash = new boolean[n]; for (int i = n-1; i > 0; i--) { if(!hash[i]) { int a = A[i]; if(a % k == 0) { int p = a / k; int j = Arrays.binarySearch(A, p); if(j >= 0 && j < i) { hash[j] = true; count--; } } } } out.print(count); } } class Input { final int SIZE = 8192; private InputStream in; private byte[] buf = new byte[SIZE]; private int last, current, total; public Input(InputStream stream) throws IOException { in = stream; last = read(); } private int read() throws IOException { if (total == -1) return -1; if (current >= total) { current = 0; total = in.read(buf); if (total <= 0) return -1; } return buf[current++]; } private void advance() throws IOException { while (true) { if (last == -1) return; if (!isValidChar(last)) last = read(); else break; } } private boolean isValidChar(int c) { return c > 32 && c < 127; } public boolean isEOF() throws IOException { advance(); return last == -1; } public String nextString() throws IOException { advance(); if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (!isValidChar(last)) break; } return s.toString(); } public String nextLine() throws IOException { if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (last == '\n' || last == -1) break; } return s.toString(); } public String nextLine(boolean ignoreIfEmpty) throws IOException { if (!ignoreIfEmpty) return nextLine(); String s = nextLine(); while (s.trim().length() == 0) s = nextLine(); return s; } public int nextInt() throws IOException { advance(); if (last == -1) throw new EOFException(); int n = 0, s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public long nextLong() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } long n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public BigInteger nextBigInt() throws IOException { return new BigInteger(nextString()); } public char nextChar() throws IOException { advance(); return (char) last; } public double nextDouble() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } double n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last) || last == '.') break; } if (last == '.') { last = read(); if (last == -1) throw new EOFException(); double m = 1; while (true) { m = m / 10; n = n + (last - '0') * m; last = read(); if (!isValidChar(last)) break; } } return n * s; } public BigDecimal nextBigDecimal() throws IOException { return new BigDecimal(nextString()); } public void close() throws IOException { in.close(); in = null; } } class Output { final int SIZE = 8192; private Writer out; private char cb[] = new char[SIZE]; private int nChars = SIZE, nextChar = 0; private char lineSeparator = '\n'; public Output(OutputStream stream) { out = new OutputStreamWriter(stream); } void flushBuffer() throws IOException { if (nextChar == 0) return; out.write(cb, 0, nextChar); nextChar = 0; } void write(int c) throws IOException { if (nextChar >= nChars) flushBuffer(); cb[nextChar++] = (char) c; } void write(String s, int off, int len) throws IOException { int b = off, t = off + len; while (b < t) { int a = nChars - nextChar, a1 = t - b; int d = a < a1 ? a : a1; s.getChars(b, b + d, cb, nextChar); b += d; nextChar += d; if (nextChar >= nChars) flushBuffer(); } } void write(String s) throws IOException { write(s, 0, s.length()); } public void print(Object obj) throws IOException { write(String.valueOf(obj)); } public void println(Object obj) throws IOException { write(String.valueOf(obj)); write(lineSeparator); } public void printf(String format, Object... obj) throws IOException { write(String.format(format, obj)); } public void close() throws IOException { flushBuffer(); out.close(); out = null; } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.math.*; import java.util.*; import java.lang.*; public class Main { static Input in; static Output out; public static void main(String[] args) throws IOException { in = new Input(System.in); out = new Output(System.out); run(); out.close(); System.exit(0); } private static void run() throws IOException { int n = in.nextInt(), k = in.nextInt(); int[] A = new int[n]; for (int i = 0; i < n; i++) { A[i] = in.nextInt(); } Arrays.sort(A); int count = n; boolean[] hash = new boolean[n]; for (int i = n-1; i > 0; i--) { if(!hash[i]) { int a = A[i]; if(a % k == 0) { int p = a / k; int j = Arrays.binarySearch(A, p); if(j >= 0 && j < i) { hash[j] = true; count--; } } } } out.print(count); } } class Input { final int SIZE = 8192; private InputStream in; private byte[] buf = new byte[SIZE]; private int last, current, total; public Input(InputStream stream) throws IOException { in = stream; last = read(); } private int read() throws IOException { if (total == -1) return -1; if (current >= total) { current = 0; total = in.read(buf); if (total <= 0) return -1; } return buf[current++]; } private void advance() throws IOException { while (true) { if (last == -1) return; if (!isValidChar(last)) last = read(); else break; } } private boolean isValidChar(int c) { return c > 32 && c < 127; } public boolean isEOF() throws IOException { advance(); return last == -1; } public String nextString() throws IOException { advance(); if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (!isValidChar(last)) break; } return s.toString(); } public String nextLine() throws IOException { if (last == -1) throw new EOFException(); StringBuilder s = new StringBuilder(); while (true) { s.appendCodePoint(last); last = read(); if (last == '\n' || last == -1) break; } return s.toString(); } public String nextLine(boolean ignoreIfEmpty) throws IOException { if (!ignoreIfEmpty) return nextLine(); String s = nextLine(); while (s.trim().length() == 0) s = nextLine(); return s; } public int nextInt() throws IOException { advance(); if (last == -1) throw new EOFException(); int n = 0, s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public long nextLong() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } long n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last)) break; } return n * s; } public BigInteger nextBigInt() throws IOException { return new BigInteger(nextString()); } public char nextChar() throws IOException { advance(); return (char) last; } public double nextDouble() throws IOException { advance(); if (last == -1) throw new EOFException(); int s = 1; if (last == '-') { s = -1; last = read(); if (last == -1) throw new EOFException(); } double n = 0; while (true) { n = n * 10 + last - '0'; last = read(); if (!isValidChar(last) || last == '.') break; } if (last == '.') { last = read(); if (last == -1) throw new EOFException(); double m = 1; while (true) { m = m / 10; n = n + (last - '0') * m; last = read(); if (!isValidChar(last)) break; } } return n * s; } public BigDecimal nextBigDecimal() throws IOException { return new BigDecimal(nextString()); } public void close() throws IOException { in.close(); in = null; } } class Output { final int SIZE = 8192; private Writer out; private char cb[] = new char[SIZE]; private int nChars = SIZE, nextChar = 0; private char lineSeparator = '\n'; public Output(OutputStream stream) { out = new OutputStreamWriter(stream); } void flushBuffer() throws IOException { if (nextChar == 0) return; out.write(cb, 0, nextChar); nextChar = 0; } void write(int c) throws IOException { if (nextChar >= nChars) flushBuffer(); cb[nextChar++] = (char) c; } void write(String s, int off, int len) throws IOException { int b = off, t = off + len; while (b < t) { int a = nChars - nextChar, a1 = t - b; int d = a < a1 ? a : a1; s.getChars(b, b + d, cb, nextChar); b += d; nextChar += d; if (nextChar >= nChars) flushBuffer(); } } void write(String s) throws IOException { write(s, 0, s.length()); } public void print(Object obj) throws IOException { write(String.valueOf(obj)); } public void println(Object obj) throws IOException { write(String.valueOf(obj)); write(lineSeparator); } public void printf(String format, Object... obj) throws IOException { write(String.format(format, obj)); } public void close() throws IOException { flushBuffer(); out.close(); out = null; } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,923
1,552
1,576
import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.util.InputMismatchException; public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); Task solver = new Task(); int testCount = 1; for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } } class Task { int n; int[] a; int[] b; public void solve(int testNumber, InputReader in, PrintWriter out) { n = in.readInt(); a = new int[n]; b = new int[n]; for (int i = 0; i < n; ++i) a[i] = b[i] = in.readInt(); sort(0, n - 1); int different = 0; for (int i = 0; i < n; ++i) if (a[i] != b[i]) ++different; out.println(different <= 2 ? "YES" : "NO"); } public void sort(int lo, int hi) { if (lo < hi) { int mid = (lo + hi) / 2; sort(lo, mid); sort(mid + 1, hi); merge(lo, mid, hi); } } public void merge(int lo, int mid, int hi) { int n1 = mid - lo + 1; int n2 = hi - (mid + 1) + 1; int[] x = new int[n1 + 1]; int[] y = new int[n2 + 1]; for (int i = 0; i < n1; ++i) x[i] = b[lo + i]; for (int j = 0; j < n2; ++j) y[j] = b[mid + 1 + j]; x[n1] = y[n2] = Integer.MAX_VALUE; for (int k = lo, i = 0, j = 0; k <= hi; ++k) b[k] = x[i] < y[j] ? x[i++] : y[j++]; } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public Long readLong() { return Long.parseLong(readString()); } public Double readDouble() { return Double.parseDouble(readString()); } public static boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.util.InputMismatchException; public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); Task solver = new Task(); int testCount = 1; for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } } class Task { int n; int[] a; int[] b; public void solve(int testNumber, InputReader in, PrintWriter out) { n = in.readInt(); a = new int[n]; b = new int[n]; for (int i = 0; i < n; ++i) a[i] = b[i] = in.readInt(); sort(0, n - 1); int different = 0; for (int i = 0; i < n; ++i) if (a[i] != b[i]) ++different; out.println(different <= 2 ? "YES" : "NO"); } public void sort(int lo, int hi) { if (lo < hi) { int mid = (lo + hi) / 2; sort(lo, mid); sort(mid + 1, hi); merge(lo, mid, hi); } } public void merge(int lo, int mid, int hi) { int n1 = mid - lo + 1; int n2 = hi - (mid + 1) + 1; int[] x = new int[n1 + 1]; int[] y = new int[n2 + 1]; for (int i = 0; i < n1; ++i) x[i] = b[lo + i]; for (int j = 0; j < n2; ++j) y[j] = b[mid + 1 + j]; x[n1] = y[n2] = Integer.MAX_VALUE; for (int k = lo, i = 0, j = 0; k <= hi; ++k) b[k] = x[i] < y[j] ? x[i++] : y[j++]; } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public Long readLong() { return Long.parseLong(readString()); } public Double readDouble() { return Double.parseDouble(readString()); } public static boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.util.InputMismatchException; public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); Task solver = new Task(); int testCount = 1; for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } } class Task { int n; int[] a; int[] b; public void solve(int testNumber, InputReader in, PrintWriter out) { n = in.readInt(); a = new int[n]; b = new int[n]; for (int i = 0; i < n; ++i) a[i] = b[i] = in.readInt(); sort(0, n - 1); int different = 0; for (int i = 0; i < n; ++i) if (a[i] != b[i]) ++different; out.println(different <= 2 ? "YES" : "NO"); } public void sort(int lo, int hi) { if (lo < hi) { int mid = (lo + hi) / 2; sort(lo, mid); sort(mid + 1, hi); merge(lo, mid, hi); } } public void merge(int lo, int mid, int hi) { int n1 = mid - lo + 1; int n2 = hi - (mid + 1) + 1; int[] x = new int[n1 + 1]; int[] y = new int[n2 + 1]; for (int i = 0; i < n1; ++i) x[i] = b[lo + i]; for (int j = 0; j < n2; ++j) y[j] = b[mid + 1 + j]; x[n1] = y[n2] = Integer.MAX_VALUE; for (int k = lo, i = 0, j = 0; k <= hi; ++k) b[k] = x[i] < y[j] ? x[i++] : y[j++]; } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuffer res = new StringBuffer(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public Long readLong() { return Long.parseLong(readString()); } public Double readDouble() { return Double.parseDouble(readString()); } public static boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,254
1,573
2,088
import java.util.Arrays; import java.util.Scanner; public class AAA { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); int max = a[0]; int ind = 0; for (int k = 1; k < n; k++) { if (a[k] > max) { max = a[k]; ind = k; } } if (max != 1) { a[ind] = 1; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } else { a[0] = 2; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class AAA { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); int max = a[0]; int ind = 0; for (int k = 1; k < n; k++) { if (a[k] > max) { max = a[k]; ind = k; } } if (max != 1) { a[ind] = 1; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } else { a[0] = 2; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class AAA { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); int max = a[0]; int ind = 0; for (int k = 1; k < n; k++) { if (a[k] > max) { max = a[k]; ind = k; } } if (max != 1) { a[ind] = 1; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } else { a[0] = 2; Arrays.sort(a); for (int i = 0; i < a.length - 1; i++) System.out.print(a[i] + " "); System.out.println(a[a.length - 1]); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
585
2,084
2,818
import java.util.*; public class subtractionn { public static void main(String[] args) { // TODO Auto-generated method stub Scanner in = new Scanner(System.in); int t; t=in.nextInt(); while(t!=0) { int a=in.nextInt(); int b=in.nextInt(); int total=0,neww=0; if(a%b==0) { System.out.println(a/b); } else if(b%a==0) { System.out.println(b/a); } else { while(a!=0 && b!=0) { if(a>b) { total=total+(a/b); a=a%b; if(a==0) { break; } } else if(b>a) { total=total+(b/a); b=b%a; if(b==0) { break; } } else { System.exit(0); } } System.out.println(total); } t--; } } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; public class subtractionn { public static void main(String[] args) { // TODO Auto-generated method stub Scanner in = new Scanner(System.in); int t; t=in.nextInt(); while(t!=0) { int a=in.nextInt(); int b=in.nextInt(); int total=0,neww=0; if(a%b==0) { System.out.println(a/b); } else if(b%a==0) { System.out.println(b/a); } else { while(a!=0 && b!=0) { if(a>b) { total=total+(a/b); a=a%b; if(a==0) { break; } } else if(b>a) { total=total+(b/a); b=b%a; if(b==0) { break; } } else { System.exit(0); } } System.out.println(total); } t--; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; public class subtractionn { public static void main(String[] args) { // TODO Auto-generated method stub Scanner in = new Scanner(System.in); int t; t=in.nextInt(); while(t!=0) { int a=in.nextInt(); int b=in.nextInt(); int total=0,neww=0; if(a%b==0) { System.out.println(a/b); } else if(b%a==0) { System.out.println(b/a); } else { while(a!=0 && b!=0) { if(a>b) { total=total+(a/b); a=a%b; if(a==0) { break; } } else if(b>a) { total=total+(b/a); b=b%a; if(b==0) { break; } } else { System.exit(0); } } System.out.println(total); } t--; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
606
2,812
1,779
import java.io.*; import java.util.*; import java.math.*; import static java.lang.Math.*; public class Main implements Runnable { class Home implements Comparable<Home> { @Override public int compareTo(Home arg0) { return st - arg0.st; } int st, end; } public void solve() throws IOException { int n = nextInt(), t = nextInt() * 2; Home[] h = new Home[n]; for(int i = 0; i < n; ++i) { int x = nextInt() * 2, a = nextInt() * 2; h[i] = new Home(); h[i].st = x - a / 2; h[i].end = x + a / 2; } Arrays.sort(h); int ans = 2; for(int i = 0; i + 1 < n; ++i) { int delta = h[i + 1].st - h[i].end; if (delta == t) ans++; if (delta > t) ans += 2; } pw.println(ans); } static final String filename = "A"; static final boolean fromConsole = true; public void run() { try { if (!fromConsole) { in = new BufferedReader(new FileReader(filename + ".in")); pw = new PrintWriter(filename + ".out"); } else { in = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); } st = new StringTokenizer(""); long st = System.currentTimeMillis(); solve(); //pw.printf("\nWorking time: %d ms\n", System.currentTimeMillis() - st); pw.close(); } catch (Exception e) { e.printStackTrace(); System.exit(-1); } } private StringTokenizer st; private BufferedReader in; private PrintWriter pw; boolean hasNext() throws IOException { while (!st.hasMoreTokens()) { String line = in.readLine(); if (line == null) { return false; } st = new StringTokenizer(line); } return st.hasMoreTokens(); } String next() throws IOException { return hasNext() ? st.nextToken() : null; } int nextInt() throws IOException { return Integer.parseInt(next()); } BigInteger nextBigInteger() throws IOException { return new BigInteger(next()); } long nextLong() throws IOException { return Long.parseLong(next()); } double nextDouble() throws IOException { return Double.parseDouble(next()); } public static void main(String[] args) { new Thread(new Main()).start(); } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; import static java.lang.Math.*; public class Main implements Runnable { class Home implements Comparable<Home> { @Override public int compareTo(Home arg0) { return st - arg0.st; } int st, end; } public void solve() throws IOException { int n = nextInt(), t = nextInt() * 2; Home[] h = new Home[n]; for(int i = 0; i < n; ++i) { int x = nextInt() * 2, a = nextInt() * 2; h[i] = new Home(); h[i].st = x - a / 2; h[i].end = x + a / 2; } Arrays.sort(h); int ans = 2; for(int i = 0; i + 1 < n; ++i) { int delta = h[i + 1].st - h[i].end; if (delta == t) ans++; if (delta > t) ans += 2; } pw.println(ans); } static final String filename = "A"; static final boolean fromConsole = true; public void run() { try { if (!fromConsole) { in = new BufferedReader(new FileReader(filename + ".in")); pw = new PrintWriter(filename + ".out"); } else { in = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); } st = new StringTokenizer(""); long st = System.currentTimeMillis(); solve(); //pw.printf("\nWorking time: %d ms\n", System.currentTimeMillis() - st); pw.close(); } catch (Exception e) { e.printStackTrace(); System.exit(-1); } } private StringTokenizer st; private BufferedReader in; private PrintWriter pw; boolean hasNext() throws IOException { while (!st.hasMoreTokens()) { String line = in.readLine(); if (line == null) { return false; } st = new StringTokenizer(line); } return st.hasMoreTokens(); } String next() throws IOException { return hasNext() ? st.nextToken() : null; } int nextInt() throws IOException { return Integer.parseInt(next()); } BigInteger nextBigInteger() throws IOException { return new BigInteger(next()); } long nextLong() throws IOException { return Long.parseLong(next()); } double nextDouble() throws IOException { return Double.parseDouble(next()); } public static void main(String[] args) { new Thread(new Main()).start(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; import static java.lang.Math.*; public class Main implements Runnable { class Home implements Comparable<Home> { @Override public int compareTo(Home arg0) { return st - arg0.st; } int st, end; } public void solve() throws IOException { int n = nextInt(), t = nextInt() * 2; Home[] h = new Home[n]; for(int i = 0; i < n; ++i) { int x = nextInt() * 2, a = nextInt() * 2; h[i] = new Home(); h[i].st = x - a / 2; h[i].end = x + a / 2; } Arrays.sort(h); int ans = 2; for(int i = 0; i + 1 < n; ++i) { int delta = h[i + 1].st - h[i].end; if (delta == t) ans++; if (delta > t) ans += 2; } pw.println(ans); } static final String filename = "A"; static final boolean fromConsole = true; public void run() { try { if (!fromConsole) { in = new BufferedReader(new FileReader(filename + ".in")); pw = new PrintWriter(filename + ".out"); } else { in = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); } st = new StringTokenizer(""); long st = System.currentTimeMillis(); solve(); //pw.printf("\nWorking time: %d ms\n", System.currentTimeMillis() - st); pw.close(); } catch (Exception e) { e.printStackTrace(); System.exit(-1); } } private StringTokenizer st; private BufferedReader in; private PrintWriter pw; boolean hasNext() throws IOException { while (!st.hasMoreTokens()) { String line = in.readLine(); if (line == null) { return false; } st = new StringTokenizer(line); } return st.hasMoreTokens(); } String next() throws IOException { return hasNext() ? st.nextToken() : null; } int nextInt() throws IOException { return Integer.parseInt(next()); } BigInteger nextBigInteger() throws IOException { return new BigInteger(next()); } long nextLong() throws IOException { return Long.parseLong(next()); } double nextDouble() throws IOException { return Double.parseDouble(next()); } public static void main(String[] args) { new Thread(new Main()).start(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
933
1,775
4,471
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; import java.io.IOException; import java.util.Comparator; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastReader in = new FastReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = Integer.parseInt(in.next()); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { public void solve(int testNumber, FastReader in, PrintWriter out) { int n = in.nextInt(); int m = in.nextInt(); E2RotateColumnsHardVersion.Column[] columns = new E2RotateColumnsHardVersion.Column[m]; for (int i = 0; i < columns.length; ++i) columns[i] = new E2RotateColumnsHardVersion.Column(new int[n]); for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { columns[j].v[i] = in.nextInt(); if (i == n - 1) columns[j].initMax(); } } Arrays.sort(columns, new Comparator<E2RotateColumnsHardVersion.Column>() { public int compare(E2RotateColumnsHardVersion.Column o1, E2RotateColumnsHardVersion.Column o2) { return o2.max - o1.max; } }); if (columns.length > n) columns = Arrays.copyOf(columns, n); long[] dp = new long[1 << n]; for (E2RotateColumnsHardVersion.Column c : columns) { long[] ndp = new long[1 << n]; System.arraycopy(dp, 0, ndp, 0, dp.length); for (int rot = 0; rot < n; ++rot) { long[] temp = new long[1 << n]; System.arraycopy(dp, 0, temp, 0, dp.length); for (int i = 0, pos = rot; i < n; ++i, ++pos) { if (pos >= n) pos = 0; int val = c.v[pos]; for (int j = 0; j < temp.length; ++j) { if ((j & (1 << i)) == 0) temp[j | (1 << i)] = Math.max(temp[j | (1 << i)], temp[j] + val); } } for (int i = 0; i < ndp.length; ++i) ndp[i] = Math.max(ndp[i], temp[i]); } dp = ndp; } out.println(dp[dp.length - 1]); } static class Column { int[] v; int max; public Column(int[] v) { this.v = v; } void initMax() { max = 0; for (int vv : v) max = Math.max(max, vv); } } } static class FastReader { private InputStream stream; private byte[] buf = new byte[8192]; private int curChar; private int pnumChars; public FastReader(InputStream stream) { this.stream = stream; } private int pread() { if (pnumChars == -1) { throw new InputMismatchException(); } if (curChar >= pnumChars) { curChar = 0; try { pnumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (pnumChars <= 0) { return -1; } } return buf[curChar++]; } public String next() { return nextString(); } public int nextInt() { int c = pread(); while (isSpaceChar(c)) c = pread(); int sgn = 1; if (c == '-') { sgn = -1; c = pread(); } int res = 0; do { if (c == ',') { c = pread(); } if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = pread(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = pread(); while (isSpaceChar(c)) c = pread(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = pread(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; import java.io.IOException; import java.util.Comparator; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastReader in = new FastReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = Integer.parseInt(in.next()); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { public void solve(int testNumber, FastReader in, PrintWriter out) { int n = in.nextInt(); int m = in.nextInt(); E2RotateColumnsHardVersion.Column[] columns = new E2RotateColumnsHardVersion.Column[m]; for (int i = 0; i < columns.length; ++i) columns[i] = new E2RotateColumnsHardVersion.Column(new int[n]); for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { columns[j].v[i] = in.nextInt(); if (i == n - 1) columns[j].initMax(); } } Arrays.sort(columns, new Comparator<E2RotateColumnsHardVersion.Column>() { public int compare(E2RotateColumnsHardVersion.Column o1, E2RotateColumnsHardVersion.Column o2) { return o2.max - o1.max; } }); if (columns.length > n) columns = Arrays.copyOf(columns, n); long[] dp = new long[1 << n]; for (E2RotateColumnsHardVersion.Column c : columns) { long[] ndp = new long[1 << n]; System.arraycopy(dp, 0, ndp, 0, dp.length); for (int rot = 0; rot < n; ++rot) { long[] temp = new long[1 << n]; System.arraycopy(dp, 0, temp, 0, dp.length); for (int i = 0, pos = rot; i < n; ++i, ++pos) { if (pos >= n) pos = 0; int val = c.v[pos]; for (int j = 0; j < temp.length; ++j) { if ((j & (1 << i)) == 0) temp[j | (1 << i)] = Math.max(temp[j | (1 << i)], temp[j] + val); } } for (int i = 0; i < ndp.length; ++i) ndp[i] = Math.max(ndp[i], temp[i]); } dp = ndp; } out.println(dp[dp.length - 1]); } static class Column { int[] v; int max; public Column(int[] v) { this.v = v; } void initMax() { max = 0; for (int vv : v) max = Math.max(max, vv); } } } static class FastReader { private InputStream stream; private byte[] buf = new byte[8192]; private int curChar; private int pnumChars; public FastReader(InputStream stream) { this.stream = stream; } private int pread() { if (pnumChars == -1) { throw new InputMismatchException(); } if (curChar >= pnumChars) { curChar = 0; try { pnumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (pnumChars <= 0) { return -1; } } return buf[curChar++]; } public String next() { return nextString(); } public int nextInt() { int c = pread(); while (isSpaceChar(c)) c = pread(); int sgn = 1; if (c == '-') { sgn = -1; c = pread(); } int res = 0; do { if (c == ',') { c = pread(); } if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = pread(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = pread(); while (isSpaceChar(c)) c = pread(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = pread(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; import java.io.IOException; import java.util.Comparator; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; FastReader in = new FastReader(inputStream); PrintWriter out = new PrintWriter(outputStream); E2RotateColumnsHardVersion solver = new E2RotateColumnsHardVersion(); int testCount = Integer.parseInt(in.next()); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class E2RotateColumnsHardVersion { public void solve(int testNumber, FastReader in, PrintWriter out) { int n = in.nextInt(); int m = in.nextInt(); E2RotateColumnsHardVersion.Column[] columns = new E2RotateColumnsHardVersion.Column[m]; for (int i = 0; i < columns.length; ++i) columns[i] = new E2RotateColumnsHardVersion.Column(new int[n]); for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { columns[j].v[i] = in.nextInt(); if (i == n - 1) columns[j].initMax(); } } Arrays.sort(columns, new Comparator<E2RotateColumnsHardVersion.Column>() { public int compare(E2RotateColumnsHardVersion.Column o1, E2RotateColumnsHardVersion.Column o2) { return o2.max - o1.max; } }); if (columns.length > n) columns = Arrays.copyOf(columns, n); long[] dp = new long[1 << n]; for (E2RotateColumnsHardVersion.Column c : columns) { long[] ndp = new long[1 << n]; System.arraycopy(dp, 0, ndp, 0, dp.length); for (int rot = 0; rot < n; ++rot) { long[] temp = new long[1 << n]; System.arraycopy(dp, 0, temp, 0, dp.length); for (int i = 0, pos = rot; i < n; ++i, ++pos) { if (pos >= n) pos = 0; int val = c.v[pos]; for (int j = 0; j < temp.length; ++j) { if ((j & (1 << i)) == 0) temp[j | (1 << i)] = Math.max(temp[j | (1 << i)], temp[j] + val); } } for (int i = 0; i < ndp.length; ++i) ndp[i] = Math.max(ndp[i], temp[i]); } dp = ndp; } out.println(dp[dp.length - 1]); } static class Column { int[] v; int max; public Column(int[] v) { this.v = v; } void initMax() { max = 0; for (int vv : v) max = Math.max(max, vv); } } } static class FastReader { private InputStream stream; private byte[] buf = new byte[8192]; private int curChar; private int pnumChars; public FastReader(InputStream stream) { this.stream = stream; } private int pread() { if (pnumChars == -1) { throw new InputMismatchException(); } if (curChar >= pnumChars) { curChar = 0; try { pnumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (pnumChars <= 0) { return -1; } } return buf[curChar++]; } public String next() { return nextString(); } public int nextInt() { int c = pread(); while (isSpaceChar(c)) c = pread(); int sgn = 1; if (c == '-') { sgn = -1; c = pread(); } int res = 0; do { if (c == ',') { c = pread(); } if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = pread(); } while (!isSpaceChar(c)); return res * sgn; } public String nextString() { int c = pread(); while (isSpaceChar(c)) c = pread(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = pread(); } while (!isSpaceChar(c)); return res.toString(); } private boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } } } </CODE> <EVALUATION_RUBRIC> - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,486
4,460
1,635
/** * Write a description of class VK2A here. * * @author (your name) * @version (a version number or a date) */ import java.util.*; public class VK2A { public static void main(String args[]) { Scanner S = new Scanner(System.in); int n = S.nextInt(); int a = S.nextInt(); int b = S.nextInt(); int[] A = new int[n]; for(int i = 0; i < n; i++) A[i] = S.nextInt(); for(int i = 0; i < n; i++) for(int j = 0; j < n - i - 1; j++) { if(A[j] < A[j + 1]) { int temp = A[j]; A[j] = A[j + 1]; A[j + 1] = temp; } } System.out.println(A[a - 1] - A[a]); } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /** * Write a description of class VK2A here. * * @author (your name) * @version (a version number or a date) */ import java.util.*; public class VK2A { public static void main(String args[]) { Scanner S = new Scanner(System.in); int n = S.nextInt(); int a = S.nextInt(); int b = S.nextInt(); int[] A = new int[n]; for(int i = 0; i < n; i++) A[i] = S.nextInt(); for(int i = 0; i < n; i++) for(int j = 0; j < n - i - 1; j++) { if(A[j] < A[j + 1]) { int temp = A[j]; A[j] = A[j + 1]; A[j + 1] = temp; } } System.out.println(A[a - 1] - A[a]); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> /** * Write a description of class VK2A here. * * @author (your name) * @version (a version number or a date) */ import java.util.*; public class VK2A { public static void main(String args[]) { Scanner S = new Scanner(System.in); int n = S.nextInt(); int a = S.nextInt(); int b = S.nextInt(); int[] A = new int[n]; for(int i = 0; i < n; i++) A[i] = S.nextInt(); for(int i = 0; i < n; i++) for(int j = 0; j < n - i - 1; j++) { if(A[j] < A[j + 1]) { int temp = A[j]; A[j] = A[j + 1]; A[j + 1] = temp; } } System.out.println(A[a - 1] - A[a]); } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
567
1,632
4,194
import java.io.*; import java.util.*; public class Main { static Scanner in; static PrintWriter out; // static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;} public static void main(String[] args) throws Exception { in = new Scanner(System.in); out = new PrintWriter(System.out); // in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); int n = in.nextInt(); double[][] p = new double[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) p[i][j] = in.nextDouble(); double[] q = new double[1 << n]; q[(1 << n) - 1] = 1; for (int mask = (1 << n) - 1; mask > 0; mask--) { int count = 0; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) count++; if (count <= 1) continue; count = count*(count - 1)/2; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) for (int s = 0; s < t; s++) if (((1 << s) & mask) != 0) { q[mask - (1 << t)] += q[mask] / count * p[s][t]; q[mask - (1 << s)] += q[mask] / count * p[t][s]; } } for (int i = 0; i < n; i++) out.print(q[1 << i] + " "); out.close(); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { static Scanner in; static PrintWriter out; // static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;} public static void main(String[] args) throws Exception { in = new Scanner(System.in); out = new PrintWriter(System.out); // in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); int n = in.nextInt(); double[][] p = new double[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) p[i][j] = in.nextDouble(); double[] q = new double[1 << n]; q[(1 << n) - 1] = 1; for (int mask = (1 << n) - 1; mask > 0; mask--) { int count = 0; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) count++; if (count <= 1) continue; count = count*(count - 1)/2; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) for (int s = 0; s < t; s++) if (((1 << s) & mask) != 0) { q[mask - (1 << t)] += q[mask] / count * p[s][t]; q[mask - (1 << s)] += q[mask] / count * p[t][s]; } } for (int i = 0; i < n; i++) out.print(q[1 << i] + " "); out.close(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { static Scanner in; static PrintWriter out; // static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;} public static void main(String[] args) throws Exception { in = new Scanner(System.in); out = new PrintWriter(System.out); // in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); int n = in.nextInt(); double[][] p = new double[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) p[i][j] = in.nextDouble(); double[] q = new double[1 << n]; q[(1 << n) - 1] = 1; for (int mask = (1 << n) - 1; mask > 0; mask--) { int count = 0; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) count++; if (count <= 1) continue; count = count*(count - 1)/2; for (int t = 0; t < n; t++) if (((1 << t) & mask) != 0) for (int s = 0; s < t; s++) if (((1 << s) & mask) != 0) { q[mask - (1 << t)] += q[mask] / count * p[s][t]; q[mask - (1 << s)] += q[mask] / count * p[t][s]; } } for (int i = 0; i < n; i++) out.print(q[1 << i] + " "); out.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
739
4,183
978
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class Cf2 { static boolean ok(long n, long k, long eatten) { long moves = n-eatten; long ans = moves*(moves+1)/2; ans -= eatten; return ans <= k; } public static void main(String[] args) { FastReader in = new FastReader(); long n = in.nextInt(); long k = in.nextInt(); long left = 0, right = n; while (left <= right) { long middle = (left+right)/2; if (ok(n, k, middle)) right = middle-1; else left=middle+1; } System.out.println(left); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class Cf2 { static boolean ok(long n, long k, long eatten) { long moves = n-eatten; long ans = moves*(moves+1)/2; ans -= eatten; return ans <= k; } public static void main(String[] args) { FastReader in = new FastReader(); long n = in.nextInt(); long k = in.nextInt(); long left = 0, right = n; while (left <= right) { long middle = (left+right)/2; if (ok(n, k, middle)) right = middle-1; else left=middle+1; } System.out.println(left); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class Cf2 { static boolean ok(long n, long k, long eatten) { long moves = n-eatten; long ans = moves*(moves+1)/2; ans -= eatten; return ans <= k; } public static void main(String[] args) { FastReader in = new FastReader(); long n = in.nextInt(); long k = in.nextInt(); long left = 0, right = n; while (left <= right) { long middle = (left+right)/2; if (ok(n, k, middle)) right = middle-1; else left=middle+1; } System.out.println(left); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
691
977
4,387
import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.StringTokenizer; public class B { static int first(int n) { int res = 0; while(n > 0 && (n & 1) == 0){ n >>= 1; res++; } return res; } public static void main(String[] args) throws Exception { Scanner bf = new Scanner(System.in); PrintWriter out = new PrintWriter(System.out); int n = bf.nextInt(), m = bf.nextInt(); ArrayList<Integer> [] adjList = new ArrayList[n]; for (int i = 0; i < adjList.length; i++) { adjList[i] = new ArrayList<Integer>(); } for (int i = 0; i < m; i++) { int u = bf.nextInt()-1, v = bf.nextInt()-1; adjList[u].add(v); adjList[v].add(u); } long [][] memo = new long[(1<<n)][n]; for (int i = 0; i < n; i++) { memo[1<<i][i] = 1; } long ans = 0; for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) == 1) continue; for (int j = 0; j < n; j++) { if((i & (1<<j)) == 0 || j == first(i)) continue; for(int v:adjList[j]) memo[i][j] += memo[i^(1<<j)][v]; } } for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) < 3) continue; int t = first(i); for (int j = 0; j < n; j++) { if(adjList[j].contains(t)) ans += memo[i][j]; } } out.println(ans/2); out.flush(); out.close(); } static class Scanner { StringTokenizer st; BufferedReader br; public Scanner(InputStream s) { br = new BufferedReader(new InputStreamReader(s)); } public Scanner(FileReader fileReader) { br = new BufferedReader(fileReader); } public String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public String nextLine() throws IOException { return br.readLine(); } public boolean ready() throws IOException { return br.ready(); } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.StringTokenizer; public class B { static int first(int n) { int res = 0; while(n > 0 && (n & 1) == 0){ n >>= 1; res++; } return res; } public static void main(String[] args) throws Exception { Scanner bf = new Scanner(System.in); PrintWriter out = new PrintWriter(System.out); int n = bf.nextInt(), m = bf.nextInt(); ArrayList<Integer> [] adjList = new ArrayList[n]; for (int i = 0; i < adjList.length; i++) { adjList[i] = new ArrayList<Integer>(); } for (int i = 0; i < m; i++) { int u = bf.nextInt()-1, v = bf.nextInt()-1; adjList[u].add(v); adjList[v].add(u); } long [][] memo = new long[(1<<n)][n]; for (int i = 0; i < n; i++) { memo[1<<i][i] = 1; } long ans = 0; for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) == 1) continue; for (int j = 0; j < n; j++) { if((i & (1<<j)) == 0 || j == first(i)) continue; for(int v:adjList[j]) memo[i][j] += memo[i^(1<<j)][v]; } } for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) < 3) continue; int t = first(i); for (int j = 0; j < n; j++) { if(adjList[j].contains(t)) ans += memo[i][j]; } } out.println(ans/2); out.flush(); out.close(); } static class Scanner { StringTokenizer st; BufferedReader br; public Scanner(InputStream s) { br = new BufferedReader(new InputStreamReader(s)); } public Scanner(FileReader fileReader) { br = new BufferedReader(fileReader); } public String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public String nextLine() throws IOException { return br.readLine(); } public boolean ready() throws IOException { return br.ready(); } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.StringTokenizer; public class B { static int first(int n) { int res = 0; while(n > 0 && (n & 1) == 0){ n >>= 1; res++; } return res; } public static void main(String[] args) throws Exception { Scanner bf = new Scanner(System.in); PrintWriter out = new PrintWriter(System.out); int n = bf.nextInt(), m = bf.nextInt(); ArrayList<Integer> [] adjList = new ArrayList[n]; for (int i = 0; i < adjList.length; i++) { adjList[i] = new ArrayList<Integer>(); } for (int i = 0; i < m; i++) { int u = bf.nextInt()-1, v = bf.nextInt()-1; adjList[u].add(v); adjList[v].add(u); } long [][] memo = new long[(1<<n)][n]; for (int i = 0; i < n; i++) { memo[1<<i][i] = 1; } long ans = 0; for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) == 1) continue; for (int j = 0; j < n; j++) { if((i & (1<<j)) == 0 || j == first(i)) continue; for(int v:adjList[j]) memo[i][j] += memo[i^(1<<j)][v]; } } for (int i = 1; i < 1<<n; i++) { if(Integer.bitCount(i) < 3) continue; int t = first(i); for (int j = 0; j < n; j++) { if(adjList[j].contains(t)) ans += memo[i][j]; } } out.println(ans/2); out.flush(); out.close(); } static class Scanner { StringTokenizer st; BufferedReader br; public Scanner(InputStream s) { br = new BufferedReader(new InputStreamReader(s)); } public Scanner(FileReader fileReader) { br = new BufferedReader(fileReader); } public String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public String nextLine() throws IOException { return br.readLine(); } public boolean ready() throws IOException { return br.ready(); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
978
4,376
3,844
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, InputReader in, OutputWriter out) { int tests = in.nextInt(); for (int t = 0; t < tests; t++) { int numLines = in.nextInt(); // greedy, maintain stack, keep stack as big as possible int stackIdx = -1; int[] stack = new int[10000]; String prev = ""; for (int x = 0; x < numLines; x++) { int depth = 0; int next = in.nextInt(); boolean found = false; for (int i = stackIdx; i >= 0; i--) { if (next == stack[i] + 1) { depth = i; found = true; break; } } if (found == true) { stackIdx = depth; stack[depth] = next; for (int i = 0; i <= depth; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else if (next == 1) { stackIdx++; stack[stackIdx] = 1; for (int i = 0; i <= stackIdx; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else { // start a new one stackIdx = 0; stack[0] = next; out.println(next); } } } } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void print(Object... objects) { for (int i = 0; i < objects.length; i++) { if (i != 0) { writer.print(' '); } writer.print(objects[i]); } } public void println() { writer.println(); } public void close() { writer.close(); } public void print(int i) { writer.print(i); } public void println(int i) { writer.println(i); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, InputReader in, OutputWriter out) { int tests = in.nextInt(); for (int t = 0; t < tests; t++) { int numLines = in.nextInt(); // greedy, maintain stack, keep stack as big as possible int stackIdx = -1; int[] stack = new int[10000]; String prev = ""; for (int x = 0; x < numLines; x++) { int depth = 0; int next = in.nextInt(); boolean found = false; for (int i = stackIdx; i >= 0; i--) { if (next == stack[i] + 1) { depth = i; found = true; break; } } if (found == true) { stackIdx = depth; stack[depth] = next; for (int i = 0; i <= depth; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else if (next == 1) { stackIdx++; stack[stackIdx] = 1; for (int i = 0; i <= stackIdx; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else { // start a new one stackIdx = 0; stack[0] = next; out.println(next); } } } } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void print(Object... objects) { for (int i = 0; i < objects.length; i++) { if (i != 0) { writer.print(' '); } writer.print(objects[i]); } } public void println() { writer.println(); } public void close() { writer.close(); } public void print(int i) { writer.print(i); } public void println(int i) { writer.println(i); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.OutputStream; import java.io.PrintWriter; import java.io.BufferedWriter; import java.io.Writer; import java.io.OutputStreamWriter; import java.util.InputMismatchException; import java.io.IOException; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); OutputWriter out = new OutputWriter(outputStream); TaskC solver = new TaskC(); solver.solve(1, in, out); out.close(); } static class TaskC { public void solve(int testNumber, InputReader in, OutputWriter out) { int tests = in.nextInt(); for (int t = 0; t < tests; t++) { int numLines = in.nextInt(); // greedy, maintain stack, keep stack as big as possible int stackIdx = -1; int[] stack = new int[10000]; String prev = ""; for (int x = 0; x < numLines; x++) { int depth = 0; int next = in.nextInt(); boolean found = false; for (int i = stackIdx; i >= 0; i--) { if (next == stack[i] + 1) { depth = i; found = true; break; } } if (found == true) { stackIdx = depth; stack[depth] = next; for (int i = 0; i <= depth; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else if (next == 1) { stackIdx++; stack[stackIdx] = 1; for (int i = 0; i <= stackIdx; i++) { if (i != 0) { out.print("."); } out.print(stack[i]); } out.println(); } else { // start a new one stackIdx = 0; stack[0] = next; out.println(next); } } } } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void print(Object... objects) { for (int i = 0; i < objects.length; i++) { if (i != 0) { writer.print(' '); } writer.print(objects[i]); } } public void println() { writer.println(); } public void close() { writer.close(); } public void print(int i) { writer.print(i); } public void println(int i) { writer.println(i); } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private InputReader.SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) { throw new InputMismatchException(); } if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) { return -1; } } return buf[curChar++]; } public int nextInt() { int c = read(); while (isSpaceChar(c)) { c = read(); } int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') { throw new InputMismatchException(); } res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public boolean isSpaceChar(int c) { if (filter != null) { return filter.isSpaceChar(c); } return isWhitespace(c); } public static boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(1): The time complexity is constant to the input size n. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,388
3,834
1,363
import java.io.*; import java.util.*; import java.math.*; public class Main { InputStream is; PrintWriter out; String INPUT = "" ; boolean local = false; int inf=0x7FFFFFFF; int MOD=(int)(1e9+7); double eps=1e-5; double PI=Math.acos(-1); void solve() { long maxn=nl(); long L=1,R=maxn,ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,mid,maxn)<1)L=mid+1; else{ R=mid-1; ans=mid; } } if(ask(1,1,ans,maxn)==1 && ask(ans+1,1,maxn,maxn)==1){ Yts1999 a1=gao(1,1,ans,maxn); Yts1999 a2=gao(ans+1,1,maxn,maxn); answer(a1,a2); }else{ L=1;R=maxn;ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,maxn,mid)<1)L=mid+1; else{ R=mid-1; ans=mid; } } Yts1999 a1=gao(1,1,maxn,ans); Yts1999 a2=gao(1,ans+1,maxn,maxn); answer(a1,a2); } } void answer(Yts1999 a1,Yts1999 a2){ out.print("!"); a1.print(); a2.print(); out.flush(); } int ask(long a,long b,long c,long d){ out.printf("? %d %d %d %d",a,b,c,d); out.println(); out.flush(); return ni(); } Yts1999 gao(long x1,long x2,long y1,long y2){ long a1=0,a2=0,a3=0,a4=0; long L,R; L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(mid,x2,y1,y2)==1){ a1=mid; L=mid+1; }else R=mid-1; } L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,mid,y2)==1){ a3=mid; R=mid-1; }else L=mid+1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,mid,y1,y2)==1){ a2=mid; L=mid+1; }else R=mid-1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,y1,mid)==1){ a4=mid; R=mid-1; }else L=mid+1; } return new Yts1999(a1,a2,a3,a4); } public class Yts1999 implements Comparable{ public long a,b,c,d; public Yts1999(long a,long b,long c,long d){ this.a=a; this.b=b; this.c=c; this.d=d; } public int compareTo(Object o) { Yts1999 to=(Yts1999)o; if(this.d<to.d) return 1; else if(this.d==to.d) return 0; else return -1; } public void print(){ out.printf(" %d %d %d %d",a,b,c,d); } } long[] exgcd(long a, long b) { if (b == 0) return new long[]{1,0,a}; long[] res = exgcd(b, a%b); long t = res[0]; res[0] = res[1]; res[1] = t; res[1] -= a/b*res[0]; return res; } long gcd(long a,long b){ return b==0?a:gcd(b,a%b); } long lcm(long a,long b){ return a*b/gcd(a,b); } private void run() { is = local? new ByteArrayInputStream(INPUT.getBytes()):System.in; out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte(){ if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns(){ int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n){ char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m){ char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n){ int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni(){ int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl(){ long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private void tr(Object... o) { if(local)System.out.println(Arrays.deepToString(o)); } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class Main { InputStream is; PrintWriter out; String INPUT = "" ; boolean local = false; int inf=0x7FFFFFFF; int MOD=(int)(1e9+7); double eps=1e-5; double PI=Math.acos(-1); void solve() { long maxn=nl(); long L=1,R=maxn,ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,mid,maxn)<1)L=mid+1; else{ R=mid-1; ans=mid; } } if(ask(1,1,ans,maxn)==1 && ask(ans+1,1,maxn,maxn)==1){ Yts1999 a1=gao(1,1,ans,maxn); Yts1999 a2=gao(ans+1,1,maxn,maxn); answer(a1,a2); }else{ L=1;R=maxn;ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,maxn,mid)<1)L=mid+1; else{ R=mid-1; ans=mid; } } Yts1999 a1=gao(1,1,maxn,ans); Yts1999 a2=gao(1,ans+1,maxn,maxn); answer(a1,a2); } } void answer(Yts1999 a1,Yts1999 a2){ out.print("!"); a1.print(); a2.print(); out.flush(); } int ask(long a,long b,long c,long d){ out.printf("? %d %d %d %d",a,b,c,d); out.println(); out.flush(); return ni(); } Yts1999 gao(long x1,long x2,long y1,long y2){ long a1=0,a2=0,a3=0,a4=0; long L,R; L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(mid,x2,y1,y2)==1){ a1=mid; L=mid+1; }else R=mid-1; } L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,mid,y2)==1){ a3=mid; R=mid-1; }else L=mid+1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,mid,y1,y2)==1){ a2=mid; L=mid+1; }else R=mid-1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,y1,mid)==1){ a4=mid; R=mid-1; }else L=mid+1; } return new Yts1999(a1,a2,a3,a4); } public class Yts1999 implements Comparable{ public long a,b,c,d; public Yts1999(long a,long b,long c,long d){ this.a=a; this.b=b; this.c=c; this.d=d; } public int compareTo(Object o) { Yts1999 to=(Yts1999)o; if(this.d<to.d) return 1; else if(this.d==to.d) return 0; else return -1; } public void print(){ out.printf(" %d %d %d %d",a,b,c,d); } } long[] exgcd(long a, long b) { if (b == 0) return new long[]{1,0,a}; long[] res = exgcd(b, a%b); long t = res[0]; res[0] = res[1]; res[1] = t; res[1] -= a/b*res[0]; return res; } long gcd(long a,long b){ return b==0?a:gcd(b,a%b); } long lcm(long a,long b){ return a*b/gcd(a,b); } private void run() { is = local? new ByteArrayInputStream(INPUT.getBytes()):System.in; out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte(){ if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns(){ int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n){ char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m){ char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n){ int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni(){ int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl(){ long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private void tr(Object... o) { if(local)System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class Main { InputStream is; PrintWriter out; String INPUT = "" ; boolean local = false; int inf=0x7FFFFFFF; int MOD=(int)(1e9+7); double eps=1e-5; double PI=Math.acos(-1); void solve() { long maxn=nl(); long L=1,R=maxn,ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,mid,maxn)<1)L=mid+1; else{ R=mid-1; ans=mid; } } if(ask(1,1,ans,maxn)==1 && ask(ans+1,1,maxn,maxn)==1){ Yts1999 a1=gao(1,1,ans,maxn); Yts1999 a2=gao(ans+1,1,maxn,maxn); answer(a1,a2); }else{ L=1;R=maxn;ans=-1; while (L<=R){ long mid=(L+R)/2; if(ask(1,1,maxn,mid)<1)L=mid+1; else{ R=mid-1; ans=mid; } } Yts1999 a1=gao(1,1,maxn,ans); Yts1999 a2=gao(1,ans+1,maxn,maxn); answer(a1,a2); } } void answer(Yts1999 a1,Yts1999 a2){ out.print("!"); a1.print(); a2.print(); out.flush(); } int ask(long a,long b,long c,long d){ out.printf("? %d %d %d %d",a,b,c,d); out.println(); out.flush(); return ni(); } Yts1999 gao(long x1,long x2,long y1,long y2){ long a1=0,a2=0,a3=0,a4=0; long L,R; L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(mid,x2,y1,y2)==1){ a1=mid; L=mid+1; }else R=mid-1; } L=x1;R=y1; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,mid,y2)==1){ a3=mid; R=mid-1; }else L=mid+1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,mid,y1,y2)==1){ a2=mid; L=mid+1; }else R=mid-1; } L=x2;R=y2; while(L<=R){ long mid=(L+R)/2; if(ask(x1,x2,y1,mid)==1){ a4=mid; R=mid-1; }else L=mid+1; } return new Yts1999(a1,a2,a3,a4); } public class Yts1999 implements Comparable{ public long a,b,c,d; public Yts1999(long a,long b,long c,long d){ this.a=a; this.b=b; this.c=c; this.d=d; } public int compareTo(Object o) { Yts1999 to=(Yts1999)o; if(this.d<to.d) return 1; else if(this.d==to.d) return 0; else return -1; } public void print(){ out.printf(" %d %d %d %d",a,b,c,d); } } long[] exgcd(long a, long b) { if (b == 0) return new long[]{1,0,a}; long[] res = exgcd(b, a%b); long t = res[0]; res[0] = res[1]; res[1] = t; res[1] -= a/b*res[0]; return res; } long gcd(long a,long b){ return b==0?a:gcd(b,a%b); } long lcm(long a,long b){ return a*b/gcd(a,b); } private void run() { is = local? new ByteArrayInputStream(INPUT.getBytes()):System.in; out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte(){ if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns(){ int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n){ char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m){ char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n){ int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni(){ int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl(){ long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private void tr(Object... o) { if(local)System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
2,127
1,361
581
import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner s = new Scanner(System.in); int n = s.nextInt(), d = s.nextInt(); int[] arr = new int[n]; for(int i = 0; i < n; i++){ arr[i] = s.nextInt(); } Arrays.sort(arr); int count = 0; for(int i = 1; i < n; i++){ int dist = arr[i] - arr[i - 1]; if(dist > 2 * d){ count += 2; }else if(dist == 2 * d){ count++; } } System.out.println(count + 2); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner s = new Scanner(System.in); int n = s.nextInt(), d = s.nextInt(); int[] arr = new int[n]; for(int i = 0; i < n; i++){ arr[i] = s.nextInt(); } Arrays.sort(arr); int count = 0; for(int i = 1; i < n; i++){ int dist = arr[i] - arr[i - 1]; if(dist > 2 * d){ count += 2; }else if(dist == 2 * d){ count++; } } System.out.println(count + 2); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner s = new Scanner(System.in); int n = s.nextInt(), d = s.nextInt(); int[] arr = new int[n]; for(int i = 0; i < n; i++){ arr[i] = s.nextInt(); } Arrays.sort(arr); int count = 0; for(int i = 1; i < n; i++){ int dist = arr[i] - arr[i - 1]; if(dist > 2 * d){ count += 2; }else if(dist == 2 * d){ count++; } } System.out.println(count + 2); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
499
580
1,694
import java.util.Scanner; public class Flatwile { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int t = sc.nextInt(); int[] c = new int[n]; int[] a = new int[n]; for (int i=0; i<n; i++){ c[i] = sc.nextInt(); a[i] = sc.nextInt(); } sort(c, a); int res = 1; double prev = Integer.MIN_VALUE; for(int i=0; i<c.length; i++){ if (c[i]-a[i]/2d - prev >=t){ res++; } if (i!=c.length-1 && c[i+1]-a[i+1]/2d-(c[i]+a[i]/2d)>t ){ res++; } prev = c[i] +a[i]/2d; } System.out.println(res); } private static void sort(int[] a, int[] b){ for(int i=0; i<a.length; i++){ for(int j=i+1; j<a.length; j++){ if (a[i]>a[j]){ swap(a, i, j); swap(b, i, j); } } } } private static void swap(int[] a, int i, int j) { int t = a[i]; a[i] = a[j]; a[j] = t; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class Flatwile { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int t = sc.nextInt(); int[] c = new int[n]; int[] a = new int[n]; for (int i=0; i<n; i++){ c[i] = sc.nextInt(); a[i] = sc.nextInt(); } sort(c, a); int res = 1; double prev = Integer.MIN_VALUE; for(int i=0; i<c.length; i++){ if (c[i]-a[i]/2d - prev >=t){ res++; } if (i!=c.length-1 && c[i+1]-a[i+1]/2d-(c[i]+a[i]/2d)>t ){ res++; } prev = c[i] +a[i]/2d; } System.out.println(res); } private static void sort(int[] a, int[] b){ for(int i=0; i<a.length; i++){ for(int j=i+1; j<a.length; j++){ if (a[i]>a[j]){ swap(a, i, j); swap(b, i, j); } } } } private static void swap(int[] a, int i, int j) { int t = a[i]; a[i] = a[j]; a[j] = t; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Scanner; public class Flatwile { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int t = sc.nextInt(); int[] c = new int[n]; int[] a = new int[n]; for (int i=0; i<n; i++){ c[i] = sc.nextInt(); a[i] = sc.nextInt(); } sort(c, a); int res = 1; double prev = Integer.MIN_VALUE; for(int i=0; i<c.length; i++){ if (c[i]-a[i]/2d - prev >=t){ res++; } if (i!=c.length-1 && c[i+1]-a[i+1]/2d-(c[i]+a[i]/2d)>t ){ res++; } prev = c[i] +a[i]/2d; } System.out.println(res); } private static void sort(int[] a, int[] b){ for(int i=0; i<a.length; i++){ for(int j=i+1; j<a.length; j++){ if (a[i]>a[j]){ swap(a, i, j); swap(b, i, j); } } } } private static void swap(int[] a, int i, int j) { int t = a[i]; a[i] = a[j]; a[j] = t; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
653
1,691
1,238
import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class A{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int mod=(int)1e9+9; long n, m, k; void run(){ n=sc.nextLong(); m=sc.nextLong(); k=sc.nextLong(); solve(); } void solve(){ long ans=0; long s=n-m; long remain=max(n-s*k, 0); // debug("remain", remain); ans=m-remain; // debug("ans", ans); long r=remain%k; ans=(ans+r)%mod; remain-=r; // debug("remain2", remain); long a=remain/k; long add=(powMod(2, a, mod)-1)*k%mod*2%mod; // debug("add", add); ans=(ans+add)%mod; // debug(ans); println(ans+""); } long powMod(long x, long k, long mod){ if(k==0){ return 1%mod; }else if(k%2==0){ return powMod(x*x%mod, k/2, mod); }else{ return x*powMod(x, k-1, mod)%mod; } } void println(String s){ System.out.println(s); } void print(String s){ System.out.print(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new A().run(); } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class A{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int mod=(int)1e9+9; long n, m, k; void run(){ n=sc.nextLong(); m=sc.nextLong(); k=sc.nextLong(); solve(); } void solve(){ long ans=0; long s=n-m; long remain=max(n-s*k, 0); // debug("remain", remain); ans=m-remain; // debug("ans", ans); long r=remain%k; ans=(ans+r)%mod; remain-=r; // debug("remain2", remain); long a=remain/k; long add=(powMod(2, a, mod)-1)*k%mod*2%mod; // debug("add", add); ans=(ans+add)%mod; // debug(ans); println(ans+""); } long powMod(long x, long k, long mod){ if(k==0){ return 1%mod; }else if(k%2==0){ return powMod(x*x%mod, k/2, mod); }else{ return x*powMod(x, k-1, mod)%mod; } } void println(String s){ System.out.println(s); } void print(String s){ System.out.print(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new A().run(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.math.*; import java.io.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class A{ Scanner sc=new Scanner(System.in); int INF=1<<28; double EPS=1e-9; int mod=(int)1e9+9; long n, m, k; void run(){ n=sc.nextLong(); m=sc.nextLong(); k=sc.nextLong(); solve(); } void solve(){ long ans=0; long s=n-m; long remain=max(n-s*k, 0); // debug("remain", remain); ans=m-remain; // debug("ans", ans); long r=remain%k; ans=(ans+r)%mod; remain-=r; // debug("remain2", remain); long a=remain/k; long add=(powMod(2, a, mod)-1)*k%mod*2%mod; // debug("add", add); ans=(ans+add)%mod; // debug(ans); println(ans+""); } long powMod(long x, long k, long mod){ if(k==0){ return 1%mod; }else if(k%2==0){ return powMod(x*x%mod, k/2, mod); }else{ return x*powMod(x, k-1, mod)%mod; } } void println(String s){ System.out.println(s); } void print(String s){ System.out.print(s); } void debug(Object... os){ System.err.println(Arrays.deepToString(os)); } public static void main(String[] args){ Locale.setDefault(Locale.US); new A().run(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
717
1,237
2,727
import java.util.Scanner; public class LCM { public static void main(String[] args) { Scanner scan = new Scanner(System.in); long n = scan.nextLong(); if (n <= 2) System.out.println(n); else if (n % 2 == 1) System.out.println(n * (n - 1) * (n - 2)); else if (n % 3 == 0) System.out.println((n - 1) * (n - 2) * (n - 3)); else System.out.println(n * (n - 1) * (n - 3)); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Scanner; public class LCM { public static void main(String[] args) { Scanner scan = new Scanner(System.in); long n = scan.nextLong(); if (n <= 2) System.out.println(n); else if (n % 2 == 1) System.out.println(n * (n - 1) * (n - 2)); else if (n % 3 == 0) System.out.println((n - 1) * (n - 2) * (n - 3)); else System.out.println(n * (n - 1) * (n - 3)); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class LCM { public static void main(String[] args) { Scanner scan = new Scanner(System.in); long n = scan.nextLong(); if (n <= 2) System.out.println(n); else if (n % 2 == 1) System.out.println(n * (n - 1) * (n - 2)); else if (n % 3 == 0) System.out.println((n - 1) * (n - 2) * (n - 3)); else System.out.println(n * (n - 1) * (n - 3)); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
494
2,721
527
import java.io.*; import java.util.Arrays; import java.util.StringJoiner; import java.util.StringTokenizer; import java.util.function.Function; public class Main { static String S; public static void main(String[] args) { FastScanner sc = new FastScanner(System.in); S = sc.next(); System.out.println(solve()); } static int solve() { int ans = -1; int time = 1; int n = S.length(); for (int i = 1; i < n*2; i++) { if( S.charAt((i-1)%n) != S.charAt(i%n) ) { time++; } else { ans = Math.max(time, ans); time = 1; } } ans = Math.max(time, ans); if( ans == n*2 ) { return n; } else { return ans; } } @SuppressWarnings("unused") static class FastScanner { private BufferedReader reader; private StringTokenizer tokenizer; FastScanner(InputStream in) { reader = new BufferedReader(new InputStreamReader(in)); tokenizer = null; } String next() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } String nextLine() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { return reader.readLine(); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken("\n"); } long nextLong() { return Long.parseLong(next()); } int nextInt() { return Integer.parseInt(next()); } int[] nextIntArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } int[] nextIntArray(int n, int delta) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt() + delta; return a; } long[] nextLongArray(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nextLong(); return a; } } static <A> void writeLines(A[] as, Function<A, String> f) { PrintWriter pw = new PrintWriter(System.out); for (A a : as) { pw.println(f.apply(a)); } pw.flush(); } static void writeLines(int[] as) { PrintWriter pw = new PrintWriter(System.out); for (int a : as) pw.println(a); pw.flush(); } static void writeLines(long[] as) { PrintWriter pw = new PrintWriter(System.out); for (long a : as) pw.println(a); pw.flush(); } static int max(int... as) { int max = Integer.MIN_VALUE; for (int a : as) max = Math.max(a, max); return max; } static int min(int... as) { int min = Integer.MAX_VALUE; for (int a : as) min = Math.min(a, min); return min; } static void debug(Object... args) { StringJoiner j = new StringJoiner(" "); for (Object arg : args) { if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg)); else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg)); else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg)); else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg)); else j.add(arg.toString()); } System.err.println(j.toString()); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.Arrays; import java.util.StringJoiner; import java.util.StringTokenizer; import java.util.function.Function; public class Main { static String S; public static void main(String[] args) { FastScanner sc = new FastScanner(System.in); S = sc.next(); System.out.println(solve()); } static int solve() { int ans = -1; int time = 1; int n = S.length(); for (int i = 1; i < n*2; i++) { if( S.charAt((i-1)%n) != S.charAt(i%n) ) { time++; } else { ans = Math.max(time, ans); time = 1; } } ans = Math.max(time, ans); if( ans == n*2 ) { return n; } else { return ans; } } @SuppressWarnings("unused") static class FastScanner { private BufferedReader reader; private StringTokenizer tokenizer; FastScanner(InputStream in) { reader = new BufferedReader(new InputStreamReader(in)); tokenizer = null; } String next() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } String nextLine() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { return reader.readLine(); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken("\n"); } long nextLong() { return Long.parseLong(next()); } int nextInt() { return Integer.parseInt(next()); } int[] nextIntArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } int[] nextIntArray(int n, int delta) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt() + delta; return a; } long[] nextLongArray(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nextLong(); return a; } } static <A> void writeLines(A[] as, Function<A, String> f) { PrintWriter pw = new PrintWriter(System.out); for (A a : as) { pw.println(f.apply(a)); } pw.flush(); } static void writeLines(int[] as) { PrintWriter pw = new PrintWriter(System.out); for (int a : as) pw.println(a); pw.flush(); } static void writeLines(long[] as) { PrintWriter pw = new PrintWriter(System.out); for (long a : as) pw.println(a); pw.flush(); } static int max(int... as) { int max = Integer.MIN_VALUE; for (int a : as) max = Math.max(a, max); return max; } static int min(int... as) { int min = Integer.MAX_VALUE; for (int a : as) min = Math.min(a, min); return min; } static void debug(Object... args) { StringJoiner j = new StringJoiner(" "); for (Object arg : args) { if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg)); else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg)); else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg)); else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg)); else j.add(arg.toString()); } System.err.println(j.toString()); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.Arrays; import java.util.StringJoiner; import java.util.StringTokenizer; import java.util.function.Function; public class Main { static String S; public static void main(String[] args) { FastScanner sc = new FastScanner(System.in); S = sc.next(); System.out.println(solve()); } static int solve() { int ans = -1; int time = 1; int n = S.length(); for (int i = 1; i < n*2; i++) { if( S.charAt((i-1)%n) != S.charAt(i%n) ) { time++; } else { ans = Math.max(time, ans); time = 1; } } ans = Math.max(time, ans); if( ans == n*2 ) { return n; } else { return ans; } } @SuppressWarnings("unused") static class FastScanner { private BufferedReader reader; private StringTokenizer tokenizer; FastScanner(InputStream in) { reader = new BufferedReader(new InputStreamReader(in)); tokenizer = null; } String next() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } String nextLine() { if (tokenizer == null || !tokenizer.hasMoreTokens()) { try { return reader.readLine(); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken("\n"); } long nextLong() { return Long.parseLong(next()); } int nextInt() { return Integer.parseInt(next()); } int[] nextIntArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } int[] nextIntArray(int n, int delta) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt() + delta; return a; } long[] nextLongArray(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nextLong(); return a; } } static <A> void writeLines(A[] as, Function<A, String> f) { PrintWriter pw = new PrintWriter(System.out); for (A a : as) { pw.println(f.apply(a)); } pw.flush(); } static void writeLines(int[] as) { PrintWriter pw = new PrintWriter(System.out); for (int a : as) pw.println(a); pw.flush(); } static void writeLines(long[] as) { PrintWriter pw = new PrintWriter(System.out); for (long a : as) pw.println(a); pw.flush(); } static int max(int... as) { int max = Integer.MIN_VALUE; for (int a : as) max = Math.max(a, max); return max; } static int min(int... as) { int min = Integer.MAX_VALUE; for (int a : as) min = Math.min(a, min); return min; } static void debug(Object... args) { StringJoiner j = new StringJoiner(" "); for (Object arg : args) { if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg)); else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg)); else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg)); else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg)); else j.add(arg.toString()); } System.err.println(j.toString()); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,183
526
2,111
import java.util.*; import java.io.*; import java.math.BigInteger; public class Problem { static int mod = (int) (1e9+7); static InputReader in; static PrintWriter out; static int[] rt; static int[] size; static void initialize(int n){ rt = new int[n + 1]; size = new int[n + 1]; for(int i = 0; i < rt.length; i++){ rt[i] = i; size[i] = 1; } } static int root(int x){ while(rt[x] != x){ rt[x] = rt[rt[x]]; x = rt[x]; } return x; } static long union(int x,int y){ int root_x = root(x); int root_y = root(y); if(root_x == root_y) return 0; long val = size[root_x] *1l* size[root_y]; if(size[root_x]<size[root_y]){ rt[root_x] = rt[root_y]; size[root_y] += size[root_x]; } else{ rt[root_y] = rt[root_x]; size[root_x] += size[root_y]; } return val; } static void solve() { in = new InputReader(System.in); out = new PrintWriter(System.out); int t = 1; while(t-- > 0){ int n = in.nextInt(); int[] arr = in.nextIntArray(n); ArrayList<Pair> list = new ArrayList<>(); for(int i = 1; i < n; i++){ int u = in.nextInt() - 1; int v = in.nextInt() - 1; list.add(new Pair(u, v, Math.max(arr[u],arr[v]))); } list.sort((p1,p2) -> Integer.compare(p1.i, p2.i)); initialize(n); long s1 = 0; for(int i = 0; i < list.size(); i++){ s1 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } for(int i = 0; i < list.size(); i++){ Pair p = list.get(i); p.i = Math.min(arr[p.x],arr[p.y]); } list.sort((p1,p2) -> -Integer.compare(p1.i, p2.i)); initialize(n); long s2 = 0; for(int i = 0; i < list.size(); i++){ s2 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } out.println(s1 - s2); } out.close(); } public static void main(String[] args) { new Thread(null ,new Runnable(){ public void run(){ try{ solve(); } catch(Exception e){ e.printStackTrace(); } } },"1",1<<26).start(); } static class Pair implements Comparable<Pair> { int x,y; int i; Pair (int x,int y) { this.x = x; this.y = y; } Pair (int x,int y, int i) { this.x = x; this.y = y; this.i = i; } public int compareTo(Pair o) { if(this.x != o.x) return -Integer.compare(this.x, o.y); return -Integer.compare(this.y,o.y); //return 0; } public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y==y; } return false; } @Override public String toString() { return x + " "+ y + " "+i; } /*public int hashCode() { return new Long(x).hashCode() * 31 + new Long(y).hashCode(); }*/ } static long add(long a,long b){ long x=(a+b); while(x>=mod) x-=mod; return x; } static long sub(long a,long b){ long x=(a-b); while(x<0) x+=mod; return x; } static long mul(long a,long b){ long x=(a*b); while(x>=mod) x-=mod; return x; } static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static long gcd(long x,long y) { if(y==0) return x; else return gcd(y,x%y); } static int gcd(int x,int y) { if(y==0) return x; else return gcd(y,x%y); } static long pow(long n,long p,long m) { long result = 1; if(p==0){ return n; } while(p!=0) { if(p%2==1) result *= n; if(result >= m) result %= m; p >>=1; n*=n; if(n >= m) n%=m; } return result; } static long pow(long n,long p) { long result = 1; if(p==0) return 1; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String readString() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.io.*; import java.math.BigInteger; public class Problem { static int mod = (int) (1e9+7); static InputReader in; static PrintWriter out; static int[] rt; static int[] size; static void initialize(int n){ rt = new int[n + 1]; size = new int[n + 1]; for(int i = 0; i < rt.length; i++){ rt[i] = i; size[i] = 1; } } static int root(int x){ while(rt[x] != x){ rt[x] = rt[rt[x]]; x = rt[x]; } return x; } static long union(int x,int y){ int root_x = root(x); int root_y = root(y); if(root_x == root_y) return 0; long val = size[root_x] *1l* size[root_y]; if(size[root_x]<size[root_y]){ rt[root_x] = rt[root_y]; size[root_y] += size[root_x]; } else{ rt[root_y] = rt[root_x]; size[root_x] += size[root_y]; } return val; } static void solve() { in = new InputReader(System.in); out = new PrintWriter(System.out); int t = 1; while(t-- > 0){ int n = in.nextInt(); int[] arr = in.nextIntArray(n); ArrayList<Pair> list = new ArrayList<>(); for(int i = 1; i < n; i++){ int u = in.nextInt() - 1; int v = in.nextInt() - 1; list.add(new Pair(u, v, Math.max(arr[u],arr[v]))); } list.sort((p1,p2) -> Integer.compare(p1.i, p2.i)); initialize(n); long s1 = 0; for(int i = 0; i < list.size(); i++){ s1 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } for(int i = 0; i < list.size(); i++){ Pair p = list.get(i); p.i = Math.min(arr[p.x],arr[p.y]); } list.sort((p1,p2) -> -Integer.compare(p1.i, p2.i)); initialize(n); long s2 = 0; for(int i = 0; i < list.size(); i++){ s2 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } out.println(s1 - s2); } out.close(); } public static void main(String[] args) { new Thread(null ,new Runnable(){ public void run(){ try{ solve(); } catch(Exception e){ e.printStackTrace(); } } },"1",1<<26).start(); } static class Pair implements Comparable<Pair> { int x,y; int i; Pair (int x,int y) { this.x = x; this.y = y; } Pair (int x,int y, int i) { this.x = x; this.y = y; this.i = i; } public int compareTo(Pair o) { if(this.x != o.x) return -Integer.compare(this.x, o.y); return -Integer.compare(this.y,o.y); //return 0; } public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y==y; } return false; } @Override public String toString() { return x + " "+ y + " "+i; } /*public int hashCode() { return new Long(x).hashCode() * 31 + new Long(y).hashCode(); }*/ } static long add(long a,long b){ long x=(a+b); while(x>=mod) x-=mod; return x; } static long sub(long a,long b){ long x=(a-b); while(x<0) x+=mod; return x; } static long mul(long a,long b){ long x=(a*b); while(x>=mod) x-=mod; return x; } static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static long gcd(long x,long y) { if(y==0) return x; else return gcd(y,x%y); } static int gcd(int x,int y) { if(y==0) return x; else return gcd(y,x%y); } static long pow(long n,long p,long m) { long result = 1; if(p==0){ return n; } while(p!=0) { if(p%2==1) result *= n; if(result >= m) result %= m; p >>=1; n*=n; if(n >= m) n%=m; } return result; } static long pow(long n,long p) { long result = 1; if(p==0) return 1; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String readString() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.*; import java.io.*; import java.math.BigInteger; public class Problem { static int mod = (int) (1e9+7); static InputReader in; static PrintWriter out; static int[] rt; static int[] size; static void initialize(int n){ rt = new int[n + 1]; size = new int[n + 1]; for(int i = 0; i < rt.length; i++){ rt[i] = i; size[i] = 1; } } static int root(int x){ while(rt[x] != x){ rt[x] = rt[rt[x]]; x = rt[x]; } return x; } static long union(int x,int y){ int root_x = root(x); int root_y = root(y); if(root_x == root_y) return 0; long val = size[root_x] *1l* size[root_y]; if(size[root_x]<size[root_y]){ rt[root_x] = rt[root_y]; size[root_y] += size[root_x]; } else{ rt[root_y] = rt[root_x]; size[root_x] += size[root_y]; } return val; } static void solve() { in = new InputReader(System.in); out = new PrintWriter(System.out); int t = 1; while(t-- > 0){ int n = in.nextInt(); int[] arr = in.nextIntArray(n); ArrayList<Pair> list = new ArrayList<>(); for(int i = 1; i < n; i++){ int u = in.nextInt() - 1; int v = in.nextInt() - 1; list.add(new Pair(u, v, Math.max(arr[u],arr[v]))); } list.sort((p1,p2) -> Integer.compare(p1.i, p2.i)); initialize(n); long s1 = 0; for(int i = 0; i < list.size(); i++){ s1 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } for(int i = 0; i < list.size(); i++){ Pair p = list.get(i); p.i = Math.min(arr[p.x],arr[p.y]); } list.sort((p1,p2) -> -Integer.compare(p1.i, p2.i)); initialize(n); long s2 = 0; for(int i = 0; i < list.size(); i++){ s2 += union(list.get(i).x, list.get(i).y) * list.get(i).i; } out.println(s1 - s2); } out.close(); } public static void main(String[] args) { new Thread(null ,new Runnable(){ public void run(){ try{ solve(); } catch(Exception e){ e.printStackTrace(); } } },"1",1<<26).start(); } static class Pair implements Comparable<Pair> { int x,y; int i; Pair (int x,int y) { this.x = x; this.y = y; } Pair (int x,int y, int i) { this.x = x; this.y = y; this.i = i; } public int compareTo(Pair o) { if(this.x != o.x) return -Integer.compare(this.x, o.y); return -Integer.compare(this.y,o.y); //return 0; } public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y==y; } return false; } @Override public String toString() { return x + " "+ y + " "+i; } /*public int hashCode() { return new Long(x).hashCode() * 31 + new Long(y).hashCode(); }*/ } static long add(long a,long b){ long x=(a+b); while(x>=mod) x-=mod; return x; } static long sub(long a,long b){ long x=(a-b); while(x<0) x+=mod; return x; } static long mul(long a,long b){ long x=(a*b); while(x>=mod) x-=mod; return x; } static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } static long gcd(long x,long y) { if(y==0) return x; else return gcd(y,x%y); } static int gcd(int x,int y) { if(y==0) return x; else return gcd(y,x%y); } static long pow(long n,long p,long m) { long result = 1; if(p==0){ return n; } while(p!=0) { if(p%2==1) result *= n; if(result >= m) result %= m; p >>=1; n*=n; if(n >= m) n%=m; } return result; } static long pow(long n,long p) { long result = 1; if(p==0) return 1; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } static class InputReader { private final InputStream stream; private final byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } public long[] nextLongArray(int n) { long a[] = new long[n]; for (int i = 0; i < n; i++) { a[i] = nextLong(); } return a; } public String readString() { int c = snext(); while (isSpaceChar(c)) { c = snext(); } StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isSpaceChar(c)); return res.toString(); } public String nextLine() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isEndOfLine(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
2,407
2,107
3,075
import java.util.Scanner; public class ToyArmy { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long n = sc.nextLong(); System.out.println(n / 2 + n); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class ToyArmy { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long n = sc.nextLong(); System.out.println(n / 2 + n); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.Scanner; public class ToyArmy { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long n = sc.nextLong(); System.out.println(n / 2 + n); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
379
3,069
1,951
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.Arrays; import java.util.Comparator; import java.util.StringTokenizer; import java.util.logging.Level; import java.util.logging.Logger; public class ProblemA { private final BufferedReader in; private final PrintStream out; private StringTokenizer tok = new StringTokenizer(""); private String nextLine = null; public static void main(String[] args) throws Exception { new ProblemA(); } private ProblemA() throws Exception { in = new BufferedReader(new InputStreamReader(System.in)); out = System.out; start(); end(); } private int nextInt() { return Integer.parseInt(nextWord()); } private String nextWord() { if (tok.hasMoreTokens()) { return tok.nextToken(); } else { while (!tok.hasMoreTokens()) { try { nextLine = in.readLine(); if (nextLine == null) { return null; } else { tok = new StringTokenizer(nextLine); } } catch (IOException ex) { Logger.getLogger(ProblemA.class.getName()).log(Level.SEVERE, null, ex); } } return tok.nextToken(); } } private void start() { int n = nextInt(); int k = nextInt(); T[] ts = new T[n]; for (int i = 0; i < n; i++) { ts[i] = new T(nextInt(), nextInt()); } Arrays.sort(ts, new Comparator<T>() { @Override public int compare(T o1, T o2) { if (o1.p > o2.p) { return -1; } if (o1.p < o2.p) { return 1; } if (o1.t < o2.t) { return -1; } if (o1.t > o2.t) { return 1; } return 0; } }); int t = ts[k - 1].t; int p = ts[k - 1].p; int res = 0; for (int i = 0; i < n; i++) { if (ts[i].p == p && ts[i].t == t) { res++; } } out.println(res); } class T { int p; int t; public T(int p, int t) { this.p = p; this.t = t; } } private void end() { out.close(); } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.Arrays; import java.util.Comparator; import java.util.StringTokenizer; import java.util.logging.Level; import java.util.logging.Logger; public class ProblemA { private final BufferedReader in; private final PrintStream out; private StringTokenizer tok = new StringTokenizer(""); private String nextLine = null; public static void main(String[] args) throws Exception { new ProblemA(); } private ProblemA() throws Exception { in = new BufferedReader(new InputStreamReader(System.in)); out = System.out; start(); end(); } private int nextInt() { return Integer.parseInt(nextWord()); } private String nextWord() { if (tok.hasMoreTokens()) { return tok.nextToken(); } else { while (!tok.hasMoreTokens()) { try { nextLine = in.readLine(); if (nextLine == null) { return null; } else { tok = new StringTokenizer(nextLine); } } catch (IOException ex) { Logger.getLogger(ProblemA.class.getName()).log(Level.SEVERE, null, ex); } } return tok.nextToken(); } } private void start() { int n = nextInt(); int k = nextInt(); T[] ts = new T[n]; for (int i = 0; i < n; i++) { ts[i] = new T(nextInt(), nextInt()); } Arrays.sort(ts, new Comparator<T>() { @Override public int compare(T o1, T o2) { if (o1.p > o2.p) { return -1; } if (o1.p < o2.p) { return 1; } if (o1.t < o2.t) { return -1; } if (o1.t > o2.t) { return 1; } return 0; } }); int t = ts[k - 1].t; int p = ts[k - 1].p; int res = 0; for (int i = 0; i < n; i++) { if (ts[i].p == p && ts[i].t == t) { res++; } } out.println(res); } class T { int p; int t; public T(int p, int t) { this.p = p; this.t = t; } } private void end() { out.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.Arrays; import java.util.Comparator; import java.util.StringTokenizer; import java.util.logging.Level; import java.util.logging.Logger; public class ProblemA { private final BufferedReader in; private final PrintStream out; private StringTokenizer tok = new StringTokenizer(""); private String nextLine = null; public static void main(String[] args) throws Exception { new ProblemA(); } private ProblemA() throws Exception { in = new BufferedReader(new InputStreamReader(System.in)); out = System.out; start(); end(); } private int nextInt() { return Integer.parseInt(nextWord()); } private String nextWord() { if (tok.hasMoreTokens()) { return tok.nextToken(); } else { while (!tok.hasMoreTokens()) { try { nextLine = in.readLine(); if (nextLine == null) { return null; } else { tok = new StringTokenizer(nextLine); } } catch (IOException ex) { Logger.getLogger(ProblemA.class.getName()).log(Level.SEVERE, null, ex); } } return tok.nextToken(); } } private void start() { int n = nextInt(); int k = nextInt(); T[] ts = new T[n]; for (int i = 0; i < n; i++) { ts[i] = new T(nextInt(), nextInt()); } Arrays.sort(ts, new Comparator<T>() { @Override public int compare(T o1, T o2) { if (o1.p > o2.p) { return -1; } if (o1.p < o2.p) { return 1; } if (o1.t < o2.t) { return -1; } if (o1.t > o2.t) { return 1; } return 0; } }); int t = ts[k - 1].t; int p = ts[k - 1].p; int res = 0; for (int i = 0; i < n; i++) { if (ts[i].p == p && ts[i].t == t) { res++; } } out.println(res); } class T { int p; int t; public T(int p, int t) { this.p = p; this.t = t; } } private void end() { out.close(); } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
910
1,947
641
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.io.StreamTokenizer; public class Main { private static StreamTokenizer in; private static PrintWriter out; private static int nextInt() throws Exception { in.nextToken(); return (int)in.nval; } private static String nextString() throws Exception { in.nextToken(); return in.sval; } public static void main(String[] args) throws Exception { in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); out = new PrintWriter(System.out); int n = nextInt(); byte f = (byte)nextInt(); byte s = (byte)nextInt(); byte t = (byte)nextInt(); boolean bf = false; boolean bs = false; boolean bt = false; if((f&1) == 0){bf = true;} if((s&1) == 0){bs = true;} if((t&1) == 0){bt = true;} //System.out.println(bf+""+bs+""+bt); if((!bf)&&bs&&bt){System.out.println(1);return;} if(bf&&(!bs)&&bt){System.out.println(2);return;} if(bf&&bs&&(!bt)){System.out.println(3);return;} if(bf&&!bs&&!bt){System.out.println(1);return;} if(!bf&&bs&&!bt){System.out.println(2);return;} if(!bf&&!bs&&bt){System.out.println(3);return;} for(int i = 4; i<=n; i++){ byte g = (byte) nextInt(); if(((g+f)&1) == 1){System.out.println(i); return;} } out.flush(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.io.StreamTokenizer; public class Main { private static StreamTokenizer in; private static PrintWriter out; private static int nextInt() throws Exception { in.nextToken(); return (int)in.nval; } private static String nextString() throws Exception { in.nextToken(); return in.sval; } public static void main(String[] args) throws Exception { in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); out = new PrintWriter(System.out); int n = nextInt(); byte f = (byte)nextInt(); byte s = (byte)nextInt(); byte t = (byte)nextInt(); boolean bf = false; boolean bs = false; boolean bt = false; if((f&1) == 0){bf = true;} if((s&1) == 0){bs = true;} if((t&1) == 0){bt = true;} //System.out.println(bf+""+bs+""+bt); if((!bf)&&bs&&bt){System.out.println(1);return;} if(bf&&(!bs)&&bt){System.out.println(2);return;} if(bf&&bs&&(!bt)){System.out.println(3);return;} if(bf&&!bs&&!bt){System.out.println(1);return;} if(!bf&&bs&&!bt){System.out.println(2);return;} if(!bf&&!bs&&bt){System.out.println(3);return;} for(int i = 4; i<=n; i++){ byte g = (byte) nextInt(); if(((g+f)&1) == 1){System.out.println(i); return;} } out.flush(); } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(1): The time complexity is constant to the input size n. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.io.StreamTokenizer; public class Main { private static StreamTokenizer in; private static PrintWriter out; private static int nextInt() throws Exception { in.nextToken(); return (int)in.nval; } private static String nextString() throws Exception { in.nextToken(); return in.sval; } public static void main(String[] args) throws Exception { in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); out = new PrintWriter(System.out); int n = nextInt(); byte f = (byte)nextInt(); byte s = (byte)nextInt(); byte t = (byte)nextInt(); boolean bf = false; boolean bs = false; boolean bt = false; if((f&1) == 0){bf = true;} if((s&1) == 0){bs = true;} if((t&1) == 0){bt = true;} //System.out.println(bf+""+bs+""+bt); if((!bf)&&bs&&bt){System.out.println(1);return;} if(bf&&(!bs)&&bt){System.out.println(2);return;} if(bf&&bs&&(!bt)){System.out.println(3);return;} if(bf&&!bs&&!bt){System.out.println(1);return;} if(!bf&&bs&&!bt){System.out.println(2);return;} if(!bf&&!bs&&bt){System.out.println(3);return;} for(int i = 4; i<=n; i++){ byte g = (byte) nextInt(); if(((g+f)&1) == 1){System.out.println(i); return;} } out.flush(); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
681
640
977
import java.io.*; import java.math.BigInteger; import java.util.*; import java.lang.*; import static java.lang.Math.*; public class Main implements Runnable { public static void main(String[] args) { new Thread(null, new Main(), "Check2", 1 << 28).start();// to increse stack size in java } static long mod = (long) (1e9 + 7); public void run() { InputReader in = new InputReader(System.in); PrintWriter w = new PrintWriter(System.out); long n = in.nextLong(); long k = in.nextLong(); long ans= 0; for (long i=1;i<=Math.min(100000,n);i++){ long left = i; long right = n-i; long sum = left * (left + 1); sum /= 2; if (sum - right ==k){ ans = right; } } w.println(ans); w.close(); } void debug(Object...args) { System.out.println(Arrays.deepToString(args)); } static class pair implements Comparable<pair>{ int a; int b; pair(int a,int b){ this.a = a; this.b = b; } public int compareTo(pair o){ if(this.b != o.b)return this.b - o.b; return this.a - o.a; } } long modinv(long a,long b) { long p=power(b,mod-2,mod); p=a%mod*p%mod; p%=mod; return p; } long power(long x,long y,long mod){ if(y==0)return 1%mod; if(y==1)return x%mod; long res=1; x=x%mod; while(y>0) { if((y%2)!=0){ res=(res*x)%mod; } y=y/2; x=(x*x)%mod; } return res; } long gcd(long a,long b){ if(b==0)return a; return gcd(b,a%b); } void sev(int a[],int n){ for(int i=2;i<=n;i++)a[i]=i; for(int i=2;i<=n;i++){ if(a[i]!=0){ for(int j=2*i;j<=n;){ a[j]=0; j=j+i; } } } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars==-1) throw new InputMismatchException(); if (curChar >= numChars){ curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if(numChars <= 0) return -1; } return buf[curChar++]; } public String nextLine() { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String stock = ""; try { stock = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return stock; } public int nextInt() { int c = read(); while(isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if(c<'0'||c>'9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.math.BigInteger; import java.util.*; import java.lang.*; import static java.lang.Math.*; public class Main implements Runnable { public static void main(String[] args) { new Thread(null, new Main(), "Check2", 1 << 28).start();// to increse stack size in java } static long mod = (long) (1e9 + 7); public void run() { InputReader in = new InputReader(System.in); PrintWriter w = new PrintWriter(System.out); long n = in.nextLong(); long k = in.nextLong(); long ans= 0; for (long i=1;i<=Math.min(100000,n);i++){ long left = i; long right = n-i; long sum = left * (left + 1); sum /= 2; if (sum - right ==k){ ans = right; } } w.println(ans); w.close(); } void debug(Object...args) { System.out.println(Arrays.deepToString(args)); } static class pair implements Comparable<pair>{ int a; int b; pair(int a,int b){ this.a = a; this.b = b; } public int compareTo(pair o){ if(this.b != o.b)return this.b - o.b; return this.a - o.a; } } long modinv(long a,long b) { long p=power(b,mod-2,mod); p=a%mod*p%mod; p%=mod; return p; } long power(long x,long y,long mod){ if(y==0)return 1%mod; if(y==1)return x%mod; long res=1; x=x%mod; while(y>0) { if((y%2)!=0){ res=(res*x)%mod; } y=y/2; x=(x*x)%mod; } return res; } long gcd(long a,long b){ if(b==0)return a; return gcd(b,a%b); } void sev(int a[],int n){ for(int i=2;i<=n;i++)a[i]=i; for(int i=2;i<=n;i++){ if(a[i]!=0){ for(int j=2*i;j<=n;){ a[j]=0; j=j+i; } } } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars==-1) throw new InputMismatchException(); if (curChar >= numChars){ curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if(numChars <= 0) return -1; } return buf[curChar++]; } public String nextLine() { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String stock = ""; try { stock = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return stock; } public int nextInt() { int c = read(); while(isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if(c<'0'||c>'9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.math.BigInteger; import java.util.*; import java.lang.*; import static java.lang.Math.*; public class Main implements Runnable { public static void main(String[] args) { new Thread(null, new Main(), "Check2", 1 << 28).start();// to increse stack size in java } static long mod = (long) (1e9 + 7); public void run() { InputReader in = new InputReader(System.in); PrintWriter w = new PrintWriter(System.out); long n = in.nextLong(); long k = in.nextLong(); long ans= 0; for (long i=1;i<=Math.min(100000,n);i++){ long left = i; long right = n-i; long sum = left * (left + 1); sum /= 2; if (sum - right ==k){ ans = right; } } w.println(ans); w.close(); } void debug(Object...args) { System.out.println(Arrays.deepToString(args)); } static class pair implements Comparable<pair>{ int a; int b; pair(int a,int b){ this.a = a; this.b = b; } public int compareTo(pair o){ if(this.b != o.b)return this.b - o.b; return this.a - o.a; } } long modinv(long a,long b) { long p=power(b,mod-2,mod); p=a%mod*p%mod; p%=mod; return p; } long power(long x,long y,long mod){ if(y==0)return 1%mod; if(y==1)return x%mod; long res=1; x=x%mod; while(y>0) { if((y%2)!=0){ res=(res*x)%mod; } y=y/2; x=(x*x)%mod; } return res; } long gcd(long a,long b){ if(b==0)return a; return gcd(b,a%b); } void sev(int a[],int n){ for(int i=2;i<=n;i++)a[i]=i; for(int i=2;i<=n;i++){ if(a[i]!=0){ for(int j=2*i;j<=n;){ a[j]=0; j=j+i; } } } } static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars==-1) throw new InputMismatchException(); if (curChar >= numChars){ curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if(numChars <= 0) return -1; } return buf[curChar++]; } public String nextLine() { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String stock = ""; try { stock = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return stock; } public int nextInt() { int c = read(); while(isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if(c<'0'||c>'9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public double nextDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, nextInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,769
976
203
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class AnnoyingPresent { //UPSOLVED public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(br.readLine()); long n = Long.parseLong(st.nextToken()) , m = Long.parseLong(st.nextToken()); long sum = 0; for(int i=0;i<m;i++){ StringTokenizer st1 = new StringTokenizer(br.readLine()); sum+= n* Long.parseLong(st1.nextToken()); Long a= Long.parseLong(st1.nextToken()); if(a < 0){ if(n % 2 == 0) sum += n*n / 4*a; else{ sum += (n/2) * (n/2+1) * a; } } else sum += (a*(n) * (n-1) / 2); } System.out.println((double)sum/n); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class AnnoyingPresent { //UPSOLVED public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(br.readLine()); long n = Long.parseLong(st.nextToken()) , m = Long.parseLong(st.nextToken()); long sum = 0; for(int i=0;i<m;i++){ StringTokenizer st1 = new StringTokenizer(br.readLine()); sum+= n* Long.parseLong(st1.nextToken()); Long a= Long.parseLong(st1.nextToken()); if(a < 0){ if(n % 2 == 0) sum += n*n / 4*a; else{ sum += (n/2) * (n/2+1) * a; } } else sum += (a*(n) * (n-1) / 2); } System.out.println((double)sum/n); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class AnnoyingPresent { //UPSOLVED public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(br.readLine()); long n = Long.parseLong(st.nextToken()) , m = Long.parseLong(st.nextToken()); long sum = 0; for(int i=0;i<m;i++){ StringTokenizer st1 = new StringTokenizer(br.readLine()); sum+= n* Long.parseLong(st1.nextToken()); Long a= Long.parseLong(st1.nextToken()); if(a < 0){ if(n % 2 == 0) sum += n*n / 4*a; else{ sum += (n/2) * (n/2+1) * a; } } else sum += (a*(n) * (n-1) / 2); } System.out.println((double)sum/n); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
557
203
4,380
import java.awt.List; import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Scanner; import java.util.StringTokenizer; public class T { static Scanner in = new Scanner(); static PrintWriter out = new PrintWriter(System.out); static boolean adj[][]; static int n, m, from; static long memo[][]; static long Num_Cycle; public static void main(String[] args) throws IOException { n = in.nextInt(); m = in.nextInt(); adj = new boolean[n][n]; memo = new long[n][1 << n]; for (int i = 0; i < m; i++) { int u = in.nextInt() - 1; int v = in.nextInt() - 1; adj[u][v] = adj[v][u] = true; } for (long arr[] : memo) { Arrays.fill(arr, -1); } Num_Cycle = 0L; for (int i = 0; i < n; i++) { from = i; Num_Cycle += dp(from, (1 << i)); } out.println(Num_Cycle / 2); out.flush(); out.close(); } static long dp(int start, int mask) { if (memo[start][mask] != -1) { return (memo[start][mask]); } long ans = 0L; if (adj[start][from] && Integer.bitCount(mask) >= 3) {// Cycle has // atleast 3 // node and 3 // edges ans++; } for (int i = from + 1; i < n; i++) { if (adj[start][i] && ((mask & (1 << i)) == 0)) { ans += dp(i, mask | (1 << i)); } } return memo[start][mask] = ans; } static class Scanner { BufferedReader br; StringTokenizer st; Scanner() { br = new BufferedReader(new InputStreamReader(System.in)); } Scanner(String file) throws FileNotFoundException { br = new BufferedReader(new FileReader(file)); } String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } String nextLine() throws IOException { return br.readLine(); } long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.awt.List; import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Scanner; import java.util.StringTokenizer; public class T { static Scanner in = new Scanner(); static PrintWriter out = new PrintWriter(System.out); static boolean adj[][]; static int n, m, from; static long memo[][]; static long Num_Cycle; public static void main(String[] args) throws IOException { n = in.nextInt(); m = in.nextInt(); adj = new boolean[n][n]; memo = new long[n][1 << n]; for (int i = 0; i < m; i++) { int u = in.nextInt() - 1; int v = in.nextInt() - 1; adj[u][v] = adj[v][u] = true; } for (long arr[] : memo) { Arrays.fill(arr, -1); } Num_Cycle = 0L; for (int i = 0; i < n; i++) { from = i; Num_Cycle += dp(from, (1 << i)); } out.println(Num_Cycle / 2); out.flush(); out.close(); } static long dp(int start, int mask) { if (memo[start][mask] != -1) { return (memo[start][mask]); } long ans = 0L; if (adj[start][from] && Integer.bitCount(mask) >= 3) {// Cycle has // atleast 3 // node and 3 // edges ans++; } for (int i = from + 1; i < n; i++) { if (adj[start][i] && ((mask & (1 << i)) == 0)) { ans += dp(i, mask | (1 << i)); } } return memo[start][mask] = ans; } static class Scanner { BufferedReader br; StringTokenizer st; Scanner() { br = new BufferedReader(new InputStreamReader(System.in)); } Scanner(String file) throws FileNotFoundException { br = new BufferedReader(new FileReader(file)); } String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } String nextLine() throws IOException { return br.readLine(); } long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.awt.List; import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Scanner; import java.util.StringTokenizer; public class T { static Scanner in = new Scanner(); static PrintWriter out = new PrintWriter(System.out); static boolean adj[][]; static int n, m, from; static long memo[][]; static long Num_Cycle; public static void main(String[] args) throws IOException { n = in.nextInt(); m = in.nextInt(); adj = new boolean[n][n]; memo = new long[n][1 << n]; for (int i = 0; i < m; i++) { int u = in.nextInt() - 1; int v = in.nextInt() - 1; adj[u][v] = adj[v][u] = true; } for (long arr[] : memo) { Arrays.fill(arr, -1); } Num_Cycle = 0L; for (int i = 0; i < n; i++) { from = i; Num_Cycle += dp(from, (1 << i)); } out.println(Num_Cycle / 2); out.flush(); out.close(); } static long dp(int start, int mask) { if (memo[start][mask] != -1) { return (memo[start][mask]); } long ans = 0L; if (adj[start][from] && Integer.bitCount(mask) >= 3) {// Cycle has // atleast 3 // node and 3 // edges ans++; } for (int i = from + 1; i < n; i++) { if (adj[start][i] && ((mask & (1 << i)) == 0)) { ans += dp(i, mask | (1 << i)); } } return memo[start][mask] = ans; } static class Scanner { BufferedReader br; StringTokenizer st; Scanner() { br = new BufferedReader(new InputStreamReader(System.in)); } Scanner(String file) throws FileNotFoundException { br = new BufferedReader(new FileReader(file)); } String next() throws IOException { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } String nextLine() throws IOException { return br.readLine(); } long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
954
4,369
2,337
import java.util.*; import java.text.*; import java.io.*; import java.math.*; public class code5 { InputStream is; PrintWriter out; static long mod=pow(10,9)+7; static int dx[]={0,0,1,-1},dy[]={1,-1,0,0}; String arr[]; long dp[][]; void solve() throws IOException { int n=ni(); int a[]=na(n); int q=ni(); boolean flag=false; for(int i=0;i<n;i++) { for(int j=0;j<i;j++) { if(a[j]>a[i]) flag^=true; } } while(q--!=0) { int l=ni()-1; int r=ni()-1; int num=(r-l+1); int tot=num*(num-1)/2; if(tot%2==1) flag^=true; if(flag) out.println("odd"); else out.println("even"); } } int sum(int i) { int sum=0; while(i!=0) { if((i%10)%2==1) sum+=i%10; i/=10; } return sum; } ArrayList<Integer> al[]; void take(int n,int m) { al=new ArrayList[n]; for(int i=0;i<n;i++) al[i]=new ArrayList<Integer>(); for(int i=0;i<m;i++) { int x=ni()-1; int y=ni()-1; al[x].add(y); al[y].add(x); } } int check(long n) { int count=0; while(n!=0) { if(n%10!=0) break; n/=10; count++; } return count; } public static int count(int x) { int num=0; while(x!=0) { x=x&(x-1); num++; } return num; } static long d, x, y; void extendedEuclid(long A, long B) { if(B == 0) { d = A; x = 1; y = 0; } else { extendedEuclid(B, A%B); long temp = x; x = y; y = temp - (A/B)*y; } } long modInverse(long A,long M) //A and M are coprime { extendedEuclid(A,M); return (x%M+M)%M; //x may be negative } public static void mergeSort(int[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(int arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); int left[] = new int[n1]; int right[] = new int[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } public static void mergeSort(long[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(long arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); long left[] = new long[n1]; long right[] = new long[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } static class Pair implements Comparable<Pair>{ int x,y,k; int i,dir; long val; Pair (int x,int y){ this.x=x; this.y=y; } public int compareTo(Pair o) { if(this.x!=o.x) return this.x-o.x; return this.y-o.y; } public String toString(){ return x+" "+y+" "+k+" "+i;} public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y == y; } return false; } public int hashCode() { return new Long(x).hashCode()*31 + new Long(y).hashCode() ; } } public static boolean isPal(String s){ for(int i=0, j=s.length()-1;i<=j;i++,j--){ if(s.charAt(i)!=s.charAt(j)) return false; } return true; } public static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } public static long gcd(long x,long y){ if(y==0) return x; else return gcd(y,x%y); } public static int gcd(int x,int y){ if(y==0) return x; return gcd(y,x%y); } public static long gcdExtended(long a,long b,long[] x){ if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } public static int abs(int a,int b){ return (int)Math.abs(a-b); } public static long abs(long a,long b){ return (long)Math.abs(a-b); } public static int max(int a,int b){ if(a>b) return a; else return b; } public static int min(int a,int b){ if(a>b) return b; else return a; } public static long max(long a,long b){ if(a>b) return a; else return b; } public static long min(long a,long b){ if(a>b) return b; else return a; } public static long pow(long n,long p,long m){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; if(result>=m) result%=m; p >>=1; n*=n; if(n>=m) n%=m; } return result; } public static long pow(long n,long p){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } public static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } void run() throws Exception { is = System.in; out = new PrintWriter(System.out); solve(); out.flush(); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { public void run() { try { new code5().run(); } catch (Exception e) { e.printStackTrace(); } } }, "1", 1 << 26).start(); //new code5().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private long[] nl(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nl(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.text.*; import java.io.*; import java.math.*; public class code5 { InputStream is; PrintWriter out; static long mod=pow(10,9)+7; static int dx[]={0,0,1,-1},dy[]={1,-1,0,0}; String arr[]; long dp[][]; void solve() throws IOException { int n=ni(); int a[]=na(n); int q=ni(); boolean flag=false; for(int i=0;i<n;i++) { for(int j=0;j<i;j++) { if(a[j]>a[i]) flag^=true; } } while(q--!=0) { int l=ni()-1; int r=ni()-1; int num=(r-l+1); int tot=num*(num-1)/2; if(tot%2==1) flag^=true; if(flag) out.println("odd"); else out.println("even"); } } int sum(int i) { int sum=0; while(i!=0) { if((i%10)%2==1) sum+=i%10; i/=10; } return sum; } ArrayList<Integer> al[]; void take(int n,int m) { al=new ArrayList[n]; for(int i=0;i<n;i++) al[i]=new ArrayList<Integer>(); for(int i=0;i<m;i++) { int x=ni()-1; int y=ni()-1; al[x].add(y); al[y].add(x); } } int check(long n) { int count=0; while(n!=0) { if(n%10!=0) break; n/=10; count++; } return count; } public static int count(int x) { int num=0; while(x!=0) { x=x&(x-1); num++; } return num; } static long d, x, y; void extendedEuclid(long A, long B) { if(B == 0) { d = A; x = 1; y = 0; } else { extendedEuclid(B, A%B); long temp = x; x = y; y = temp - (A/B)*y; } } long modInverse(long A,long M) //A and M are coprime { extendedEuclid(A,M); return (x%M+M)%M; //x may be negative } public static void mergeSort(int[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(int arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); int left[] = new int[n1]; int right[] = new int[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } public static void mergeSort(long[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(long arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); long left[] = new long[n1]; long right[] = new long[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } static class Pair implements Comparable<Pair>{ int x,y,k; int i,dir; long val; Pair (int x,int y){ this.x=x; this.y=y; } public int compareTo(Pair o) { if(this.x!=o.x) return this.x-o.x; return this.y-o.y; } public String toString(){ return x+" "+y+" "+k+" "+i;} public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y == y; } return false; } public int hashCode() { return new Long(x).hashCode()*31 + new Long(y).hashCode() ; } } public static boolean isPal(String s){ for(int i=0, j=s.length()-1;i<=j;i++,j--){ if(s.charAt(i)!=s.charAt(j)) return false; } return true; } public static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } public static long gcd(long x,long y){ if(y==0) return x; else return gcd(y,x%y); } public static int gcd(int x,int y){ if(y==0) return x; return gcd(y,x%y); } public static long gcdExtended(long a,long b,long[] x){ if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } public static int abs(int a,int b){ return (int)Math.abs(a-b); } public static long abs(long a,long b){ return (long)Math.abs(a-b); } public static int max(int a,int b){ if(a>b) return a; else return b; } public static int min(int a,int b){ if(a>b) return b; else return a; } public static long max(long a,long b){ if(a>b) return a; else return b; } public static long min(long a,long b){ if(a>b) return b; else return a; } public static long pow(long n,long p,long m){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; if(result>=m) result%=m; p >>=1; n*=n; if(n>=m) n%=m; } return result; } public static long pow(long n,long p){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } public static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } void run() throws Exception { is = System.in; out = new PrintWriter(System.out); solve(); out.flush(); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { public void run() { try { new code5().run(); } catch (Exception e) { e.printStackTrace(); } } }, "1", 1 << 26).start(); //new code5().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private long[] nl(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nl(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.*; import java.text.*; import java.io.*; import java.math.*; public class code5 { InputStream is; PrintWriter out; static long mod=pow(10,9)+7; static int dx[]={0,0,1,-1},dy[]={1,-1,0,0}; String arr[]; long dp[][]; void solve() throws IOException { int n=ni(); int a[]=na(n); int q=ni(); boolean flag=false; for(int i=0;i<n;i++) { for(int j=0;j<i;j++) { if(a[j]>a[i]) flag^=true; } } while(q--!=0) { int l=ni()-1; int r=ni()-1; int num=(r-l+1); int tot=num*(num-1)/2; if(tot%2==1) flag^=true; if(flag) out.println("odd"); else out.println("even"); } } int sum(int i) { int sum=0; while(i!=0) { if((i%10)%2==1) sum+=i%10; i/=10; } return sum; } ArrayList<Integer> al[]; void take(int n,int m) { al=new ArrayList[n]; for(int i=0;i<n;i++) al[i]=new ArrayList<Integer>(); for(int i=0;i<m;i++) { int x=ni()-1; int y=ni()-1; al[x].add(y); al[y].add(x); } } int check(long n) { int count=0; while(n!=0) { if(n%10!=0) break; n/=10; count++; } return count; } public static int count(int x) { int num=0; while(x!=0) { x=x&(x-1); num++; } return num; } static long d, x, y; void extendedEuclid(long A, long B) { if(B == 0) { d = A; x = 1; y = 0; } else { extendedEuclid(B, A%B); long temp = x; x = y; y = temp - (A/B)*y; } } long modInverse(long A,long M) //A and M are coprime { extendedEuclid(A,M); return (x%M+M)%M; //x may be negative } public static void mergeSort(int[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(int arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); int left[] = new int[n1]; int right[] = new int[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } public static void mergeSort(long[] arr, int l ,int r){ if((r-l)>=1){ int mid = (l+r)/2; mergeSort(arr,l,mid); mergeSort(arr,mid+1,r); merge(arr,l,r,mid); } } public static void merge(long arr[], int l, int r, int mid){ int n1 = (mid-l+1), n2 = (r-mid); long left[] = new long[n1]; long right[] = new long[n2]; for(int i =0 ;i<n1;i++) left[i] = arr[l+i]; for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i]; int i =0, j =0, k = l; while(i<n1 && j<n2){ if(left[i]>right[j]){ arr[k++] = right[j++]; } else{ arr[k++] = left[i++]; } } while(i<n1) arr[k++] = left[i++]; while(j<n2) arr[k++] = right[j++]; } static class Pair implements Comparable<Pair>{ int x,y,k; int i,dir; long val; Pair (int x,int y){ this.x=x; this.y=y; } public int compareTo(Pair o) { if(this.x!=o.x) return this.x-o.x; return this.y-o.y; } public String toString(){ return x+" "+y+" "+k+" "+i;} public boolean equals(Object o) { if (o instanceof Pair) { Pair p = (Pair)o; return p.x == x && p.y == y; } return false; } public int hashCode() { return new Long(x).hashCode()*31 + new Long(y).hashCode() ; } } public static boolean isPal(String s){ for(int i=0, j=s.length()-1;i<=j;i++,j--){ if(s.charAt(i)!=s.charAt(j)) return false; } return true; } public static String rev(String s){ StringBuilder sb=new StringBuilder(s); sb.reverse(); return sb.toString(); } public static long gcd(long x,long y){ if(y==0) return x; else return gcd(y,x%y); } public static int gcd(int x,int y){ if(y==0) return x; return gcd(y,x%y); } public static long gcdExtended(long a,long b,long[] x){ if(a==0){ x[0]=0; x[1]=1; return b; } long[] y=new long[2]; long gcd=gcdExtended(b%a, a, y); x[0]=y[1]-(b/a)*y[0]; x[1]=y[0]; return gcd; } public static int abs(int a,int b){ return (int)Math.abs(a-b); } public static long abs(long a,long b){ return (long)Math.abs(a-b); } public static int max(int a,int b){ if(a>b) return a; else return b; } public static int min(int a,int b){ if(a>b) return b; else return a; } public static long max(long a,long b){ if(a>b) return a; else return b; } public static long min(long a,long b){ if(a>b) return b; else return a; } public static long pow(long n,long p,long m){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; if(result>=m) result%=m; p >>=1; n*=n; if(n>=m) n%=m; } return result; } public static long pow(long n,long p){ long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; p >>=1; n*=n; } return result; } public static void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } void run() throws Exception { is = System.in; out = new PrintWriter(System.out); solve(); out.flush(); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { public void run() { try { new code5().run(); } catch (Exception e) { e.printStackTrace(); } } }, "1", 1 << 26).start(); //new code5().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private long[] nl(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nl(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
3,181
2,332
430
import java.io.*; import java.util.*; public class Main { private static void solve(InputReader in, OutputWriter out) { int n = in.nextInt(); List<List<Integer>> g = new ArrayList<>(n + 1); for (int i = 0; i < n + 1; i++) { g.add(new LinkedList<>()); } int degree1 = 0, degree2 = 0, root = 0; for (int i = 0; i < n - 1; i++) { int a = in.nextInt(); int b = in.nextInt(); g.get(a).add(b); g.get(b).add(a); if (g.get(a).size() > degree1) { if (a == root) { degree1 = g.get(a).size(); } else { degree2 = degree1; degree1 = g.get(a).size(); root = a; } } else if (g.get(a).size() > degree2) { degree2 = g.get(a).size(); } if (g.get(b).size() > degree1) { if (b == root) { degree1 = g.get(b).size(); } else { degree2 = degree1; degree1 = g.get(b).size(); root = b; } } else if (g.get(b).size() > degree2) { degree2 = g.get(b).size(); } } if (degree2 > 2) { out.print("No"); } else { out.println("Yes"); List<Integer> leaves = new LinkedList<>(); for (int i = 1; i <= n; i++) { if (i != root) { if (g.get(i).size() == 1) { leaves.add(i); } } } out.println(leaves.size()); for (int i : leaves) { out.println(root + " " + i); } } } private static void shuffleArray(int[] array) { int index; Random random = new Random(); for (int i = array.length - 1; i > 0; i--) { index = random.nextInt(i + 1); if (index != i) { array[index] ^= array[i]; array[i] ^= array[index]; array[index] ^= array[i]; } } } public static void main(String[] args) { InputReader in = new InputReader(System.in); OutputWriter out = new OutputWriter(System.out); solve(in, out); in.close(); out.close(); } private static class InputReader { private BufferedReader br; private StringTokenizer st; InputReader(InputStream is) { br = new BufferedReader(new InputStreamReader(is)); st = null; } String nextLine() { String line = null; try { line = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return line; } String next() { while (st == null || !st.hasMoreTokens()) { String line = nextLine(); if (line == null) return null; st = new StringTokenizer(line); } return st.nextToken(); } byte nextByte() { return Byte.parseByte(next()); } short nextShort() { return Short.parseShort(next()); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } void close() { try { br.close(); } catch (IOException e) { e.printStackTrace(); } } } private static class OutputWriter { BufferedWriter bw; OutputWriter(OutputStream os) { bw = new BufferedWriter(new OutputStreamWriter(os)); } void print(int i) { print(Integer.toString(i)); } void println(int i) { println(Integer.toString(i)); } void print(long l) { print(Long.toString(l)); } void println(long l) { println(Long.toString(l)); } void print(double d) { print(Double.toString(d)); } void println(double d) { println(Double.toString(d)); } void print(boolean b) { print(Boolean.toString(b)); } void println(boolean b) { println(Boolean.toString(b)); } void print(char c) { try { bw.write(c); } catch (IOException e) { e.printStackTrace(); } } void println(char c) { println(Character.toString(c)); } void print(String s) { try { bw.write(s); } catch (IOException e) { e.printStackTrace(); } } void println(String s) { print(s); print('\n'); } void close() { try { bw.close(); } catch (IOException e) { e.printStackTrace(); } } } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { private static void solve(InputReader in, OutputWriter out) { int n = in.nextInt(); List<List<Integer>> g = new ArrayList<>(n + 1); for (int i = 0; i < n + 1; i++) { g.add(new LinkedList<>()); } int degree1 = 0, degree2 = 0, root = 0; for (int i = 0; i < n - 1; i++) { int a = in.nextInt(); int b = in.nextInt(); g.get(a).add(b); g.get(b).add(a); if (g.get(a).size() > degree1) { if (a == root) { degree1 = g.get(a).size(); } else { degree2 = degree1; degree1 = g.get(a).size(); root = a; } } else if (g.get(a).size() > degree2) { degree2 = g.get(a).size(); } if (g.get(b).size() > degree1) { if (b == root) { degree1 = g.get(b).size(); } else { degree2 = degree1; degree1 = g.get(b).size(); root = b; } } else if (g.get(b).size() > degree2) { degree2 = g.get(b).size(); } } if (degree2 > 2) { out.print("No"); } else { out.println("Yes"); List<Integer> leaves = new LinkedList<>(); for (int i = 1; i <= n; i++) { if (i != root) { if (g.get(i).size() == 1) { leaves.add(i); } } } out.println(leaves.size()); for (int i : leaves) { out.println(root + " " + i); } } } private static void shuffleArray(int[] array) { int index; Random random = new Random(); for (int i = array.length - 1; i > 0; i--) { index = random.nextInt(i + 1); if (index != i) { array[index] ^= array[i]; array[i] ^= array[index]; array[index] ^= array[i]; } } } public static void main(String[] args) { InputReader in = new InputReader(System.in); OutputWriter out = new OutputWriter(System.out); solve(in, out); in.close(); out.close(); } private static class InputReader { private BufferedReader br; private StringTokenizer st; InputReader(InputStream is) { br = new BufferedReader(new InputStreamReader(is)); st = null; } String nextLine() { String line = null; try { line = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return line; } String next() { while (st == null || !st.hasMoreTokens()) { String line = nextLine(); if (line == null) return null; st = new StringTokenizer(line); } return st.nextToken(); } byte nextByte() { return Byte.parseByte(next()); } short nextShort() { return Short.parseShort(next()); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } void close() { try { br.close(); } catch (IOException e) { e.printStackTrace(); } } } private static class OutputWriter { BufferedWriter bw; OutputWriter(OutputStream os) { bw = new BufferedWriter(new OutputStreamWriter(os)); } void print(int i) { print(Integer.toString(i)); } void println(int i) { println(Integer.toString(i)); } void print(long l) { print(Long.toString(l)); } void println(long l) { println(Long.toString(l)); } void print(double d) { print(Double.toString(d)); } void println(double d) { println(Double.toString(d)); } void print(boolean b) { print(Boolean.toString(b)); } void println(boolean b) { println(Boolean.toString(b)); } void print(char c) { try { bw.write(c); } catch (IOException e) { e.printStackTrace(); } } void println(char c) { println(Character.toString(c)); } void print(String s) { try { bw.write(s); } catch (IOException e) { e.printStackTrace(); } } void println(String s) { print(s); print('\n'); } void close() { try { bw.close(); } catch (IOException e) { e.printStackTrace(); } } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { private static void solve(InputReader in, OutputWriter out) { int n = in.nextInt(); List<List<Integer>> g = new ArrayList<>(n + 1); for (int i = 0; i < n + 1; i++) { g.add(new LinkedList<>()); } int degree1 = 0, degree2 = 0, root = 0; for (int i = 0; i < n - 1; i++) { int a = in.nextInt(); int b = in.nextInt(); g.get(a).add(b); g.get(b).add(a); if (g.get(a).size() > degree1) { if (a == root) { degree1 = g.get(a).size(); } else { degree2 = degree1; degree1 = g.get(a).size(); root = a; } } else if (g.get(a).size() > degree2) { degree2 = g.get(a).size(); } if (g.get(b).size() > degree1) { if (b == root) { degree1 = g.get(b).size(); } else { degree2 = degree1; degree1 = g.get(b).size(); root = b; } } else if (g.get(b).size() > degree2) { degree2 = g.get(b).size(); } } if (degree2 > 2) { out.print("No"); } else { out.println("Yes"); List<Integer> leaves = new LinkedList<>(); for (int i = 1; i <= n; i++) { if (i != root) { if (g.get(i).size() == 1) { leaves.add(i); } } } out.println(leaves.size()); for (int i : leaves) { out.println(root + " " + i); } } } private static void shuffleArray(int[] array) { int index; Random random = new Random(); for (int i = array.length - 1; i > 0; i--) { index = random.nextInt(i + 1); if (index != i) { array[index] ^= array[i]; array[i] ^= array[index]; array[index] ^= array[i]; } } } public static void main(String[] args) { InputReader in = new InputReader(System.in); OutputWriter out = new OutputWriter(System.out); solve(in, out); in.close(); out.close(); } private static class InputReader { private BufferedReader br; private StringTokenizer st; InputReader(InputStream is) { br = new BufferedReader(new InputStreamReader(is)); st = null; } String nextLine() { String line = null; try { line = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return line; } String next() { while (st == null || !st.hasMoreTokens()) { String line = nextLine(); if (line == null) return null; st = new StringTokenizer(line); } return st.nextToken(); } byte nextByte() { return Byte.parseByte(next()); } short nextShort() { return Short.parseShort(next()); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } void close() { try { br.close(); } catch (IOException e) { e.printStackTrace(); } } } private static class OutputWriter { BufferedWriter bw; OutputWriter(OutputStream os) { bw = new BufferedWriter(new OutputStreamWriter(os)); } void print(int i) { print(Integer.toString(i)); } void println(int i) { println(Integer.toString(i)); } void print(long l) { print(Long.toString(l)); } void println(long l) { println(Long.toString(l)); } void print(double d) { print(Double.toString(d)); } void println(double d) { println(Double.toString(d)); } void print(boolean b) { print(Boolean.toString(b)); } void println(boolean b) { println(Boolean.toString(b)); } void print(char c) { try { bw.write(c); } catch (IOException e) { e.printStackTrace(); } } void println(char c) { println(Character.toString(c)); } void print(String s) { try { bw.write(s); } catch (IOException e) { e.printStackTrace(); } } void println(String s) { print(s); print('\n'); } void close() { try { bw.close(); } catch (IOException e) { e.printStackTrace(); } } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,438
429
3,200
import java.util.Scanner; public class A { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String ns = sc.next(); sc.close(); int n1 = Integer.parseInt(ns); int n2 = Integer.parseInt(ns.substring(0, ns.length() - 1)); int n3 = Integer.parseInt(ns.substring(0, ns.length() - 2) + ns.substring(ns.length() - 1)); int max = n1; max = (n2 > max) ? (n2) : (max); max = (n3 > max) ? (n3) : (max); System.out.println(max); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Scanner; public class A { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String ns = sc.next(); sc.close(); int n1 = Integer.parseInt(ns); int n2 = Integer.parseInt(ns.substring(0, ns.length() - 1)); int n3 = Integer.parseInt(ns.substring(0, ns.length() - 2) + ns.substring(ns.length() - 1)); int max = n1; max = (n2 > max) ? (n2) : (max); max = (n3 > max) ? (n3) : (max); System.out.println(max); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Scanner; public class A { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String ns = sc.next(); sc.close(); int n1 = Integer.parseInt(ns); int n2 = Integer.parseInt(ns.substring(0, ns.length() - 1)); int n3 = Integer.parseInt(ns.substring(0, ns.length() - 2) + ns.substring(ns.length() - 1)); int max = n1; max = (n2 > max) ? (n2) : (max); max = (n3 > max) ? (n3) : (max); System.out.println(max); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The running time increases with the logarithm of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
483
3,194
4,168
import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ public class Ideone { static double p[][]; static double dp[]; static int n; public static int BitCount(int u) { int uCount; uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111); return ((uCount + (uCount >> 3)) & 030707070707) % 63; } public static double f(int mask) { if (dp[mask] > -0.5) return dp[mask]; dp[mask] = 0; int ones = BitCount(mask); double pairs = (((ones * (ones + 1))) >> 1); //System.out.println(pairs); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if ((mask & (1 << i)) != 0 && (mask & (1 << j)) == 0) dp[mask] += f(mask | (1 << j)) * p[i][j] / pairs; } } return dp[mask]; } public static void main(String[] args) throws NumberFormatException, IOException { BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); n = Integer.parseInt(bf.readLine()); p = new double[n][n]; for (int i = 0; i < n; i++) { StringTokenizer st = new StringTokenizer(bf.readLine()); for (int j = 0; j < n; j++) { p[i][j] = Double.parseDouble(st.nextToken()); } } dp = new double[1 << n]; Arrays.fill(dp, -1.0); dp[(1 << n) - 1] = 1.; for (int i = 0; i < n - 1; i++) { System.out.print(f(1 << i) + " "); } System.out.println(f((1 << (n - 1)))); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ public class Ideone { static double p[][]; static double dp[]; static int n; public static int BitCount(int u) { int uCount; uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111); return ((uCount + (uCount >> 3)) & 030707070707) % 63; } public static double f(int mask) { if (dp[mask] > -0.5) return dp[mask]; dp[mask] = 0; int ones = BitCount(mask); double pairs = (((ones * (ones + 1))) >> 1); //System.out.println(pairs); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if ((mask & (1 << i)) != 0 && (mask & (1 << j)) == 0) dp[mask] += f(mask | (1 << j)) * p[i][j] / pairs; } } return dp[mask]; } public static void main(String[] args) throws NumberFormatException, IOException { BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); n = Integer.parseInt(bf.readLine()); p = new double[n][n]; for (int i = 0; i < n; i++) { StringTokenizer st = new StringTokenizer(bf.readLine()); for (int j = 0; j < n; j++) { p[i][j] = Double.parseDouble(st.nextToken()); } } dp = new double[1 << n]; Arrays.fill(dp, -1.0); dp[(1 << n) - 1] = 1.; for (int i = 0; i < n - 1; i++) { System.out.print(f(1 << i) + " "); } System.out.println(f((1 << (n - 1)))); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ public class Ideone { static double p[][]; static double dp[]; static int n; public static int BitCount(int u) { int uCount; uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111); return ((uCount + (uCount >> 3)) & 030707070707) % 63; } public static double f(int mask) { if (dp[mask] > -0.5) return dp[mask]; dp[mask] = 0; int ones = BitCount(mask); double pairs = (((ones * (ones + 1))) >> 1); //System.out.println(pairs); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if ((mask & (1 << i)) != 0 && (mask & (1 << j)) == 0) dp[mask] += f(mask | (1 << j)) * p[i][j] / pairs; } } return dp[mask]; } public static void main(String[] args) throws NumberFormatException, IOException { BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); n = Integer.parseInt(bf.readLine()); p = new double[n][n]; for (int i = 0; i < n; i++) { StringTokenizer st = new StringTokenizer(bf.readLine()); for (int j = 0; j < n; j++) { p[i][j] = Double.parseDouble(st.nextToken()); } } dp = new double[1 << n]; Arrays.fill(dp, -1.0); dp[(1 << n) - 1] = 1.; for (int i = 0; i < n - 1; i++) { System.out.print(f(1 << i) + " "); } System.out.println(f((1 << (n - 1)))); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
845
4,157
1,581
import java.awt.geom.*; import java.io.*; import java.math.*; import java.util.*; import java.util.regex.*; import static java.lang.Math.*; public class A { public A() throws Exception { int n = in.nextInt(); int[] arr = new int[n]; for (int i=0; i<n; i++) arr[i] = in.nextInt(); int[] arr2 = arr.clone(); Arrays.sort(arr2); int diff = 0; for (int i=0; i<n; i++) { if (arr2[i]!=arr[i]) diff++; } if (diff<=2) System.out.println("YES"); else System.out.println("NO"); } Scanner in = new Scanner(System.in); StringBuilder buf = new StringBuilder(); public static void main(String[] args) throws Exception { // {{{ new A(); } // }}} public static void debug(Object... arr) { // {{{ System.err.println(Arrays.deepToString(arr)); } // }}} public static class Scanner { // {{{ BufferedReader br; String line; StringTokenizer st; public Scanner(InputStream in) { br = new BufferedReader(new InputStreamReader(in)); } public boolean hasNext() throws IOException { while ((st==null||!st.hasMoreTokens())&&(line=br.readLine())!=null) st = new StringTokenizer(line); return st.hasMoreTokens(); } public String next() throws IOException { if (hasNext()) return st.nextToken(); throw new NoSuchElementException(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public double nextDouble() throws IOException { return Double.parseDouble(next()); } } // }}} }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.awt.geom.*; import java.io.*; import java.math.*; import java.util.*; import java.util.regex.*; import static java.lang.Math.*; public class A { public A() throws Exception { int n = in.nextInt(); int[] arr = new int[n]; for (int i=0; i<n; i++) arr[i] = in.nextInt(); int[] arr2 = arr.clone(); Arrays.sort(arr2); int diff = 0; for (int i=0; i<n; i++) { if (arr2[i]!=arr[i]) diff++; } if (diff<=2) System.out.println("YES"); else System.out.println("NO"); } Scanner in = new Scanner(System.in); StringBuilder buf = new StringBuilder(); public static void main(String[] args) throws Exception { // {{{ new A(); } // }}} public static void debug(Object... arr) { // {{{ System.err.println(Arrays.deepToString(arr)); } // }}} public static class Scanner { // {{{ BufferedReader br; String line; StringTokenizer st; public Scanner(InputStream in) { br = new BufferedReader(new InputStreamReader(in)); } public boolean hasNext() throws IOException { while ((st==null||!st.hasMoreTokens())&&(line=br.readLine())!=null) st = new StringTokenizer(line); return st.hasMoreTokens(); } public String next() throws IOException { if (hasNext()) return st.nextToken(); throw new NoSuchElementException(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public double nextDouble() throws IOException { return Double.parseDouble(next()); } } // }}} } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.awt.geom.*; import java.io.*; import java.math.*; import java.util.*; import java.util.regex.*; import static java.lang.Math.*; public class A { public A() throws Exception { int n = in.nextInt(); int[] arr = new int[n]; for (int i=0; i<n; i++) arr[i] = in.nextInt(); int[] arr2 = arr.clone(); Arrays.sort(arr2); int diff = 0; for (int i=0; i<n; i++) { if (arr2[i]!=arr[i]) diff++; } if (diff<=2) System.out.println("YES"); else System.out.println("NO"); } Scanner in = new Scanner(System.in); StringBuilder buf = new StringBuilder(); public static void main(String[] args) throws Exception { // {{{ new A(); } // }}} public static void debug(Object... arr) { // {{{ System.err.println(Arrays.deepToString(arr)); } // }}} public static class Scanner { // {{{ BufferedReader br; String line; StringTokenizer st; public Scanner(InputStream in) { br = new BufferedReader(new InputStreamReader(in)); } public boolean hasNext() throws IOException { while ((st==null||!st.hasMoreTokens())&&(line=br.readLine())!=null) st = new StringTokenizer(line); return st.hasMoreTokens(); } public String next() throws IOException { if (hasNext()) return st.nextToken(); throw new NoSuchElementException(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public double nextDouble() throws IOException { return Double.parseDouble(next()); } } // }}} } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
732
1,578
1,320
import java.math.BigInteger; import java.util.*; import java.io.*; public class Contest { public static void main(String[] args)throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] s=br.readLine().split(" "); BigInteger x = new BigInteger(s[0]); BigInteger k = new BigInteger(s[1]); BigInteger mod = new BigInteger(String.valueOf((int) (Math.pow(10, 9) + 7))); BigInteger two = new BigInteger("2"); BigInteger interm = two.modPow(k, mod); BigInteger res = interm.multiply(x).add(interm.multiply(x)).subtract(interm).add(BigInteger.ONE).mod(mod); if(x.equals(BigInteger.ZERO)) { System.out.println("0"); return; } System.out.println(res); } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.math.BigInteger; import java.util.*; import java.io.*; public class Contest { public static void main(String[] args)throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] s=br.readLine().split(" "); BigInteger x = new BigInteger(s[0]); BigInteger k = new BigInteger(s[1]); BigInteger mod = new BigInteger(String.valueOf((int) (Math.pow(10, 9) + 7))); BigInteger two = new BigInteger("2"); BigInteger interm = two.modPow(k, mod); BigInteger res = interm.multiply(x).add(interm.multiply(x)).subtract(interm).add(BigInteger.ONE).mod(mod); if(x.equals(BigInteger.ZERO)) { System.out.println("0"); return; } System.out.println(res); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.math.BigInteger; import java.util.*; import java.io.*; public class Contest { public static void main(String[] args)throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] s=br.readLine().split(" "); BigInteger x = new BigInteger(s[0]); BigInteger k = new BigInteger(s[1]); BigInteger mod = new BigInteger(String.valueOf((int) (Math.pow(10, 9) + 7))); BigInteger two = new BigInteger("2"); BigInteger interm = two.modPow(k, mod); BigInteger res = interm.multiply(x).add(interm.multiply(x)).subtract(interm).add(BigInteger.ONE).mod(mod); if(x.equals(BigInteger.ZERO)) { System.out.println("0"); return; } System.out.println(res); } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
509
1,318
351
import java.io.*; import java.util.*; public class Solution { static BufferedReader br=null; static PrintWriter pw=null; static StringTokenizer st=null; static String FILENAME=""; void nline(){ try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } int ni(){ while(st==null || !st.hasMoreTokens()) nline(); return Integer.parseInt(st.nextToken()); } long nl(){ while(st==null || !st.hasMoreTokens()) nline(); return Long.parseLong(st.nextToken()); } double nd(){ while(st==null || !st.hasMoreTokens()) nline(); return Double.parseDouble(st.nextToken()); } String ns(){ while(st==null || !st.hasMoreTokens()) nline(); return st.nextToken(); } String nstr(){ String s=""; try { s=br.readLine(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } return s; } public void solve(){ int n = ni(); String s = nstr(); int[] a = new int[n]; for (int i=0;i<n;i++) a[i]=s.charAt(i)=='H'?0:1; int hc = 0; for (int i=0;i<n;i++) if (a[i]==0) hc++; int min = 1000000; for (int ss=0;ss<n;ss++) { int rc = 0; for (int i=0;i<hc;i++) if (a[(i+ss)%n]==0) rc++; if (hc-rc<min) min=hc-rc; } pw.print(min); } public void run(){ solve(); pw.close(); } public static void main(String[] args) { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); new Solution().run(); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Solution { static BufferedReader br=null; static PrintWriter pw=null; static StringTokenizer st=null; static String FILENAME=""; void nline(){ try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } int ni(){ while(st==null || !st.hasMoreTokens()) nline(); return Integer.parseInt(st.nextToken()); } long nl(){ while(st==null || !st.hasMoreTokens()) nline(); return Long.parseLong(st.nextToken()); } double nd(){ while(st==null || !st.hasMoreTokens()) nline(); return Double.parseDouble(st.nextToken()); } String ns(){ while(st==null || !st.hasMoreTokens()) nline(); return st.nextToken(); } String nstr(){ String s=""; try { s=br.readLine(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } return s; } public void solve(){ int n = ni(); String s = nstr(); int[] a = new int[n]; for (int i=0;i<n;i++) a[i]=s.charAt(i)=='H'?0:1; int hc = 0; for (int i=0;i<n;i++) if (a[i]==0) hc++; int min = 1000000; for (int ss=0;ss<n;ss++) { int rc = 0; for (int i=0;i<hc;i++) if (a[(i+ss)%n]==0) rc++; if (hc-rc<min) min=hc-rc; } pw.print(min); } public void run(){ solve(); pw.close(); } public static void main(String[] args) { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); new Solution().run(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Solution { static BufferedReader br=null; static PrintWriter pw=null; static StringTokenizer st=null; static String FILENAME=""; void nline(){ try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } int ni(){ while(st==null || !st.hasMoreTokens()) nline(); return Integer.parseInt(st.nextToken()); } long nl(){ while(st==null || !st.hasMoreTokens()) nline(); return Long.parseLong(st.nextToken()); } double nd(){ while(st==null || !st.hasMoreTokens()) nline(); return Double.parseDouble(st.nextToken()); } String ns(){ while(st==null || !st.hasMoreTokens()) nline(); return st.nextToken(); } String nstr(){ String s=""; try { s=br.readLine(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } return s; } public void solve(){ int n = ni(); String s = nstr(); int[] a = new int[n]; for (int i=0;i<n;i++) a[i]=s.charAt(i)=='H'?0:1; int hc = 0; for (int i=0;i<n;i++) if (a[i]==0) hc++; int min = 1000000; for (int ss=0;ss<n;ss++) { int rc = 0; for (int i=0;i<hc;i++) if (a[(i+ss)%n]==0) rc++; if (hc-rc<min) min=hc-rc; } pw.print(min); } public void run(){ solve(); pw.close(); } public static void main(String[] args) { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(System.out); new Solution().run(); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
777
350
3,262
import java.io.PrintWriter; import java.util.Scanner; public class HexadecimalTheorem { /** * @param args */ public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); PrintWriter out = new PrintWriter(System.out); out.printf("%d %d %d%n", 0, 0, n); out.flush(); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.PrintWriter; import java.util.Scanner; public class HexadecimalTheorem { /** * @param args */ public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); PrintWriter out = new PrintWriter(System.out); out.printf("%d %d %d%n", 0, 0, n); out.flush(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.PrintWriter; import java.util.Scanner; public class HexadecimalTheorem { /** * @param args */ public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); PrintWriter out = new PrintWriter(System.out); out.printf("%d %d %d%n", 0, 0, n); out.flush(); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
439
3,256
2,292
//package round455; import java.io.ByteArrayInputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; public class C { InputStream is; PrintWriter out; String INPUT = ""; void solve() { int n = ni(); int mod = 1000000007; long[] dp = new long[5005]; dp[0] = 1; for(int i = 0;i < n;i++){ char c = nc(); if(c == 's'){ if(i < n-1){ for(int j = 5003;j >= 0;j--){ dp[j] += dp[j+1]; if(dp[j] >= mod)dp[j] -= mod; } } }else{ for(int j = 5003;j >= 0;j--){ dp[j+1] = dp[j]; } dp[0] = 0; } } long ans = 0; for(int i = 0;i < 5005;i++)ans += dp[i]; out.println(ans % mod); } void run() throws Exception { is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes()); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new C().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> //package round455; import java.io.ByteArrayInputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; public class C { InputStream is; PrintWriter out; String INPUT = ""; void solve() { int n = ni(); int mod = 1000000007; long[] dp = new long[5005]; dp[0] = 1; for(int i = 0;i < n;i++){ char c = nc(); if(c == 's'){ if(i < n-1){ for(int j = 5003;j >= 0;j--){ dp[j] += dp[j+1]; if(dp[j] >= mod)dp[j] -= mod; } } }else{ for(int j = 5003;j >= 0;j--){ dp[j+1] = dp[j]; } dp[0] = 0; } } long ans = 0; for(int i = 0;i < 5005;i++)ans += dp[i]; out.println(ans % mod); } void run() throws Exception { is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes()); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new C().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package round455; import java.io.ByteArrayInputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; public class C { InputStream is; PrintWriter out; String INPUT = ""; void solve() { int n = ni(); int mod = 1000000007; long[] dp = new long[5005]; dp[0] = 1; for(int i = 0;i < n;i++){ char c = nc(); if(c == 's'){ if(i < n-1){ for(int j = 5003;j >= 0;j--){ dp[j] += dp[j+1]; if(dp[j] >= mod)dp[j] -= mod; } } }else{ for(int j = 5003;j >= 0;j--){ dp[j+1] = dp[j]; } dp[0] = 0; } } long ans = 0; for(int i = 0;i < 5005;i++)ans += dp[i]; out.println(ans % mod); } void run() throws Exception { is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes()); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new C().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,431
2,287
4,066
import java.awt.Point; import java.util.Arrays; import java.util.Scanner; public class C { static int[] DP; static Point[] Next; static int[][] pair; static int[] single; static int n; public static int get(int mask) { if (mask + 1 == (1 << n)) return 0; if (DP[mask] != -1) return DP[mask]; int x = 0; for (;; x++) if ((mask & (1 << x)) == 0) break; int min = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { if ((mask & (1 << i)) != 0 || i == x) continue; int temp = pair[x][i] + get(mask | (1 << i) | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, i); } } int temp = single[x] + get(mask | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, -1); } return DP[mask] = min; } public static void main(String[] args) { Scanner in = new Scanner(System.in); Point start = new Point(in.nextInt(), in.nextInt()); n = in.nextInt(); Point[] A = new Point[n]; for (int i = 0; i < n; i++) A[i] = new Point(in.nextInt(), in.nextInt()); DP = new int[1 << n]; Next = new Point[1 << n]; Arrays.fill(DP, -1); pair = new int[n][n]; single = new int[n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { int dx1 = A[i].x - start.x; int dy1 = A[i].y - start.y; int dx2 = A[j].x - A[i].x; int dy2 = A[j].y - A[i].y; int dx3 = A[j].x - start.x; int dy3 = A[j].y - start.y; pair[i][j] = dx1 * dx1 + dy1 * dy1 + dx2 * dx2 + dy2 * dy2 + dx3 * dx3 + dy3 * dy3; single[i] = 2 * (dx1 * dx1 + dy1 * dy1); } int ans = get(0); System.out.println(ans); int mask = 0; while (mask + 1 != (1 << n)) { Point temp = Next[mask]; if (temp.y == -1) System.out.print("0 " + (temp.x + 1) + " "); else { System.out .print("0 " + (temp.x + 1) + " " + (temp.y + 1) + " "); mask |= (1 << temp.y); } mask |= (1 << temp.x); } System.out.println("0"); } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.awt.Point; import java.util.Arrays; import java.util.Scanner; public class C { static int[] DP; static Point[] Next; static int[][] pair; static int[] single; static int n; public static int get(int mask) { if (mask + 1 == (1 << n)) return 0; if (DP[mask] != -1) return DP[mask]; int x = 0; for (;; x++) if ((mask & (1 << x)) == 0) break; int min = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { if ((mask & (1 << i)) != 0 || i == x) continue; int temp = pair[x][i] + get(mask | (1 << i) | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, i); } } int temp = single[x] + get(mask | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, -1); } return DP[mask] = min; } public static void main(String[] args) { Scanner in = new Scanner(System.in); Point start = new Point(in.nextInt(), in.nextInt()); n = in.nextInt(); Point[] A = new Point[n]; for (int i = 0; i < n; i++) A[i] = new Point(in.nextInt(), in.nextInt()); DP = new int[1 << n]; Next = new Point[1 << n]; Arrays.fill(DP, -1); pair = new int[n][n]; single = new int[n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { int dx1 = A[i].x - start.x; int dy1 = A[i].y - start.y; int dx2 = A[j].x - A[i].x; int dy2 = A[j].y - A[i].y; int dx3 = A[j].x - start.x; int dy3 = A[j].y - start.y; pair[i][j] = dx1 * dx1 + dy1 * dy1 + dx2 * dx2 + dy2 * dy2 + dx3 * dx3 + dy3 * dy3; single[i] = 2 * (dx1 * dx1 + dy1 * dy1); } int ans = get(0); System.out.println(ans); int mask = 0; while (mask + 1 != (1 << n)) { Point temp = Next[mask]; if (temp.y == -1) System.out.print("0 " + (temp.x + 1) + " "); else { System.out .print("0 " + (temp.x + 1) + " " + (temp.y + 1) + " "); mask |= (1 << temp.y); } mask |= (1 << temp.x); } System.out.println("0"); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.awt.Point; import java.util.Arrays; import java.util.Scanner; public class C { static int[] DP; static Point[] Next; static int[][] pair; static int[] single; static int n; public static int get(int mask) { if (mask + 1 == (1 << n)) return 0; if (DP[mask] != -1) return DP[mask]; int x = 0; for (;; x++) if ((mask & (1 << x)) == 0) break; int min = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { if ((mask & (1 << i)) != 0 || i == x) continue; int temp = pair[x][i] + get(mask | (1 << i) | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, i); } } int temp = single[x] + get(mask | (1 << x)); if (temp < min) { min = temp; Next[mask] = new Point(x, -1); } return DP[mask] = min; } public static void main(String[] args) { Scanner in = new Scanner(System.in); Point start = new Point(in.nextInt(), in.nextInt()); n = in.nextInt(); Point[] A = new Point[n]; for (int i = 0; i < n; i++) A[i] = new Point(in.nextInt(), in.nextInt()); DP = new int[1 << n]; Next = new Point[1 << n]; Arrays.fill(DP, -1); pair = new int[n][n]; single = new int[n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { int dx1 = A[i].x - start.x; int dy1 = A[i].y - start.y; int dx2 = A[j].x - A[i].x; int dy2 = A[j].y - A[i].y; int dx3 = A[j].x - start.x; int dy3 = A[j].y - start.y; pair[i][j] = dx1 * dx1 + dy1 * dy1 + dx2 * dx2 + dy2 * dy2 + dx3 * dx3 + dy3 * dy3; single[i] = 2 * (dx1 * dx1 + dy1 * dy1); } int ans = get(0); System.out.println(ans); int mask = 0; while (mask + 1 != (1 << n)) { Point temp = Next[mask]; if (temp.y == -1) System.out.print("0 " + (temp.x + 1) + " "); else { System.out .print("0 " + (temp.x + 1) + " " + (temp.y + 1) + " "); mask |= (1 << temp.y); } mask |= (1 << temp.x); } System.out.println("0"); } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,037
4,055
2,548
import java.io.BufferedOutputStream; import java.io.PrintWriter; import java.util.*; public class E1180D { public static void main(String args[]) { Scanner sc = new Scanner(System.in); PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out)); int n = sc.nextInt(); int m = sc.nextInt(); // Move from both ends, Time limit exceeded on test 6 for (int i= 1; i<= m/2; i++) { // String s = ""; int i2 = m -i + 1; // the other end of i // i is left row, i2 is right row for (int j = 1; j <= n ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i); pw.println(j2+ " " + i2); // s += j + " " + i + "\n" + j2+ " " + i2 + "\n"; } // out.print(s); } // if n is odd, there is one line in the middle if (m % 2 == 1) { int i2 = m /2 + 1; // this is the middle column for (int j = 1; j <= n/2 ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i2); pw.println(j2+ " " + i2); } if (n %2 == 1) { int j = n /2 + 1; pw.println(j + " " + i2); } } pw.flush(); pw.close(); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedOutputStream; import java.io.PrintWriter; import java.util.*; public class E1180D { public static void main(String args[]) { Scanner sc = new Scanner(System.in); PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out)); int n = sc.nextInt(); int m = sc.nextInt(); // Move from both ends, Time limit exceeded on test 6 for (int i= 1; i<= m/2; i++) { // String s = ""; int i2 = m -i + 1; // the other end of i // i is left row, i2 is right row for (int j = 1; j <= n ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i); pw.println(j2+ " " + i2); // s += j + " " + i + "\n" + j2+ " " + i2 + "\n"; } // out.print(s); } // if n is odd, there is one line in the middle if (m % 2 == 1) { int i2 = m /2 + 1; // this is the middle column for (int j = 1; j <= n/2 ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i2); pw.println(j2+ " " + i2); } if (n %2 == 1) { int j = n /2 + 1; pw.println(j + " " + i2); } } pw.flush(); pw.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedOutputStream; import java.io.PrintWriter; import java.util.*; public class E1180D { public static void main(String args[]) { Scanner sc = new Scanner(System.in); PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out)); int n = sc.nextInt(); int m = sc.nextInt(); // Move from both ends, Time limit exceeded on test 6 for (int i= 1; i<= m/2; i++) { // String s = ""; int i2 = m -i + 1; // the other end of i // i is left row, i2 is right row for (int j = 1; j <= n ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i); pw.println(j2+ " " + i2); // s += j + " " + i + "\n" + j2+ " " + i2 + "\n"; } // out.print(s); } // if n is odd, there is one line in the middle if (m % 2 == 1) { int i2 = m /2 + 1; // this is the middle column for (int j = 1; j <= n/2 ; j++) { int j2 = n - j + 1; // start with (i,j), then go thru all the cell with (,i) and (,i2) pw.println(j + " " + i2); pw.println(j2+ " " + i2); } if (n %2 == 1) { int j = n /2 + 1; pw.println(j + " " + i2); } } pw.flush(); pw.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
778
2,542
900
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class B { public static void main(String[] args) throws IOException { InputReader in = new InputReader(); int n = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); if (k > n) { System.out.println(-1 + " " + -1); return; } int[] v = new int[100010]; int cnt = 0; for (int i = 0; i < k; i++) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; } int i = k; while (cnt < k && i < n) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; i++; } if (cnt != k) { System.out.println(-1 + " " + -1); } else { int st = 0; while (st < n && st < i && v[a[st]] > 1) { v[a[st]]--; st++; } System.out.println((st+1) + " " + (i)); } } static class InputReader { BufferedReader in; StringTokenizer st; public InputReader() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); st = new StringTokenizer(in.readLine()); } public String next() throws IOException { while (!st.hasMoreElements()) st = new StringTokenizer(in.readLine()); return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class B { public static void main(String[] args) throws IOException { InputReader in = new InputReader(); int n = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); if (k > n) { System.out.println(-1 + " " + -1); return; } int[] v = new int[100010]; int cnt = 0; for (int i = 0; i < k; i++) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; } int i = k; while (cnt < k && i < n) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; i++; } if (cnt != k) { System.out.println(-1 + " " + -1); } else { int st = 0; while (st < n && st < i && v[a[st]] > 1) { v[a[st]]--; st++; } System.out.println((st+1) + " " + (i)); } } static class InputReader { BufferedReader in; StringTokenizer st; public InputReader() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); st = new StringTokenizer(in.readLine()); } public String next() throws IOException { while (!st.hasMoreElements()) st = new StringTokenizer(in.readLine()); return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; public class B { public static void main(String[] args) throws IOException { InputReader in = new InputReader(); int n = in.nextInt(); int k = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = in.nextInt(); if (k > n) { System.out.println(-1 + " " + -1); return; } int[] v = new int[100010]; int cnt = 0; for (int i = 0; i < k; i++) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; } int i = k; while (cnt < k && i < n) { if (v[a[i]] == 0) { cnt++; } v[a[i]]++; i++; } if (cnt != k) { System.out.println(-1 + " " + -1); } else { int st = 0; while (st < n && st < i && v[a[st]] > 1) { v[a[st]]--; st++; } System.out.println((st+1) + " " + (i)); } } static class InputReader { BufferedReader in; StringTokenizer st; public InputReader() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); st = new StringTokenizer(in.readLine()); } public String next() throws IOException { while (!st.hasMoreElements()) st = new StringTokenizer(in.readLine()); return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); } } } </CODE> <EVALUATION_RUBRIC> - O(1): The execution time is unaffected by the size of the input n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
759
899
2,558
import java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws Exception { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int n=Integer.parseInt(br.readLine()); String s=br.readLine(); String ss[]=s.split(" "); int arr[]=new int[n]; for(int i=0;i<n;i++) arr[i]=Integer.parseInt(ss[i]); Arrays.sort(arr); int coun=0,coun2=0; for(int i=arr[0],k=0;k<n;) { for(int j=k;j<n;j++) { if(arr[j]%i==0) { arr[j]=-1; coun2++; } } Arrays.sort(arr); k=coun2; coun++; if(coun2<n) i=arr[coun2]; else break; } System.out.println(coun); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws Exception { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int n=Integer.parseInt(br.readLine()); String s=br.readLine(); String ss[]=s.split(" "); int arr[]=new int[n]; for(int i=0;i<n;i++) arr[i]=Integer.parseInt(ss[i]); Arrays.sort(arr); int coun=0,coun2=0; for(int i=arr[0],k=0;k<n;) { for(int j=k;j<n;j++) { if(arr[j]%i==0) { arr[j]=-1; coun2++; } } Arrays.sort(arr); k=coun2; coun++; if(coun2<n) i=arr[coun2]; else break; } System.out.println(coun); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws Exception { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int n=Integer.parseInt(br.readLine()); String s=br.readLine(); String ss[]=s.split(" "); int arr[]=new int[n]; for(int i=0;i<n;i++) arr[i]=Integer.parseInt(ss[i]); Arrays.sort(arr); int coun=0,coun2=0; for(int i=arr[0],k=0;k<n;) { for(int j=k;j<n;j++) { if(arr[j]%i==0) { arr[j]=-1; coun2++; } } Arrays.sort(arr); k=coun2; coun++; if(coun2<n) i=arr[coun2]; else break; } System.out.println(coun); } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
536
2,552
421
import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; import java.util.List; import java.util.Map; import java.util.Scanner; import java.util.Set; import java.util.TreeSet; public class Main { int work(int x){ if(x%2==0)return x+1; else return x-1; } static int N = 200050; class Node implements Comparable <Node>{ int x, id; Node(int x, int id){ this.x = x; this.id = id; } public int compareTo(Node o){ return Integer.compare(x, o.x); } public String toString(){ return id + "=" + x; } } class Edge{ int from, to, nex; Edge (int from, int to, int nex){ this.from = from; this.to = to; this.nex = nex; } } Edge[] edge = new Edge[N*10]; int[] head = new int[N]; int edgenum; void addedge(int u, int v){ Edge E = new Edge(u, v, head[u]); edge[edgenum] = E; head[u] = edgenum ++; } int n; int[] p = new int[N], ans = new int[N]; int a, b, max; Map<Integer, Integer> map = new HashMap(); boolean match(int x, int y, int col){ int P = map.get(x); if(map.containsKey(y-x) == false) return false; int Q = map.get(y - x); if(ans[Q] == -1 || x * 2 == y){ ans[Q] = ans[P] = col; } else { if(match(a+b-2*y+x, y, col)) ans[Q] = ans[P] = col; else return false; } return true; } boolean solve(){ if(max >= a && max >= b)return false; for(int i = 1; i <= n; i++) if(ans[i] == -1) { if(match(p[i], a, 0)==false && match(p[i], b, 1) == false) return false; } return true; } void init(){ n = cin.nextInt(); a = cin.nextInt(); b = cin.nextInt(); max = 0; for(int i = 1; i <= n; i++){ ans[i] = -1; p[i] = cin.nextInt(); map.put(p[i], i); if(p[i] > max) max = p[i]; } } public void work(){ init(); if(solve()){ out.println("YES"); for(int i = 1; i <= n; i++)out.print(ans[i]+" "); out.println(); } else out.println("NO"); } Main() { cin = new Scanner(System.in); out = new PrintWriter(System.out); } public static void main(String[] args) { Main e = new Main(); e.work(); out.close(); } public Scanner cin; public static PrintWriter out; } /* http://blog.csdn.net/keshuai19940722/article/details/39528801 */
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; import java.util.List; import java.util.Map; import java.util.Scanner; import java.util.Set; import java.util.TreeSet; public class Main { int work(int x){ if(x%2==0)return x+1; else return x-1; } static int N = 200050; class Node implements Comparable <Node>{ int x, id; Node(int x, int id){ this.x = x; this.id = id; } public int compareTo(Node o){ return Integer.compare(x, o.x); } public String toString(){ return id + "=" + x; } } class Edge{ int from, to, nex; Edge (int from, int to, int nex){ this.from = from; this.to = to; this.nex = nex; } } Edge[] edge = new Edge[N*10]; int[] head = new int[N]; int edgenum; void addedge(int u, int v){ Edge E = new Edge(u, v, head[u]); edge[edgenum] = E; head[u] = edgenum ++; } int n; int[] p = new int[N], ans = new int[N]; int a, b, max; Map<Integer, Integer> map = new HashMap(); boolean match(int x, int y, int col){ int P = map.get(x); if(map.containsKey(y-x) == false) return false; int Q = map.get(y - x); if(ans[Q] == -1 || x * 2 == y){ ans[Q] = ans[P] = col; } else { if(match(a+b-2*y+x, y, col)) ans[Q] = ans[P] = col; else return false; } return true; } boolean solve(){ if(max >= a && max >= b)return false; for(int i = 1; i <= n; i++) if(ans[i] == -1) { if(match(p[i], a, 0)==false && match(p[i], b, 1) == false) return false; } return true; } void init(){ n = cin.nextInt(); a = cin.nextInt(); b = cin.nextInt(); max = 0; for(int i = 1; i <= n; i++){ ans[i] = -1; p[i] = cin.nextInt(); map.put(p[i], i); if(p[i] > max) max = p[i]; } } public void work(){ init(); if(solve()){ out.println("YES"); for(int i = 1; i <= n; i++)out.print(ans[i]+" "); out.println(); } else out.println("NO"); } Main() { cin = new Scanner(System.in); out = new PrintWriter(System.out); } public static void main(String[] args) { Main e = new Main(); e.work(); out.close(); } public Scanner cin; public static PrintWriter out; } /* http://blog.csdn.net/keshuai19940722/article/details/39528801 */ </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; import java.util.List; import java.util.Map; import java.util.Scanner; import java.util.Set; import java.util.TreeSet; public class Main { int work(int x){ if(x%2==0)return x+1; else return x-1; } static int N = 200050; class Node implements Comparable <Node>{ int x, id; Node(int x, int id){ this.x = x; this.id = id; } public int compareTo(Node o){ return Integer.compare(x, o.x); } public String toString(){ return id + "=" + x; } } class Edge{ int from, to, nex; Edge (int from, int to, int nex){ this.from = from; this.to = to; this.nex = nex; } } Edge[] edge = new Edge[N*10]; int[] head = new int[N]; int edgenum; void addedge(int u, int v){ Edge E = new Edge(u, v, head[u]); edge[edgenum] = E; head[u] = edgenum ++; } int n; int[] p = new int[N], ans = new int[N]; int a, b, max; Map<Integer, Integer> map = new HashMap(); boolean match(int x, int y, int col){ int P = map.get(x); if(map.containsKey(y-x) == false) return false; int Q = map.get(y - x); if(ans[Q] == -1 || x * 2 == y){ ans[Q] = ans[P] = col; } else { if(match(a+b-2*y+x, y, col)) ans[Q] = ans[P] = col; else return false; } return true; } boolean solve(){ if(max >= a && max >= b)return false; for(int i = 1; i <= n; i++) if(ans[i] == -1) { if(match(p[i], a, 0)==false && match(p[i], b, 1) == false) return false; } return true; } void init(){ n = cin.nextInt(); a = cin.nextInt(); b = cin.nextInt(); max = 0; for(int i = 1; i <= n; i++){ ans[i] = -1; p[i] = cin.nextInt(); map.put(p[i], i); if(p[i] > max) max = p[i]; } } public void work(){ init(); if(solve()){ out.println("YES"); for(int i = 1; i <= n; i++)out.print(ans[i]+" "); out.println(); } else out.println("NO"); } Main() { cin = new Scanner(System.in); out = new PrintWriter(System.out); } public static void main(String[] args) { Main e = new Main(); e.work(); out.close(); } public Scanner cin; public static PrintWriter out; } /* http://blog.csdn.net/keshuai19940722/article/details/39528801 */ </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,099
420
4,476
import java.io.*; import java.util.*; public class Sol2{ final public static int MXN = (1<<21); public static int good[][]; public static void main(String[] args) throws IOException{ FastIO sc = new FastIO(System.in); PrintWriter out = new PrintWriter(System.out); int n = sc.nextInt(); int m = sc.nextInt(); int num[][] = new int[m][m]; String str = sc.next(); for(int i=0; i<str.length()-1; i++) { int a = str.charAt(i)-'a'; int b = str.charAt(i+1)-'a'; num[a][b]++; num[b][a]++; } int lowbit[] = new int[MXN]; int dp[] = new int[MXN]; for(int i=0; i<MXN; i++) { dp[i] = Integer.MAX_VALUE; } dp[0] = 0; good = new int[MXN][m]; for(int msk = 0; msk<(1<<m); msk++) { for(int i=0; i<m; i++) { int low = Integer.numberOfTrailingZeros(Integer.lowestOneBit(msk)); if(low==32) low = 0; good[msk][i] = good[msk^(1<<low)][i] + num[i][low]; } } for(int msk = 0; msk<(1<<m); msk++) { int bits = Integer.bitCount(msk)+1; for(int i=0; i<m; i++) { if((msk&(1<<i))!=0) continue; int nxt = msk|(1<<i); dp[nxt] = Math.min(dp[nxt], dp[msk] + bits*(good[msk][i]-good[((1<<m)-1)^nxt][i])); } } out.println(dp[(1<<m)-1]); out.close(); } static void shuffle(int[] a) { Random get = new Random(); for (int i = 0; i < a.length; i++) { int r = get.nextInt(a.length); int temp = a[i]; a[i] = a[r]; a[r] = temp; } } static class FastIO { // Is your Fast I/O being bad? InputStream dis; byte[] buffer = new byte[1 << 17]; int pointer = 0; public FastIO(String fileName) throws IOException { dis = new FileInputStream(fileName); } public FastIO(InputStream is) throws IOException { dis = is; } int nextInt() throws IOException { int ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } long nextLong() throws IOException { long ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } byte nextByte() throws IOException { if (pointer == buffer.length) { dis.read(buffer, 0, buffer.length); pointer = 0; } return buffer[pointer++]; } String next() throws IOException { StringBuffer ret = new StringBuffer(); byte b; do { b = nextByte(); } while (b <= ' '); while (b > ' ') { ret.appendCodePoint(b); b = nextByte(); } return ret.toString(); } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; public class Sol2{ final public static int MXN = (1<<21); public static int good[][]; public static void main(String[] args) throws IOException{ FastIO sc = new FastIO(System.in); PrintWriter out = new PrintWriter(System.out); int n = sc.nextInt(); int m = sc.nextInt(); int num[][] = new int[m][m]; String str = sc.next(); for(int i=0; i<str.length()-1; i++) { int a = str.charAt(i)-'a'; int b = str.charAt(i+1)-'a'; num[a][b]++; num[b][a]++; } int lowbit[] = new int[MXN]; int dp[] = new int[MXN]; for(int i=0; i<MXN; i++) { dp[i] = Integer.MAX_VALUE; } dp[0] = 0; good = new int[MXN][m]; for(int msk = 0; msk<(1<<m); msk++) { for(int i=0; i<m; i++) { int low = Integer.numberOfTrailingZeros(Integer.lowestOneBit(msk)); if(low==32) low = 0; good[msk][i] = good[msk^(1<<low)][i] + num[i][low]; } } for(int msk = 0; msk<(1<<m); msk++) { int bits = Integer.bitCount(msk)+1; for(int i=0; i<m; i++) { if((msk&(1<<i))!=0) continue; int nxt = msk|(1<<i); dp[nxt] = Math.min(dp[nxt], dp[msk] + bits*(good[msk][i]-good[((1<<m)-1)^nxt][i])); } } out.println(dp[(1<<m)-1]); out.close(); } static void shuffle(int[] a) { Random get = new Random(); for (int i = 0; i < a.length; i++) { int r = get.nextInt(a.length); int temp = a[i]; a[i] = a[r]; a[r] = temp; } } static class FastIO { // Is your Fast I/O being bad? InputStream dis; byte[] buffer = new byte[1 << 17]; int pointer = 0; public FastIO(String fileName) throws IOException { dis = new FileInputStream(fileName); } public FastIO(InputStream is) throws IOException { dis = is; } int nextInt() throws IOException { int ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } long nextLong() throws IOException { long ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } byte nextByte() throws IOException { if (pointer == buffer.length) { dis.read(buffer, 0, buffer.length); pointer = 0; } return buffer[pointer++]; } String next() throws IOException { StringBuffer ret = new StringBuffer(); byte b; do { b = nextByte(); } while (b <= ' '); while (b > ' ') { ret.appendCodePoint(b); b = nextByte(); } return ret.toString(); } } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Sol2{ final public static int MXN = (1<<21); public static int good[][]; public static void main(String[] args) throws IOException{ FastIO sc = new FastIO(System.in); PrintWriter out = new PrintWriter(System.out); int n = sc.nextInt(); int m = sc.nextInt(); int num[][] = new int[m][m]; String str = sc.next(); for(int i=0; i<str.length()-1; i++) { int a = str.charAt(i)-'a'; int b = str.charAt(i+1)-'a'; num[a][b]++; num[b][a]++; } int lowbit[] = new int[MXN]; int dp[] = new int[MXN]; for(int i=0; i<MXN; i++) { dp[i] = Integer.MAX_VALUE; } dp[0] = 0; good = new int[MXN][m]; for(int msk = 0; msk<(1<<m); msk++) { for(int i=0; i<m; i++) { int low = Integer.numberOfTrailingZeros(Integer.lowestOneBit(msk)); if(low==32) low = 0; good[msk][i] = good[msk^(1<<low)][i] + num[i][low]; } } for(int msk = 0; msk<(1<<m); msk++) { int bits = Integer.bitCount(msk)+1; for(int i=0; i<m; i++) { if((msk&(1<<i))!=0) continue; int nxt = msk|(1<<i); dp[nxt] = Math.min(dp[nxt], dp[msk] + bits*(good[msk][i]-good[((1<<m)-1)^nxt][i])); } } out.println(dp[(1<<m)-1]); out.close(); } static void shuffle(int[] a) { Random get = new Random(); for (int i = 0; i < a.length; i++) { int r = get.nextInt(a.length); int temp = a[i]; a[i] = a[r]; a[r] = temp; } } static class FastIO { // Is your Fast I/O being bad? InputStream dis; byte[] buffer = new byte[1 << 17]; int pointer = 0; public FastIO(String fileName) throws IOException { dis = new FileInputStream(fileName); } public FastIO(InputStream is) throws IOException { dis = is; } int nextInt() throws IOException { int ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } long nextLong() throws IOException { long ret = 0; byte b; do { b = nextByte(); } while (b <= ' '); boolean negative = false; if (b == '-') { negative = true; b = nextByte(); } while (b >= '0' && b <= '9') { ret = 10 * ret + b - '0'; b = nextByte(); } return (negative) ? -ret : ret; } byte nextByte() throws IOException { if (pointer == buffer.length) { dis.read(buffer, 0, buffer.length); pointer = 0; } return buffer[pointer++]; } String next() throws IOException { StringBuffer ret = new StringBuffer(); byte b; do { b = nextByte(); } while (b <= ' '); while (b > ' ') { ret.appendCodePoint(b); b = nextByte(); } return ret.toString(); } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,319
4,465
1,015
import java.util.Scanner; public class DigitsSequence2 { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); long index = scanner.nextLong(); solution1(index); } static void solution1(Long index){ int i = 1; // group number long len = 9; // the max length long max = 9; // the max integer while(len < index){ long tmp = 9 * (long) Math.pow(10, i); i++; len += i * tmp; max += tmp; } long diff = len - index; // the digit number between index and len long laterCount = diff / i; // the number after index int remainder = (int) (diff % i); long current = max - laterCount; // the number of the index int k = i - 1 - remainder; System.out.println(String.valueOf(current).charAt(k)); } }
O(log(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Scanner; public class DigitsSequence2 { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); long index = scanner.nextLong(); solution1(index); } static void solution1(Long index){ int i = 1; // group number long len = 9; // the max length long max = 9; // the max integer while(len < index){ long tmp = 9 * (long) Math.pow(10, i); i++; len += i * tmp; max += tmp; } long diff = len - index; // the digit number between index and len long laterCount = diff / i; // the number after index int remainder = (int) (diff % i); long current = max - laterCount; // the number of the index int k = i - 1 - remainder; System.out.println(String.valueOf(current).charAt(k)); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class DigitsSequence2 { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); long index = scanner.nextLong(); solution1(index); } static void solution1(Long index){ int i = 1; // group number long len = 9; // the max length long max = 9; // the max integer while(len < index){ long tmp = 9 * (long) Math.pow(10, i); i++; len += i * tmp; max += tmp; } long diff = len - index; // the digit number between index and len long laterCount = diff / i; // the number after index int remainder = (int) (diff % i); long current = max - laterCount; // the number of the index int k = i - 1 - remainder; System.out.println(String.valueOf(current).charAt(k)); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
551
1,014
2,408
/* Keep solving problems. */ import java.util.*; import java.io.*; public class CFA { BufferedReader br; PrintWriter out; StringTokenizer st; boolean eof; final long MOD = 1000L * 1000L * 1000L + 7; int[] dx = {0, -1, 0, 1}; int[] dy = {1, 0, -1, 0}; void solve() throws IOException { int n = nextInt(); int[] arr = nextIntArr(n); int cnt = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] > arr[j]) { cnt++; } } } int m = nextInt(); boolean[] vis = new boolean[n]; for (int i = 0; i < m; i++) { int l = nextInt() - 1; int r = nextInt() - 1; int len = r - l + 1; if (len * (len - 1) / 2 % 2 != 0) { cnt++; } if (cnt % 2 != 0) { outln("odd"); } else { outln("even"); } } } void shuffle(int[] a) { int n = a.length; for(int i = 0; i < n; i++) { int r = i + (int) (Math.random() * (n - i)); int tmp = a[i]; a[i] = a[r]; a[r] = tmp; } } long gcd(long a, long b) { while(a != 0 && b != 0) { long c = b; b = a % b; a = c; } return a + b; } private void outln(Object o) { out.println(o); } private void out(Object o) { out.print(o); } public CFA() throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.close(); } public static void main(String[] args) throws IOException { new CFA(); } public long[] nextLongArr(int n) throws IOException{ long[] res = new long[n]; for(int i = 0; i < n; i++) res[i] = nextLong(); return res; } public int[] nextIntArr(int n) throws IOException { int[] res = new int[n]; for(int i = 0; i < n; i++) res[i] = nextInt(); return res; } public String nextToken() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (Exception e) { eof = true; return null; } } return st.nextToken(); } public String nextString() { try { return br.readLine(); } catch (IOException e) { eof = true; return null; } } public int nextInt() throws IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* Keep solving problems. */ import java.util.*; import java.io.*; public class CFA { BufferedReader br; PrintWriter out; StringTokenizer st; boolean eof; final long MOD = 1000L * 1000L * 1000L + 7; int[] dx = {0, -1, 0, 1}; int[] dy = {1, 0, -1, 0}; void solve() throws IOException { int n = nextInt(); int[] arr = nextIntArr(n); int cnt = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] > arr[j]) { cnt++; } } } int m = nextInt(); boolean[] vis = new boolean[n]; for (int i = 0; i < m; i++) { int l = nextInt() - 1; int r = nextInt() - 1; int len = r - l + 1; if (len * (len - 1) / 2 % 2 != 0) { cnt++; } if (cnt % 2 != 0) { outln("odd"); } else { outln("even"); } } } void shuffle(int[] a) { int n = a.length; for(int i = 0; i < n; i++) { int r = i + (int) (Math.random() * (n - i)); int tmp = a[i]; a[i] = a[r]; a[r] = tmp; } } long gcd(long a, long b) { while(a != 0 && b != 0) { long c = b; b = a % b; a = c; } return a + b; } private void outln(Object o) { out.println(o); } private void out(Object o) { out.print(o); } public CFA() throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.close(); } public static void main(String[] args) throws IOException { new CFA(); } public long[] nextLongArr(int n) throws IOException{ long[] res = new long[n]; for(int i = 0; i < n; i++) res[i] = nextLong(); return res; } public int[] nextIntArr(int n) throws IOException { int[] res = new int[n]; for(int i = 0; i < n; i++) res[i] = nextInt(); return res; } public String nextToken() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (Exception e) { eof = true; return null; } } return st.nextToken(); } public String nextString() { try { return br.readLine(); } catch (IOException e) { eof = true; return null; } } public int nextInt() throws IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> /* Keep solving problems. */ import java.util.*; import java.io.*; public class CFA { BufferedReader br; PrintWriter out; StringTokenizer st; boolean eof; final long MOD = 1000L * 1000L * 1000L + 7; int[] dx = {0, -1, 0, 1}; int[] dy = {1, 0, -1, 0}; void solve() throws IOException { int n = nextInt(); int[] arr = nextIntArr(n); int cnt = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] > arr[j]) { cnt++; } } } int m = nextInt(); boolean[] vis = new boolean[n]; for (int i = 0; i < m; i++) { int l = nextInt() - 1; int r = nextInt() - 1; int len = r - l + 1; if (len * (len - 1) / 2 % 2 != 0) { cnt++; } if (cnt % 2 != 0) { outln("odd"); } else { outln("even"); } } } void shuffle(int[] a) { int n = a.length; for(int i = 0; i < n; i++) { int r = i + (int) (Math.random() * (n - i)); int tmp = a[i]; a[i] = a[r]; a[r] = tmp; } } long gcd(long a, long b) { while(a != 0 && b != 0) { long c = b; b = a % b; a = c; } return a + b; } private void outln(Object o) { out.println(o); } private void out(Object o) { out.print(o); } public CFA() throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); solve(); out.close(); } public static void main(String[] args) throws IOException { new CFA(); } public long[] nextLongArr(int n) throws IOException{ long[] res = new long[n]; for(int i = 0; i < n; i++) res[i] = nextLong(); return res; } public int[] nextIntArr(int n) throws IOException { int[] res = new int[n]; for(int i = 0; i < n; i++) res[i] = nextInt(); return res; } public String nextToken() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (Exception e) { eof = true; return null; } } return st.nextToken(); } public String nextString() { try { return br.readLine(); } catch (IOException e) { eof = true; return null; } } public int nextInt() throws IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws IOException { return Double.parseDouble(nextToken()); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,122
2,403
765
import java.io.*; import java.util.*; import java.text.*; import java.math.*; public class main { public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); public static PrintWriter pw = new PrintWriter(System.out); public static String line; public static StringTokenizer st; public static ArrayList<ArrayList<Integer>> adjList; public static int[] dr = {-1, 0, 1, 0, -1, 1, 1, -1}; public static int[] dc = {0, 1, 0, -1, 1, 1, -1, -1}; public static void main(String[] args) throws Exception{ int N = Integer.parseInt(br.readLine()); char[] A = br.readLine().toCharArray(); HashSet<Character> cndD = new HashSet<Character>(); for(int i = 0; i < N; i++){ cndD.add(A[i]); } int cnt = cndD.size(); int a = 0; int b = 0; int ans = (1 << 30); HashMap<Character, Integer> d = new HashMap<Character, Integer>(); while(b < N){ if(d.containsKey(A[b])){ if(A[a] == A[b]){ a++; while(d.get(A[a]) > 1){ d.put(A[a], d.get(A[a])-1); a++; } } else{ d.put(A[b], d.get(A[b])+1); } } else{ d.put(A[b], 1); } if(d.size() == cnt){ ans = Math.min(b-a+1, ans); } b++; } pw.print(ans + "\n"); pw.close(); br.close(); } } class Pair implements Comparable<Pair>{ public int a, b; Pair(int a, int b){ this.a = a; this.b = b; } @Override public int compareTo(Pair P){ if(P.b != this.b){ return b - P.b; } else if(P.a == this.a){ return a - P.a; } else{ return 0; } } public String toString(){ return a + " " + b; } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.text.*; import java.math.*; public class main { public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); public static PrintWriter pw = new PrintWriter(System.out); public static String line; public static StringTokenizer st; public static ArrayList<ArrayList<Integer>> adjList; public static int[] dr = {-1, 0, 1, 0, -1, 1, 1, -1}; public static int[] dc = {0, 1, 0, -1, 1, 1, -1, -1}; public static void main(String[] args) throws Exception{ int N = Integer.parseInt(br.readLine()); char[] A = br.readLine().toCharArray(); HashSet<Character> cndD = new HashSet<Character>(); for(int i = 0; i < N; i++){ cndD.add(A[i]); } int cnt = cndD.size(); int a = 0; int b = 0; int ans = (1 << 30); HashMap<Character, Integer> d = new HashMap<Character, Integer>(); while(b < N){ if(d.containsKey(A[b])){ if(A[a] == A[b]){ a++; while(d.get(A[a]) > 1){ d.put(A[a], d.get(A[a])-1); a++; } } else{ d.put(A[b], d.get(A[b])+1); } } else{ d.put(A[b], 1); } if(d.size() == cnt){ ans = Math.min(b-a+1, ans); } b++; } pw.print(ans + "\n"); pw.close(); br.close(); } } class Pair implements Comparable<Pair>{ public int a, b; Pair(int a, int b){ this.a = a; this.b = b; } @Override public int compareTo(Pair P){ if(P.b != this.b){ return b - P.b; } else if(P.a == this.a){ return a - P.a; } else{ return 0; } } public String toString(){ return a + " " + b; } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.text.*; import java.math.*; public class main { public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); public static PrintWriter pw = new PrintWriter(System.out); public static String line; public static StringTokenizer st; public static ArrayList<ArrayList<Integer>> adjList; public static int[] dr = {-1, 0, 1, 0, -1, 1, 1, -1}; public static int[] dc = {0, 1, 0, -1, 1, 1, -1, -1}; public static void main(String[] args) throws Exception{ int N = Integer.parseInt(br.readLine()); char[] A = br.readLine().toCharArray(); HashSet<Character> cndD = new HashSet<Character>(); for(int i = 0; i < N; i++){ cndD.add(A[i]); } int cnt = cndD.size(); int a = 0; int b = 0; int ans = (1 << 30); HashMap<Character, Integer> d = new HashMap<Character, Integer>(); while(b < N){ if(d.containsKey(A[b])){ if(A[a] == A[b]){ a++; while(d.get(A[a]) > 1){ d.put(A[a], d.get(A[a])-1); a++; } } else{ d.put(A[b], d.get(A[b])+1); } } else{ d.put(A[b], 1); } if(d.size() == cnt){ ans = Math.min(b-a+1, ans); } b++; } pw.print(ans + "\n"); pw.close(); br.close(); } } class Pair implements Comparable<Pair>{ public int a, b; Pair(int a, int b){ this.a = a; this.b = b; } @Override public int compareTo(Pair P){ if(P.b != this.b){ return b - P.b; } else if(P.a == this.a){ return a - P.a; } else{ return 0; } } public String toString(){ return a + " " + b; } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
816
764
3,731
import java.io.*; import java.util.*; public class Main { static final int primeCount = 452; static final int[] prime = new int[primeCount]; static void build_prime() { boolean[] notPrime = new boolean[3200]; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; for (int j = i * i; j < 3200; j += i) { notPrime[j] = true; } } int count = 0; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; prime[count++] = i; } } private static void run(Reader in, PrintWriter out) throws IOException { int n = in.nextInt(); int m = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = getReal(in.nextInt()); } int[] pre = new int[n]; for (int i = 0; i < n; i++) pre[i] = -1; TreeMap<Integer, Integer> exist = new TreeMap<>(); for (int i = 0; i < n; i++) { Integer result = exist.get(a[i]); if (result != null) { pre[i] = result; } exist.put(a[i], i); } int[][] left = new int[m + 1][n]; for (int i = 0; i <= m; i++) { int start = 0; PriorityQueue<Integer> inSame = new PriorityQueue<>(); for (int j = 0; j < n; j++) { if (pre[j] >= start) { inSame.add(pre[j]); if (inSame.size() > i) { start = inSame.poll() + 1; } } left[i][j] = start; } } int[][] dp = new int[n][m + 1]; for (int i = 0; i < n; i++) { for (int j = 0; j <= m; j++) { if (j == 0) dp[i][0] = Integer.MAX_VALUE; else dp[i][j] = dp[i][j - 1]; for (int k = 0; k <= j; k++) { if (left[k][i] == 0) { dp[i][j] = 1; } else { dp[i][j] = Math.min(dp[i][j], dp[left[k][i] - 1][j - k] + 1); } } } } out.println(dp[n - 1][m]); } static int getReal(int x) { int result = 1; for (int i = 0; i < primeCount; i++) { if (x % prime[i] == 0) { int count = 0; while (x % prime[i] == 0) { count++; x /= prime[i]; } if (count % 2 == 1) { result *= prime[i]; } } } result *= x; return result; } public static void main(String[] args) throws IOException { Reader in = new Reader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); build_prime(); int t = in.nextInt(); for (int i = 0; i < t; i++) { run(in, out); } out.flush(); in.close(); out.close(); } static class Reader { BufferedReader reader; StringTokenizer st; Reader(InputStreamReader stream) { reader = new BufferedReader(stream, 32768); st = null; } void close() throws IOException { reader.close(); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } String nextLine() throws IOException { return reader.readLine(); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { static final int primeCount = 452; static final int[] prime = new int[primeCount]; static void build_prime() { boolean[] notPrime = new boolean[3200]; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; for (int j = i * i; j < 3200; j += i) { notPrime[j] = true; } } int count = 0; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; prime[count++] = i; } } private static void run(Reader in, PrintWriter out) throws IOException { int n = in.nextInt(); int m = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = getReal(in.nextInt()); } int[] pre = new int[n]; for (int i = 0; i < n; i++) pre[i] = -1; TreeMap<Integer, Integer> exist = new TreeMap<>(); for (int i = 0; i < n; i++) { Integer result = exist.get(a[i]); if (result != null) { pre[i] = result; } exist.put(a[i], i); } int[][] left = new int[m + 1][n]; for (int i = 0; i <= m; i++) { int start = 0; PriorityQueue<Integer> inSame = new PriorityQueue<>(); for (int j = 0; j < n; j++) { if (pre[j] >= start) { inSame.add(pre[j]); if (inSame.size() > i) { start = inSame.poll() + 1; } } left[i][j] = start; } } int[][] dp = new int[n][m + 1]; for (int i = 0; i < n; i++) { for (int j = 0; j <= m; j++) { if (j == 0) dp[i][0] = Integer.MAX_VALUE; else dp[i][j] = dp[i][j - 1]; for (int k = 0; k <= j; k++) { if (left[k][i] == 0) { dp[i][j] = 1; } else { dp[i][j] = Math.min(dp[i][j], dp[left[k][i] - 1][j - k] + 1); } } } } out.println(dp[n - 1][m]); } static int getReal(int x) { int result = 1; for (int i = 0; i < primeCount; i++) { if (x % prime[i] == 0) { int count = 0; while (x % prime[i] == 0) { count++; x /= prime[i]; } if (count % 2 == 1) { result *= prime[i]; } } } result *= x; return result; } public static void main(String[] args) throws IOException { Reader in = new Reader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); build_prime(); int t = in.nextInt(); for (int i = 0; i < t; i++) { run(in, out); } out.flush(); in.close(); out.close(); } static class Reader { BufferedReader reader; StringTokenizer st; Reader(InputStreamReader stream) { reader = new BufferedReader(stream, 32768); st = null; } void close() throws IOException { reader.close(); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } String nextLine() throws IOException { return reader.readLine(); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.*; import java.util.*; public class Main { static final int primeCount = 452; static final int[] prime = new int[primeCount]; static void build_prime() { boolean[] notPrime = new boolean[3200]; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; for (int j = i * i; j < 3200; j += i) { notPrime[j] = true; } } int count = 0; for (int i = 2; i < 3200; i++) { if (notPrime[i]) continue; prime[count++] = i; } } private static void run(Reader in, PrintWriter out) throws IOException { int n = in.nextInt(); int m = in.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; i++) { a[i] = getReal(in.nextInt()); } int[] pre = new int[n]; for (int i = 0; i < n; i++) pre[i] = -1; TreeMap<Integer, Integer> exist = new TreeMap<>(); for (int i = 0; i < n; i++) { Integer result = exist.get(a[i]); if (result != null) { pre[i] = result; } exist.put(a[i], i); } int[][] left = new int[m + 1][n]; for (int i = 0; i <= m; i++) { int start = 0; PriorityQueue<Integer> inSame = new PriorityQueue<>(); for (int j = 0; j < n; j++) { if (pre[j] >= start) { inSame.add(pre[j]); if (inSame.size() > i) { start = inSame.poll() + 1; } } left[i][j] = start; } } int[][] dp = new int[n][m + 1]; for (int i = 0; i < n; i++) { for (int j = 0; j <= m; j++) { if (j == 0) dp[i][0] = Integer.MAX_VALUE; else dp[i][j] = dp[i][j - 1]; for (int k = 0; k <= j; k++) { if (left[k][i] == 0) { dp[i][j] = 1; } else { dp[i][j] = Math.min(dp[i][j], dp[left[k][i] - 1][j - k] + 1); } } } } out.println(dp[n - 1][m]); } static int getReal(int x) { int result = 1; for (int i = 0; i < primeCount; i++) { if (x % prime[i] == 0) { int count = 0; while (x % prime[i] == 0) { count++; x /= prime[i]; } if (count % 2 == 1) { result *= prime[i]; } } } result *= x; return result; } public static void main(String[] args) throws IOException { Reader in = new Reader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); build_prime(); int t = in.nextInt(); for (int i = 0; i < t; i++) { run(in, out); } out.flush(); in.close(); out.close(); } static class Reader { BufferedReader reader; StringTokenizer st; Reader(InputStreamReader stream) { reader = new BufferedReader(stream, 32768); st = null; } void close() throws IOException { reader.close(); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } String nextLine() throws IOException { return reader.readLine(); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } } } </CODE> <EVALUATION_RUBRIC> - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,317
3,723
550
import java.util.*;import java.io.*;import java.math.*; public class Main { public static void process()throws IOException { long n=nl(); if(n%2==0) { long a1=(long)Math.sqrt(n); long a2=(long)Math.sqrt(n/2); if(a1*a1==n || a2*a2*2==n) {pn("YES");return;} } pn("NO"); } static AnotherReader sc; static PrintWriter out; public static void main(String[]args)throws IOException { boolean oj = System.getProperty("ONLINE_JUDGE") != null; if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);} else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");} int t=1; t=ni(); while(t-- > 0) {process();} out.flush();out.close(); } static void pn(Object o){out.println(o);} static void p(Object o){out.print(o);} static void pni(Object o){out.println(o);out.flush();} static int ni()throws IOException{return sc.nextInt();} static long nl()throws IOException{return sc.nextLong();} static double nd()throws IOException{return sc.nextDouble();} static String nln()throws IOException{return sc.nextLine();} static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;} static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;} static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));} ///////////////////////////////////////////////////////////////////////////////////////////////////////// static class AnotherReader{BufferedReader br; StringTokenizer st; AnotherReader()throws FileNotFoundException{ br=new BufferedReader(new InputStreamReader(System.in));} AnotherReader(int a)throws FileNotFoundException{ br = new BufferedReader(new FileReader("input.txt"));} String next()throws IOException{ while (st == null || !st.hasMoreElements()) {try{ st = new StringTokenizer(br.readLine());} catch (IOException e){ e.printStackTrace(); }} return st.nextToken(); } int nextInt() throws IOException{ return Integer.parseInt(next());} long nextLong() throws IOException {return Long.parseLong(next());} double nextDouble()throws IOException { return Double.parseDouble(next()); } String nextLine() throws IOException{ String str = ""; try{ str = br.readLine();} catch (IOException e){ e.printStackTrace();} return str;}} ///////////////////////////////////////////////////////////////////////////////////////////////////////////// }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.util.*;import java.io.*;import java.math.*; public class Main { public static void process()throws IOException { long n=nl(); if(n%2==0) { long a1=(long)Math.sqrt(n); long a2=(long)Math.sqrt(n/2); if(a1*a1==n || a2*a2*2==n) {pn("YES");return;} } pn("NO"); } static AnotherReader sc; static PrintWriter out; public static void main(String[]args)throws IOException { boolean oj = System.getProperty("ONLINE_JUDGE") != null; if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);} else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");} int t=1; t=ni(); while(t-- > 0) {process();} out.flush();out.close(); } static void pn(Object o){out.println(o);} static void p(Object o){out.print(o);} static void pni(Object o){out.println(o);out.flush();} static int ni()throws IOException{return sc.nextInt();} static long nl()throws IOException{return sc.nextLong();} static double nd()throws IOException{return sc.nextDouble();} static String nln()throws IOException{return sc.nextLine();} static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;} static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;} static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));} ///////////////////////////////////////////////////////////////////////////////////////////////////////// static class AnotherReader{BufferedReader br; StringTokenizer st; AnotherReader()throws FileNotFoundException{ br=new BufferedReader(new InputStreamReader(System.in));} AnotherReader(int a)throws FileNotFoundException{ br = new BufferedReader(new FileReader("input.txt"));} String next()throws IOException{ while (st == null || !st.hasMoreElements()) {try{ st = new StringTokenizer(br.readLine());} catch (IOException e){ e.printStackTrace(); }} return st.nextToken(); } int nextInt() throws IOException{ return Integer.parseInt(next());} long nextLong() throws IOException {return Long.parseLong(next());} double nextDouble()throws IOException { return Double.parseDouble(next()); } String nextLine() throws IOException{ String str = ""; try{ str = br.readLine();} catch (IOException e){ e.printStackTrace();} return str;}} ///////////////////////////////////////////////////////////////////////////////////////////////////////////// } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*;import java.io.*;import java.math.*; public class Main { public static void process()throws IOException { long n=nl(); if(n%2==0) { long a1=(long)Math.sqrt(n); long a2=(long)Math.sqrt(n/2); if(a1*a1==n || a2*a2*2==n) {pn("YES");return;} } pn("NO"); } static AnotherReader sc; static PrintWriter out; public static void main(String[]args)throws IOException { boolean oj = System.getProperty("ONLINE_JUDGE") != null; if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);} else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");} int t=1; t=ni(); while(t-- > 0) {process();} out.flush();out.close(); } static void pn(Object o){out.println(o);} static void p(Object o){out.print(o);} static void pni(Object o){out.println(o);out.flush();} static int ni()throws IOException{return sc.nextInt();} static long nl()throws IOException{return sc.nextLong();} static double nd()throws IOException{return sc.nextDouble();} static String nln()throws IOException{return sc.nextLine();} static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;} static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;} static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);} static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));} ///////////////////////////////////////////////////////////////////////////////////////////////////////// static class AnotherReader{BufferedReader br; StringTokenizer st; AnotherReader()throws FileNotFoundException{ br=new BufferedReader(new InputStreamReader(System.in));} AnotherReader(int a)throws FileNotFoundException{ br = new BufferedReader(new FileReader("input.txt"));} String next()throws IOException{ while (st == null || !st.hasMoreElements()) {try{ st = new StringTokenizer(br.readLine());} catch (IOException e){ e.printStackTrace(); }} return st.nextToken(); } int nextInt() throws IOException{ return Integer.parseInt(next());} long nextLong() throws IOException {return Long.parseLong(next());} double nextDouble()throws IOException { return Double.parseDouble(next()); } String nextLine() throws IOException{ String str = ""; try{ str = br.readLine();} catch (IOException e){ e.printStackTrace();} return str;}} ///////////////////////////////////////////////////////////////////////////////////////////////////////////// } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
979
549
3,339
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; import java.util.LinkedList; import java.util.StringTokenizer; public class GeorgeAndInterestingGraph { public static void main(String[] args) { MyScanner sc = new MyScanner(); int N = sc.nextInt(); int M = sc.nextInt(); int[] edgeFrom = new int[M]; int[] edgeTo = new int[M]; for (int i = 0; i < M; i++) { edgeFrom[i] = sc.nextInt(); edgeTo[i] = sc.nextInt(); } int best = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { boolean[][] mat = makeAdjMat(N, edgeFrom, edgeTo); best = Math.min(best, count(mat, i, M)); } System.out.println(best); } public static int count(boolean[][] mat, int center, int M) { // int N = mat.length; // int M = mat[0].length; // // int centerConnect = (mat[center][center]) ? 0 : 1; // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } // // int[][] adjMat = new int[2 * N + 2][2 * N + 2]; // for (int i = 0; i < N; i++) { // for (int j = 0; j < N; j++) { // int idx = N + j; // adjMat[i][idx] = (mat[i][j]) ? 1 : 0; // } // } // int s = 2 * N; // int t = 2 * N + 1; // for (int i = 0; i < N; i++) { // adjMat[s][i] = 1; // } // for (int i = N; i < 2 * N; i++) { // adjMat[i][t] = 1; // } // // int matches = fordFulkerson(adjMat, s, t); // // return centerConnect + matches; int N = mat.length; // int M = mat[0].length; int cntWithI = (mat[center][center]) ? 1 : 0; for (int i = 0; i < N; i++) { if (i != center) { if (mat[i][center]) { cntWithI++; } if (mat[center][i]) { cntWithI++; } } mat[i][center] = false; mat[center][i] = false; } int other = M - cntWithI; // int centerConnect = (mat[center][center]) ? 0 : 1; // // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } int matches = bipartiteMatching(mat); return (2 * N - 1 - cntWithI + other - matches + N - 1 - matches); // return (centerConnect + N - 1 - matches); } public static boolean[][] makeAdjMat(int N, int[] edgeFrom, int[] edgeTo) { boolean[][] mat = new boolean[N][N]; for (int i = 0; i < edgeFrom.length; i++) { int from = edgeFrom[i] - 1; int to = edgeTo[i] - 1; mat[from][to] = true; } return mat; } /** * Returns true if there is a path from the source 's' to the sink 't' in the residual graph. * Also fills parent[] to store the path. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static boolean fordFulkersonHelper(int[][] resid, int s, int t, int[] parent) { int V = resid.length; boolean[] visited = new boolean[V]; LinkedList<Integer> q = new LinkedList<Integer>(); q.push(s); visited[s] = true; parent[s] = -1; while (!q.isEmpty()) { int u = q.pop(); for (int v = 0; v < V; v++) { if (!visited[v] && resid[u][v] > 0) { q.push(v); parent[v] = u; visited[v] = true; } } } return visited[t]; } /** * Returns the maximum flow from 's' to 't' in the given graph. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static int fordFulkerson(int[][] graph, int s, int t) { int V = graph.length; int[][] resid = new int[V][V]; int[] parent = new int[V]; int maxFlow = 0; for (int u = 0; u < V; u++) { for (int v = 0; v < V; v++) { resid[u][v] = graph[u][v]; } } while (fordFulkersonHelper(resid, s, t, parent)) { int pathFlow = Integer.MAX_VALUE; for (int v = t; v != s; v = parent[v]) { int u = parent[v]; pathFlow = Math.min(pathFlow, resid[u][v]); } for (int v = t; v != s; v = parent[v]) { int u = parent[v]; resid[u][v] -= pathFlow; resid[v][u] += pathFlow; } maxFlow += pathFlow; } return maxFlow; } /** * Returns true if a matching for vertex 'u' is possible. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static boolean bipartiteMatchingHelper(boolean[][] bpGraph, int u, boolean[] seen, int[] matchR) { int N = bpGraph[0].length; for (int v = 0; v < N; v++) { if (bpGraph[u][v] && !seen[v]) { seen[v] = true; if (matchR[v] < 0 || bipartiteMatchingHelper(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchIJ, int[] matchJI) { int ans = bipartiteMatching(bpGraph, matchJI); for (int i = 0; i < matchJI.length; i++) { matchIJ[i] = -1; } for (int j = 0; j < matchJI.length; j++) { int i = matchJI[j]; if (i >= 0) { matchIJ[i] = j; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchJI) { int M = bpGraph.length; int N = bpGraph[0].length; for (int i = 0; i < N; i++) { matchJI[i] = -1; } int ans = 0; for (int u = 0; u < M; u++) { boolean[] seen = new boolean[N]; if (bipartiteMatchingHelper(bpGraph, u, seen, matchJI)) { ans++; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Overload of the bipartiteMatching function without output parameters. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph) { int N = bpGraph[0].length; int[] matchJI = new int[N]; return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. */ public static int bipartiteMatching(int[][] intGraph) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchIJ, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchIJ, matchJI); } /** * Converts an integer adjacency matrix of 1's and 0's to a boolean adjacency matrix. * Useful with bipartiteMatching, which takes adjancency matrix of boolean[][] as input (instead of int[][]). */ public static boolean[][] intToBooleanAdjMat(int[][] mat) { int M = mat.length; int N = mat[0].length; boolean[][] bMat = new boolean[M][N]; for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { bMat[i][j] = (mat[i][j] != 0); } } return bMat; } public static class MyScanner { BufferedReader br; StringTokenizer st; public MyScanner() { br = new BufferedReader(new InputStreamReader(System.in)); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; import java.util.LinkedList; import java.util.StringTokenizer; public class GeorgeAndInterestingGraph { public static void main(String[] args) { MyScanner sc = new MyScanner(); int N = sc.nextInt(); int M = sc.nextInt(); int[] edgeFrom = new int[M]; int[] edgeTo = new int[M]; for (int i = 0; i < M; i++) { edgeFrom[i] = sc.nextInt(); edgeTo[i] = sc.nextInt(); } int best = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { boolean[][] mat = makeAdjMat(N, edgeFrom, edgeTo); best = Math.min(best, count(mat, i, M)); } System.out.println(best); } public static int count(boolean[][] mat, int center, int M) { // int N = mat.length; // int M = mat[0].length; // // int centerConnect = (mat[center][center]) ? 0 : 1; // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } // // int[][] adjMat = new int[2 * N + 2][2 * N + 2]; // for (int i = 0; i < N; i++) { // for (int j = 0; j < N; j++) { // int idx = N + j; // adjMat[i][idx] = (mat[i][j]) ? 1 : 0; // } // } // int s = 2 * N; // int t = 2 * N + 1; // for (int i = 0; i < N; i++) { // adjMat[s][i] = 1; // } // for (int i = N; i < 2 * N; i++) { // adjMat[i][t] = 1; // } // // int matches = fordFulkerson(adjMat, s, t); // // return centerConnect + matches; int N = mat.length; // int M = mat[0].length; int cntWithI = (mat[center][center]) ? 1 : 0; for (int i = 0; i < N; i++) { if (i != center) { if (mat[i][center]) { cntWithI++; } if (mat[center][i]) { cntWithI++; } } mat[i][center] = false; mat[center][i] = false; } int other = M - cntWithI; // int centerConnect = (mat[center][center]) ? 0 : 1; // // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } int matches = bipartiteMatching(mat); return (2 * N - 1 - cntWithI + other - matches + N - 1 - matches); // return (centerConnect + N - 1 - matches); } public static boolean[][] makeAdjMat(int N, int[] edgeFrom, int[] edgeTo) { boolean[][] mat = new boolean[N][N]; for (int i = 0; i < edgeFrom.length; i++) { int from = edgeFrom[i] - 1; int to = edgeTo[i] - 1; mat[from][to] = true; } return mat; } /** * Returns true if there is a path from the source 's' to the sink 't' in the residual graph. * Also fills parent[] to store the path. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static boolean fordFulkersonHelper(int[][] resid, int s, int t, int[] parent) { int V = resid.length; boolean[] visited = new boolean[V]; LinkedList<Integer> q = new LinkedList<Integer>(); q.push(s); visited[s] = true; parent[s] = -1; while (!q.isEmpty()) { int u = q.pop(); for (int v = 0; v < V; v++) { if (!visited[v] && resid[u][v] > 0) { q.push(v); parent[v] = u; visited[v] = true; } } } return visited[t]; } /** * Returns the maximum flow from 's' to 't' in the given graph. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static int fordFulkerson(int[][] graph, int s, int t) { int V = graph.length; int[][] resid = new int[V][V]; int[] parent = new int[V]; int maxFlow = 0; for (int u = 0; u < V; u++) { for (int v = 0; v < V; v++) { resid[u][v] = graph[u][v]; } } while (fordFulkersonHelper(resid, s, t, parent)) { int pathFlow = Integer.MAX_VALUE; for (int v = t; v != s; v = parent[v]) { int u = parent[v]; pathFlow = Math.min(pathFlow, resid[u][v]); } for (int v = t; v != s; v = parent[v]) { int u = parent[v]; resid[u][v] -= pathFlow; resid[v][u] += pathFlow; } maxFlow += pathFlow; } return maxFlow; } /** * Returns true if a matching for vertex 'u' is possible. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static boolean bipartiteMatchingHelper(boolean[][] bpGraph, int u, boolean[] seen, int[] matchR) { int N = bpGraph[0].length; for (int v = 0; v < N; v++) { if (bpGraph[u][v] && !seen[v]) { seen[v] = true; if (matchR[v] < 0 || bipartiteMatchingHelper(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchIJ, int[] matchJI) { int ans = bipartiteMatching(bpGraph, matchJI); for (int i = 0; i < matchJI.length; i++) { matchIJ[i] = -1; } for (int j = 0; j < matchJI.length; j++) { int i = matchJI[j]; if (i >= 0) { matchIJ[i] = j; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchJI) { int M = bpGraph.length; int N = bpGraph[0].length; for (int i = 0; i < N; i++) { matchJI[i] = -1; } int ans = 0; for (int u = 0; u < M; u++) { boolean[] seen = new boolean[N]; if (bipartiteMatchingHelper(bpGraph, u, seen, matchJI)) { ans++; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Overload of the bipartiteMatching function without output parameters. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph) { int N = bpGraph[0].length; int[] matchJI = new int[N]; return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. */ public static int bipartiteMatching(int[][] intGraph) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchIJ, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchIJ, matchJI); } /** * Converts an integer adjacency matrix of 1's and 0's to a boolean adjacency matrix. * Useful with bipartiteMatching, which takes adjancency matrix of boolean[][] as input (instead of int[][]). */ public static boolean[][] intToBooleanAdjMat(int[][] mat) { int M = mat.length; int N = mat[0].length; boolean[][] bMat = new boolean[M][N]; for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { bMat[i][j] = (mat[i][j] != 0); } } return bMat; } public static class MyScanner { BufferedReader br; StringTokenizer st; public MyScanner() { br = new BufferedReader(new InputStreamReader(System.in)); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; import java.util.LinkedList; import java.util.StringTokenizer; public class GeorgeAndInterestingGraph { public static void main(String[] args) { MyScanner sc = new MyScanner(); int N = sc.nextInt(); int M = sc.nextInt(); int[] edgeFrom = new int[M]; int[] edgeTo = new int[M]; for (int i = 0; i < M; i++) { edgeFrom[i] = sc.nextInt(); edgeTo[i] = sc.nextInt(); } int best = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { boolean[][] mat = makeAdjMat(N, edgeFrom, edgeTo); best = Math.min(best, count(mat, i, M)); } System.out.println(best); } public static int count(boolean[][] mat, int center, int M) { // int N = mat.length; // int M = mat[0].length; // // int centerConnect = (mat[center][center]) ? 0 : 1; // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } // // int[][] adjMat = new int[2 * N + 2][2 * N + 2]; // for (int i = 0; i < N; i++) { // for (int j = 0; j < N; j++) { // int idx = N + j; // adjMat[i][idx] = (mat[i][j]) ? 1 : 0; // } // } // int s = 2 * N; // int t = 2 * N + 1; // for (int i = 0; i < N; i++) { // adjMat[s][i] = 1; // } // for (int i = N; i < 2 * N; i++) { // adjMat[i][t] = 1; // } // // int matches = fordFulkerson(adjMat, s, t); // // return centerConnect + matches; int N = mat.length; // int M = mat[0].length; int cntWithI = (mat[center][center]) ? 1 : 0; for (int i = 0; i < N; i++) { if (i != center) { if (mat[i][center]) { cntWithI++; } if (mat[center][i]) { cntWithI++; } } mat[i][center] = false; mat[center][i] = false; } int other = M - cntWithI; // int centerConnect = (mat[center][center]) ? 0 : 1; // // for (int i = 0; i < N; i++) { // if (i != center) { // if (!mat[i][center]) { // centerConnect++; // } // if (!mat[center][i]) { // centerConnect++; // } // } // mat[i][center] = false; // mat[center][i] = false; // } int matches = bipartiteMatching(mat); return (2 * N - 1 - cntWithI + other - matches + N - 1 - matches); // return (centerConnect + N - 1 - matches); } public static boolean[][] makeAdjMat(int N, int[] edgeFrom, int[] edgeTo) { boolean[][] mat = new boolean[N][N]; for (int i = 0; i < edgeFrom.length; i++) { int from = edgeFrom[i] - 1; int to = edgeTo[i] - 1; mat[from][to] = true; } return mat; } /** * Returns true if there is a path from the source 's' to the sink 't' in the residual graph. * Also fills parent[] to store the path. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static boolean fordFulkersonHelper(int[][] resid, int s, int t, int[] parent) { int V = resid.length; boolean[] visited = new boolean[V]; LinkedList<Integer> q = new LinkedList<Integer>(); q.push(s); visited[s] = true; parent[s] = -1; while (!q.isEmpty()) { int u = q.pop(); for (int v = 0; v < V; v++) { if (!visited[v] && resid[u][v] > 0) { q.push(v); parent[v] = u; visited[v] = true; } } } return visited[t]; } /** * Returns the maximum flow from 's' to 't' in the given graph. * See here for more info: http://www.geeksforgeeks.org/ford-fulkerson-algorithm-for-maximum-flow-problem/ */ public static int fordFulkerson(int[][] graph, int s, int t) { int V = graph.length; int[][] resid = new int[V][V]; int[] parent = new int[V]; int maxFlow = 0; for (int u = 0; u < V; u++) { for (int v = 0; v < V; v++) { resid[u][v] = graph[u][v]; } } while (fordFulkersonHelper(resid, s, t, parent)) { int pathFlow = Integer.MAX_VALUE; for (int v = t; v != s; v = parent[v]) { int u = parent[v]; pathFlow = Math.min(pathFlow, resid[u][v]); } for (int v = t; v != s; v = parent[v]) { int u = parent[v]; resid[u][v] -= pathFlow; resid[v][u] += pathFlow; } maxFlow += pathFlow; } return maxFlow; } /** * Returns true if a matching for vertex 'u' is possible. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static boolean bipartiteMatchingHelper(boolean[][] bpGraph, int u, boolean[] seen, int[] matchR) { int N = bpGraph[0].length; for (int v = 0; v < N; v++) { if (bpGraph[u][v] && !seen[v]) { seen[v] = true; if (matchR[v] < 0 || bipartiteMatchingHelper(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchIJ, int[] matchJI) { int ans = bipartiteMatching(bpGraph, matchJI); for (int i = 0; i < matchJI.length; i++) { matchIJ[i] = -1; } for (int j = 0; j < matchJI.length; j++) { int i = matchJI[j]; if (i >= 0) { matchIJ[i] = j; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Note: bpGraph[i][j] = true if there is an edge from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph, int[] matchJI) { int M = bpGraph.length; int N = bpGraph[0].length; for (int i = 0; i < N; i++) { matchJI[i] = -1; } int ans = 0; for (int u = 0; u < M; u++) { boolean[] seen = new boolean[N]; if (bipartiteMatchingHelper(bpGraph, u, seen, matchJI)) { ans++; } } return ans; } /** * Returns the maximum bipartite matching from an an adjacency matrix. * Overload of the bipartiteMatching function without output parameters. * See here for more info: http://www.geeksforgeeks.org/maximum-bipartite-matching/ */ public static int bipartiteMatching(boolean[][] bpGraph) { int N = bpGraph[0].length; int[] matchJI = new int[N]; return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. */ public static int bipartiteMatching(int[][] intGraph) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchJI); } /** * Overload of the bipartiteMatching function taking an adjacency matrix of int[][] instead of boolean[][]. * Note: matchIJ (array of length M) is an output variable containing the matchings, such that matchIJ[i] = j means that there is a match from i to j. * Note: matchJI (array of length N) is an output variable containing the matchings, such that matchJI[j] = i means that there is a match from i to j. */ public static int bipartiteMatching(int[][] intGraph, int[] matchIJ, int[] matchJI) { boolean[][] bpGraph = intToBooleanAdjMat(intGraph); return bipartiteMatching(bpGraph, matchIJ, matchJI); } /** * Converts an integer adjacency matrix of 1's and 0's to a boolean adjacency matrix. * Useful with bipartiteMatching, which takes adjancency matrix of boolean[][] as input (instead of int[][]). */ public static boolean[][] intToBooleanAdjMat(int[][] mat) { int M = mat.length; int N = mat[0].length; boolean[][] bMat = new boolean[M][N]; for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { bMat[i][j] = (mat[i][j] != 0); } } return bMat; } public static class MyScanner { BufferedReader br; StringTokenizer st; public MyScanner() { br = new BufferedReader(new InputStreamReader(System.in)); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(1): The time complexity is constant to the input size n. - O(n^2): The time complexity grows proportionally to the square of the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(log(n)): The time complexity increases logarithmically in relation to the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
3,319
3,333
2,242
/* * PDPM IIITDM Jabalpur * Asutosh Rana */ import java.util.*; import java.io.*; import java.math.*; public class Main { long MOD = 1000000007; InputReader in; BufferedReader br; PrintWriter out; public static void main(String[] args) throws java.lang.Exception { Main solver = new Main(); solver.in = new InputReader(System.in); solver.br = new BufferedReader(new InputStreamReader(System.in)); solver.out = new PrintWriter(System.out); solver.solve(); solver.out.flush(); solver.out.close(); } int[] A; int N; public void solve() { int tc = 1;//in.readInt(); for (int cas = 1; cas <= tc; cas++) { N = in.readInt(); A = new int[N]; for(int i =0;i<A.length;i++){ String str = in.readString(); if(str.equals("f")) A[i] = 1; } long[][] dp = new long[N+1][N+1]; for(int i=0;i<N;i++){ if(i==0){ dp[i][0] = 1; } else if(A[i-1]!=1){ dp[i][N] = dp[i-1][N]; for(int j=N-1;j>=0;j--){ /* sum of all positions from */ dp[i][j] = (dp[i-1][j] + dp[i][j+1])%MOD; } } else{ for(int j=1;j<=N;j++){ dp[i][j] = dp[i-1][j-1]%MOD; } } } long res = 0; for(int i=0;i<=N;i++){ res = (res + dp[N-1][i])%MOD; } out.println(res); } } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readInt(int[] A) { for (int i = 0; i < A.length; i++) A[i] = readInt(); } public long readLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readLong(long[] A) { for (int i = 0; i < A.length; i++) A[i] = readLong(); } public double readDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public char[] readCharA() { return readString().toCharArray(); } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> /* * PDPM IIITDM Jabalpur * Asutosh Rana */ import java.util.*; import java.io.*; import java.math.*; public class Main { long MOD = 1000000007; InputReader in; BufferedReader br; PrintWriter out; public static void main(String[] args) throws java.lang.Exception { Main solver = new Main(); solver.in = new InputReader(System.in); solver.br = new BufferedReader(new InputStreamReader(System.in)); solver.out = new PrintWriter(System.out); solver.solve(); solver.out.flush(); solver.out.close(); } int[] A; int N; public void solve() { int tc = 1;//in.readInt(); for (int cas = 1; cas <= tc; cas++) { N = in.readInt(); A = new int[N]; for(int i =0;i<A.length;i++){ String str = in.readString(); if(str.equals("f")) A[i] = 1; } long[][] dp = new long[N+1][N+1]; for(int i=0;i<N;i++){ if(i==0){ dp[i][0] = 1; } else if(A[i-1]!=1){ dp[i][N] = dp[i-1][N]; for(int j=N-1;j>=0;j--){ /* sum of all positions from */ dp[i][j] = (dp[i-1][j] + dp[i][j+1])%MOD; } } else{ for(int j=1;j<=N;j++){ dp[i][j] = dp[i-1][j-1]%MOD; } } } long res = 0; for(int i=0;i<=N;i++){ res = (res + dp[N-1][i])%MOD; } out.println(res); } } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readInt(int[] A) { for (int i = 0; i < A.length; i++) A[i] = readInt(); } public long readLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readLong(long[] A) { for (int i = 0; i < A.length; i++) A[i] = readLong(); } public double readDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public char[] readCharA() { return readString().toCharArray(); } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } </CODE> <EVALUATION_RUBRIC> - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* * PDPM IIITDM Jabalpur * Asutosh Rana */ import java.util.*; import java.io.*; import java.math.*; public class Main { long MOD = 1000000007; InputReader in; BufferedReader br; PrintWriter out; public static void main(String[] args) throws java.lang.Exception { Main solver = new Main(); solver.in = new InputReader(System.in); solver.br = new BufferedReader(new InputStreamReader(System.in)); solver.out = new PrintWriter(System.out); solver.solve(); solver.out.flush(); solver.out.close(); } int[] A; int N; public void solve() { int tc = 1;//in.readInt(); for (int cas = 1; cas <= tc; cas++) { N = in.readInt(); A = new int[N]; for(int i =0;i<A.length;i++){ String str = in.readString(); if(str.equals("f")) A[i] = 1; } long[][] dp = new long[N+1][N+1]; for(int i=0;i<N;i++){ if(i==0){ dp[i][0] = 1; } else if(A[i-1]!=1){ dp[i][N] = dp[i-1][N]; for(int j=N-1;j>=0;j--){ /* sum of all positions from */ dp[i][j] = (dp[i-1][j] + dp[i][j+1])%MOD; } } else{ for(int j=1;j<=N;j++){ dp[i][j] = dp[i-1][j-1]%MOD; } } } long res = 0; for(int i=0;i<=N;i++){ res = (res + dp[N-1][i])%MOD; } out.println(res); } } } class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int readInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readInt(int[] A) { for (int i = 0; i < A.length; i++) A[i] = readInt(); } public long readLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public void readLong(long[] A) { for (int i = 0; i < A.length; i++) A[i] = readLong(); } public double readDouble() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } double res = 0; while (!isSpaceChar(c) && c != '.') { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } if (c == '.') { c = read(); double m = 1; while (!isSpaceChar(c)) { if (c == 'e' || c == 'E') return res * Math.pow(10, readInt()); if (c < '0' || c > '9') throw new InputMismatchException(); m /= 10; res += (c - '0') * m; c = read(); } } return res * sgn; } public char[] readCharA() { return readString().toCharArray(); } public String readString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return readString(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,657
2,238
607
import java.io.*; import java.math.BigDecimal; import java.math.BigInteger; import java.math.RoundingMode; import java.util.*; import java.util.Map.Entry; public class Main { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; private PrintWriter pw; private long mod = 998244353; private StringBuilder ans_sb; private int sqrt; private void soln() { int n = nextInt(); int d = nextInt(); int[] arr = nextIntArray(n); int cnt = 2; for(int i=0;i<n-1;i++) { int a1 = arr[i]; int a2 = arr[i+1]; a1 += d; a2 -= d; if(a1 < a2) { cnt+=2; }else if(a1==a2) cnt++; } pw.println(cnt); } private class Pair implements Comparable<Pair>{ int x,i; public Pair(int a, int b) { x = a; i = b; } @Override public int compareTo( Pair o) { return this.x - o.x; } } private int cc = 0; private void dfs(int c, int p, LinkedList<Integer>[] tree, int[] t, int[] tin, int[] tout, int[] arr) { //debug(c); t[cc] = arr[c]; tin[c] = cc++; Iterator<Integer> it = tree[c].listIterator(); while(it.hasNext()) { int x = it.next(); if(x != p) { dfs(x, c, tree,t,tin,tout,arr); } } tout[c] = cc; } public class Segment { private Node[] tree; private int[] lazy; private int size; private int n; private int[] base; private class Node { private int on; private int off; } public Segment(int n, int[] arr) { this.base=arr; int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); size = 2 * (int) Math.pow(2, x) - 1; tree = new Node[size]; lazy = new int[size]; this.n = n; //this.set = set1; build(0, 0, n - 1); } public void build(int id, int l, int r) { if (l == r) { tree[id] = new Node(); if(base[l] == 1) tree[id].on++; else tree[id].off++; return; } int mid = ((l + r)>>1); build((id<<1)|1, l, mid); build((id<<1)+2, mid + 1, r); tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); //System.out.println(l+" "+r+" "+tree[id].l+" "+tree[id].r+" "+tree[id].ans); } public Node merge(Node n1, Node n2) { Node ret = new Node(); if(n1 == null && n2 == null) return null; else if(n1 == null) { ret.on = n2.on; ret.off = n2.off; } else if(n2 == null) { ret.on = n1.on; ret.off = n1.off; } else { ret.on = n1.on+n2.on; ret.off = n2.off + n1.off; } return ret; } public int query(int l, int r) { Node ret = queryUtil(l, r, 0, 0, n - 1); if(ret == null) { return 0; } else return ret.on; } private Node queryUtil(int x, int y, int id, int l, int r) { if (l > y || x > r) return null; if (x <= l && r <= y) { //shift(id,l,r); return tree[id]; } int mid = ((l + r)>>1); shift(id,l,r); Node q1 = queryUtil(x, y, (id<<1)|1, l, mid); Node q2 = queryUtil(x, y, (id<<1)+2, mid + 1, r); return merge(q1, q2); } public void update(int x, int y, int c) { update1(x, y, c, 0, 0, n-1); } private void update1(int x, int y, int colour, int id, int l, int r) { //System.out.println(l+" "+r+" "+x); if (x > r || y < l) return; if (x <= l && r <= y) { lazy[id]++; // lazy[id] %= 2; switchNode(tree[id]); return; } int mid = ((l + r)>>1); //shift(id); if(y<=mid) update1(x, y, colour, (id<<1)|1, l, mid); else if(x>mid) update1(x, y, colour, (id<<1)+2, mid + 1, r); else { update1(x, y, colour, (id<<1)|1, l, mid); update1(x, y, colour, (id<<1)+2, mid + 1, r); } tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); } private void shift(int id,int l, int r) { lazy[id] %= 2; if(lazy[id] != 0) { if(l != r) { lazy[(id<<1)|1] += lazy[id]; lazy[(id<<1)+2] += lazy[id]; switchNode(tree[(id<<1)+2]); switchNode(tree[(id<<1)|1]); } //switchNode(tree[(id<<1)+2]); lazy[id] = 0; } } private void switchNode(Node d) { d.on ^= d.off; d.off ^= d.on; d.on ^= d.off; } } private void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } private long pow(long a, long b, long c) { if (b == 0) return 1; long p = pow(a, b / 2, c); p = (p * p) % c; return (b % 2 == 0) ? p : (a * p) % c; } private long gcd(long n, long l) { if (l == 0) return n; return gcd(l, n % l); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { @Override public void run() { new Main().solve(); } }, "1", 1 << 26).start(); //new Main().solve(); } public StringBuilder solve() { InputReader(System.in); /* * try { InputReader(new FileInputStream("C:\\Users\\hardik\\Desktop\\in.txt")); * } catch(FileNotFoundException e) {} */ pw = new PrintWriter(System.out); // ans_sb = new StringBuilder(); soln(); pw.close(); // System.out.println(ans_sb); return ans_sb; } public void InputReader(InputStream stream1) { stream = stream1; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } private int nextInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private String nextToken() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private String nextLine() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isEndOfLine(c)); return res.toString(); } private int[] nextIntArray(int n) { int[] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i] = nextInt(); } return arr; } private long[] nextLongArray(int n) { long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = nextLong(); } return arr; } private void pArray(int[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private void pArray(long[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return isWhitespace(c); } private char nextChar() { int c = read(); while (isSpaceChar(c)) c = read(); char c1 = (char) c; while (!isSpaceChar(c)) c = read(); return c1; } private interface SpaceCharFilter { public boolean isSpaceChar(int ch); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.math.BigDecimal; import java.math.BigInteger; import java.math.RoundingMode; import java.util.*; import java.util.Map.Entry; public class Main { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; private PrintWriter pw; private long mod = 998244353; private StringBuilder ans_sb; private int sqrt; private void soln() { int n = nextInt(); int d = nextInt(); int[] arr = nextIntArray(n); int cnt = 2; for(int i=0;i<n-1;i++) { int a1 = arr[i]; int a2 = arr[i+1]; a1 += d; a2 -= d; if(a1 < a2) { cnt+=2; }else if(a1==a2) cnt++; } pw.println(cnt); } private class Pair implements Comparable<Pair>{ int x,i; public Pair(int a, int b) { x = a; i = b; } @Override public int compareTo( Pair o) { return this.x - o.x; } } private int cc = 0; private void dfs(int c, int p, LinkedList<Integer>[] tree, int[] t, int[] tin, int[] tout, int[] arr) { //debug(c); t[cc] = arr[c]; tin[c] = cc++; Iterator<Integer> it = tree[c].listIterator(); while(it.hasNext()) { int x = it.next(); if(x != p) { dfs(x, c, tree,t,tin,tout,arr); } } tout[c] = cc; } public class Segment { private Node[] tree; private int[] lazy; private int size; private int n; private int[] base; private class Node { private int on; private int off; } public Segment(int n, int[] arr) { this.base=arr; int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); size = 2 * (int) Math.pow(2, x) - 1; tree = new Node[size]; lazy = new int[size]; this.n = n; //this.set = set1; build(0, 0, n - 1); } public void build(int id, int l, int r) { if (l == r) { tree[id] = new Node(); if(base[l] == 1) tree[id].on++; else tree[id].off++; return; } int mid = ((l + r)>>1); build((id<<1)|1, l, mid); build((id<<1)+2, mid + 1, r); tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); //System.out.println(l+" "+r+" "+tree[id].l+" "+tree[id].r+" "+tree[id].ans); } public Node merge(Node n1, Node n2) { Node ret = new Node(); if(n1 == null && n2 == null) return null; else if(n1 == null) { ret.on = n2.on; ret.off = n2.off; } else if(n2 == null) { ret.on = n1.on; ret.off = n1.off; } else { ret.on = n1.on+n2.on; ret.off = n2.off + n1.off; } return ret; } public int query(int l, int r) { Node ret = queryUtil(l, r, 0, 0, n - 1); if(ret == null) { return 0; } else return ret.on; } private Node queryUtil(int x, int y, int id, int l, int r) { if (l > y || x > r) return null; if (x <= l && r <= y) { //shift(id,l,r); return tree[id]; } int mid = ((l + r)>>1); shift(id,l,r); Node q1 = queryUtil(x, y, (id<<1)|1, l, mid); Node q2 = queryUtil(x, y, (id<<1)+2, mid + 1, r); return merge(q1, q2); } public void update(int x, int y, int c) { update1(x, y, c, 0, 0, n-1); } private void update1(int x, int y, int colour, int id, int l, int r) { //System.out.println(l+" "+r+" "+x); if (x > r || y < l) return; if (x <= l && r <= y) { lazy[id]++; // lazy[id] %= 2; switchNode(tree[id]); return; } int mid = ((l + r)>>1); //shift(id); if(y<=mid) update1(x, y, colour, (id<<1)|1, l, mid); else if(x>mid) update1(x, y, colour, (id<<1)+2, mid + 1, r); else { update1(x, y, colour, (id<<1)|1, l, mid); update1(x, y, colour, (id<<1)+2, mid + 1, r); } tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); } private void shift(int id,int l, int r) { lazy[id] %= 2; if(lazy[id] != 0) { if(l != r) { lazy[(id<<1)|1] += lazy[id]; lazy[(id<<1)+2] += lazy[id]; switchNode(tree[(id<<1)+2]); switchNode(tree[(id<<1)|1]); } //switchNode(tree[(id<<1)+2]); lazy[id] = 0; } } private void switchNode(Node d) { d.on ^= d.off; d.off ^= d.on; d.on ^= d.off; } } private void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } private long pow(long a, long b, long c) { if (b == 0) return 1; long p = pow(a, b / 2, c); p = (p * p) % c; return (b % 2 == 0) ? p : (a * p) % c; } private long gcd(long n, long l) { if (l == 0) return n; return gcd(l, n % l); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { @Override public void run() { new Main().solve(); } }, "1", 1 << 26).start(); //new Main().solve(); } public StringBuilder solve() { InputReader(System.in); /* * try { InputReader(new FileInputStream("C:\\Users\\hardik\\Desktop\\in.txt")); * } catch(FileNotFoundException e) {} */ pw = new PrintWriter(System.out); // ans_sb = new StringBuilder(); soln(); pw.close(); // System.out.println(ans_sb); return ans_sb; } public void InputReader(InputStream stream1) { stream = stream1; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } private int nextInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private String nextToken() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private String nextLine() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isEndOfLine(c)); return res.toString(); } private int[] nextIntArray(int n) { int[] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i] = nextInt(); } return arr; } private long[] nextLongArray(int n) { long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = nextLong(); } return arr; } private void pArray(int[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private void pArray(long[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return isWhitespace(c); } private char nextChar() { int c = read(); while (isSpaceChar(c)) c = read(); char c1 = (char) c; while (!isSpaceChar(c)) c = read(); return c1; } private interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.math.BigDecimal; import java.math.BigInteger; import java.math.RoundingMode; import java.util.*; import java.util.Map.Entry; public class Main { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; private PrintWriter pw; private long mod = 998244353; private StringBuilder ans_sb; private int sqrt; private void soln() { int n = nextInt(); int d = nextInt(); int[] arr = nextIntArray(n); int cnt = 2; for(int i=0;i<n-1;i++) { int a1 = arr[i]; int a2 = arr[i+1]; a1 += d; a2 -= d; if(a1 < a2) { cnt+=2; }else if(a1==a2) cnt++; } pw.println(cnt); } private class Pair implements Comparable<Pair>{ int x,i; public Pair(int a, int b) { x = a; i = b; } @Override public int compareTo( Pair o) { return this.x - o.x; } } private int cc = 0; private void dfs(int c, int p, LinkedList<Integer>[] tree, int[] t, int[] tin, int[] tout, int[] arr) { //debug(c); t[cc] = arr[c]; tin[c] = cc++; Iterator<Integer> it = tree[c].listIterator(); while(it.hasNext()) { int x = it.next(); if(x != p) { dfs(x, c, tree,t,tin,tout,arr); } } tout[c] = cc; } public class Segment { private Node[] tree; private int[] lazy; private int size; private int n; private int[] base; private class Node { private int on; private int off; } public Segment(int n, int[] arr) { this.base=arr; int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); size = 2 * (int) Math.pow(2, x) - 1; tree = new Node[size]; lazy = new int[size]; this.n = n; //this.set = set1; build(0, 0, n - 1); } public void build(int id, int l, int r) { if (l == r) { tree[id] = new Node(); if(base[l] == 1) tree[id].on++; else tree[id].off++; return; } int mid = ((l + r)>>1); build((id<<1)|1, l, mid); build((id<<1)+2, mid + 1, r); tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); //System.out.println(l+" "+r+" "+tree[id].l+" "+tree[id].r+" "+tree[id].ans); } public Node merge(Node n1, Node n2) { Node ret = new Node(); if(n1 == null && n2 == null) return null; else if(n1 == null) { ret.on = n2.on; ret.off = n2.off; } else if(n2 == null) { ret.on = n1.on; ret.off = n1.off; } else { ret.on = n1.on+n2.on; ret.off = n2.off + n1.off; } return ret; } public int query(int l, int r) { Node ret = queryUtil(l, r, 0, 0, n - 1); if(ret == null) { return 0; } else return ret.on; } private Node queryUtil(int x, int y, int id, int l, int r) { if (l > y || x > r) return null; if (x <= l && r <= y) { //shift(id,l,r); return tree[id]; } int mid = ((l + r)>>1); shift(id,l,r); Node q1 = queryUtil(x, y, (id<<1)|1, l, mid); Node q2 = queryUtil(x, y, (id<<1)+2, mid + 1, r); return merge(q1, q2); } public void update(int x, int y, int c) { update1(x, y, c, 0, 0, n-1); } private void update1(int x, int y, int colour, int id, int l, int r) { //System.out.println(l+" "+r+" "+x); if (x > r || y < l) return; if (x <= l && r <= y) { lazy[id]++; // lazy[id] %= 2; switchNode(tree[id]); return; } int mid = ((l + r)>>1); //shift(id); if(y<=mid) update1(x, y, colour, (id<<1)|1, l, mid); else if(x>mid) update1(x, y, colour, (id<<1)+2, mid + 1, r); else { update1(x, y, colour, (id<<1)|1, l, mid); update1(x, y, colour, (id<<1)+2, mid + 1, r); } tree[id] = merge(tree[(id<<1)|1], tree[(id<<1)+2]); } private void shift(int id,int l, int r) { lazy[id] %= 2; if(lazy[id] != 0) { if(l != r) { lazy[(id<<1)|1] += lazy[id]; lazy[(id<<1)+2] += lazy[id]; switchNode(tree[(id<<1)+2]); switchNode(tree[(id<<1)|1]); } //switchNode(tree[(id<<1)+2]); lazy[id] = 0; } } private void switchNode(Node d) { d.on ^= d.off; d.off ^= d.on; d.on ^= d.off; } } private void debug(Object... o) { System.out.println(Arrays.deepToString(o)); } private long pow(long a, long b, long c) { if (b == 0) return 1; long p = pow(a, b / 2, c); p = (p * p) % c; return (b % 2 == 0) ? p : (a * p) % c; } private long gcd(long n, long l) { if (l == 0) return n; return gcd(l, n % l); } public static void main(String[] args) throws Exception { new Thread(null, new Runnable() { @Override public void run() { new Main().solve(); } }, "1", 1 << 26).start(); //new Main().solve(); } public StringBuilder solve() { InputReader(System.in); /* * try { InputReader(new FileInputStream("C:\\Users\\hardik\\Desktop\\in.txt")); * } catch(FileNotFoundException e) {} */ pw = new PrintWriter(System.out); // ans_sb = new StringBuilder(); soln(); pw.close(); // System.out.println(ans_sb); return ans_sb; } public void InputReader(InputStream stream1) { stream = stream1; } private boolean isWhitespace(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } private boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } private int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } private int nextInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } private String nextToken() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } private String nextLine() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isEndOfLine(c)); return res.toString(); } private int[] nextIntArray(int n) { int[] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i] = nextInt(); } return arr; } private long[] nextLongArray(int n) { long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = nextLong(); } return arr; } private void pArray(int[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private void pArray(long[] arr) { for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); } System.out.println(); return; } private boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return isWhitespace(c); } private char nextChar() { int c = read(); while (isSpaceChar(c)) c = read(); char c1 = (char) c; while (!isSpaceChar(c)) c = read(); return c1; } private interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
2,978
606
18
import java.util.Arrays; import java.util.Scanner; public class Solution { private static int[] a; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), m = sc.nextInt(); a = new int[101]; for (int i = 0; i < m; i++) { int type = sc.nextInt(); a[type] = a[type] + 1; } int lo=1, hi=100, max=0; while (lo <= hi) { int mid = lo + (hi - lo)/2; if (check(n, mid)) { max = mid; lo = mid+1; } else { hi = mid -1; } } System.out.println(max); } public static boolean check(int n, int target) { int result = 0; for (int i=0; i <a.length; i++) { result = result + (a[i] / target); } if (result >= n) {return true;} return false; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Solution { private static int[] a; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), m = sc.nextInt(); a = new int[101]; for (int i = 0; i < m; i++) { int type = sc.nextInt(); a[type] = a[type] + 1; } int lo=1, hi=100, max=0; while (lo <= hi) { int mid = lo + (hi - lo)/2; if (check(n, mid)) { max = mid; lo = mid+1; } else { hi = mid -1; } } System.out.println(max); } public static boolean check(int n, int target) { int result = 0; for (int i=0; i <a.length; i++) { result = result + (a[i] / target); } if (result >= n) {return true;} return false; } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Solution { private static int[] a; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), m = sc.nextInt(); a = new int[101]; for (int i = 0; i < m; i++) { int type = sc.nextInt(); a[type] = a[type] + 1; } int lo=1, hi=100, max=0; while (lo <= hi) { int mid = lo + (hi - lo)/2; if (check(n, mid)) { max = mid; lo = mid+1; } else { hi = mid -1; } } System.out.println(max); } public static boolean check(int n, int target) { int result = 0; for (int i=0; i <a.length; i++) { result = result + (a[i] / target); } if (result >= n) {return true;} return false; } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(1): The time complexity is constant to the input size n. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
576
18
3,118
import java.io.*; import java.util.*; import java.math.*; public class D5 { static int a, v, l, d; static double w; static double afterMark( int s, double w) { if (2 * s * a > v * v - w * w) { // ���� ��������� v. ����� �� return (v - w) * 1.0 / a + (s - (v * v - w * w) * 1.0 / (2 * a)) / v; } else { // ��������� double megav = Math.sqrt((2 * a * s + w * w) * 1.0); return (megav - w) / a; } } public static void main(String args[]) throws IOException { boolean online = System.getProperty("ONLINE_JUDGE") != null; Scanner in = online ? new Scanner(System.in) : new Scanner(new FileReader("input.txt")); PrintWriter out = online ? new PrintWriter(System.out) : new PrintWriter(new FileWriter("output.txt")); a = in.nextInt(); v = in.nextInt(); l = in.nextInt(); d = in.nextInt(); w = (double) in.nextInt(); double t,t1,t2; if (v > w) { // ����� ��������� ������, ����� ����� ��� // ������� ������ �� ����� if (2 * d * a > 2 * v * v - w * w) { // ���� ��������� v. ����� �� t1 = (2 * v - w) * 1.0 / a + (d - (2 * v * v - w * w) * 1.0 / (2 * a)) / v; } else if (2 * d * a > w * w) { // ���� �� ���������, �� ���� � ���� - ��������� double topv = Math.sqrt(d * a + w * w * 1.0 / 2); t1 = (2 * topv - w) * 1.0 / a; } else { // ���� ����������� �� ������ ����� - �����, ��������� t1 = Math.sqrt(2 * d * 1.0 / a); w = Math.sqrt(2 * a * d * 1.0); } // ������� ������ ����� ����� t2 = afterMark(l - d, w); t = t1 + t2; } else { t = afterMark(l, 0.0); } out.println(t); out.flush(); return; } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class D5 { static int a, v, l, d; static double w; static double afterMark( int s, double w) { if (2 * s * a > v * v - w * w) { // ���� ��������� v. ����� �� return (v - w) * 1.0 / a + (s - (v * v - w * w) * 1.0 / (2 * a)) / v; } else { // ��������� double megav = Math.sqrt((2 * a * s + w * w) * 1.0); return (megav - w) / a; } } public static void main(String args[]) throws IOException { boolean online = System.getProperty("ONLINE_JUDGE") != null; Scanner in = online ? new Scanner(System.in) : new Scanner(new FileReader("input.txt")); PrintWriter out = online ? new PrintWriter(System.out) : new PrintWriter(new FileWriter("output.txt")); a = in.nextInt(); v = in.nextInt(); l = in.nextInt(); d = in.nextInt(); w = (double) in.nextInt(); double t,t1,t2; if (v > w) { // ����� ��������� ������, ����� ����� ��� // ������� ������ �� ����� if (2 * d * a > 2 * v * v - w * w) { // ���� ��������� v. ����� �� t1 = (2 * v - w) * 1.0 / a + (d - (2 * v * v - w * w) * 1.0 / (2 * a)) / v; } else if (2 * d * a > w * w) { // ���� �� ���������, �� ���� � ���� - ��������� double topv = Math.sqrt(d * a + w * w * 1.0 / 2); t1 = (2 * topv - w) * 1.0 / a; } else { // ���� ����������� �� ������ ����� - �����, ��������� t1 = Math.sqrt(2 * d * 1.0 / a); w = Math.sqrt(2 * a * d * 1.0); } // ������� ������ ����� ����� t2 = afterMark(l - d, w); t = t1 + t2; } else { t = afterMark(l, 0.0); } out.println(t); out.flush(); return; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.*; import java.math.*; public class D5 { static int a, v, l, d; static double w; static double afterMark( int s, double w) { if (2 * s * a > v * v - w * w) { // ���� ��������� v. ����� �� return (v - w) * 1.0 / a + (s - (v * v - w * w) * 1.0 / (2 * a)) / v; } else { // ��������� double megav = Math.sqrt((2 * a * s + w * w) * 1.0); return (megav - w) / a; } } public static void main(String args[]) throws IOException { boolean online = System.getProperty("ONLINE_JUDGE") != null; Scanner in = online ? new Scanner(System.in) : new Scanner(new FileReader("input.txt")); PrintWriter out = online ? new PrintWriter(System.out) : new PrintWriter(new FileWriter("output.txt")); a = in.nextInt(); v = in.nextInt(); l = in.nextInt(); d = in.nextInt(); w = (double) in.nextInt(); double t,t1,t2; if (v > w) { // ����� ��������� ������, ����� ����� ��� // ������� ������ �� ����� if (2 * d * a > 2 * v * v - w * w) { // ���� ��������� v. ����� �� t1 = (2 * v - w) * 1.0 / a + (d - (2 * v * v - w * w) * 1.0 / (2 * a)) / v; } else if (2 * d * a > w * w) { // ���� �� ���������, �� ���� � ���� - ��������� double topv = Math.sqrt(d * a + w * w * 1.0 / 2); t1 = (2 * topv - w) * 1.0 / a; } else { // ���� ����������� �� ������ ����� - �����, ��������� t1 = Math.sqrt(2 * d * 1.0 / a); w = Math.sqrt(2 * a * d * 1.0); } // ������� ������ ����� ����� t2 = afterMark(l - d, w); t = t1 + t2; } else { t = afterMark(l, 0.0); } out.println(t); out.flush(); return; } } </CODE> <EVALUATION_RUBRIC> - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
938
3,112
4,048
import java.io.*; import java.util.ArrayList; import java.util.Arrays; import java.util.List; import java.util.StringTokenizer; public class realfast implements Runnable { private static final int INF = (int) 1e9; int time[]= new int[15]; int g[]= new int [15]; public void solve() throws IOException { int n = readInt(); int t = readInt(); for(int i=0;i<n;i++) { time[i]=readInt(); g[i]=readInt(); } long dp[][]= new long [(int)Math.pow(2,n)][4]; for(int i =0;i<(int)Math.pow(2,n);i++) { for(int j=0;j<=3;j++) dp[i][j]=-1; } long val = cal(dp,0,0,0,t); out.println(val); } public long cal(long dp[][], int mask , int genre , int t ,int T ) { int val = dp.length; if(t>T) { return 0; } if(t==T) { return 1; } if(dp[mask][genre]!=-1) return dp[mask][genre]; dp[mask][genre]=0; int i=1; int count=0; while(i<val) { if((i&mask)==0&&g[count]!=genre) { dp[mask][genre] = ((dp[mask][genre]%1000000007)+ cal(dp,mask|i,g[count],t+time[count],T)%1000000007)%1000000007; } i=2*i; count++; } return dp[mask][genre]; } /////////////////////////////////////////////////////////////////////////////////////////////////////////////////// public static void main(String[] args) { new Thread(null, new realfast(), "", 128 * (1L << 20)).start(); } private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null; private BufferedReader reader; private StringTokenizer tokenizer; private PrintWriter out; @Override public void run() { try { if (ONLINE_JUDGE || !new File("input.txt").exists()) { reader = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } else { reader = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter("output.txt"); } solve(); } catch (IOException e) { throw new RuntimeException(e); } finally { try { reader.close(); } catch (IOException e) { // nothing } out.close(); } } private String readString() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } @SuppressWarnings("unused") private int readInt() throws IOException { return Integer.parseInt(readString()); } @SuppressWarnings("unused") private long readLong() throws IOException { return Long.parseLong(readString()); } @SuppressWarnings("unused") private double readDouble() throws IOException { return Double.parseDouble(readString()); } } class edge implements Comparable<edge>{ int u ; int v ; int val; edge(int u1, int v1 , int val1) { this.u=u1; this.v=v1; this.val=val1; } public int compareTo(edge e) { return this.val-e.val; } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.ArrayList; import java.util.Arrays; import java.util.List; import java.util.StringTokenizer; public class realfast implements Runnable { private static final int INF = (int) 1e9; int time[]= new int[15]; int g[]= new int [15]; public void solve() throws IOException { int n = readInt(); int t = readInt(); for(int i=0;i<n;i++) { time[i]=readInt(); g[i]=readInt(); } long dp[][]= new long [(int)Math.pow(2,n)][4]; for(int i =0;i<(int)Math.pow(2,n);i++) { for(int j=0;j<=3;j++) dp[i][j]=-1; } long val = cal(dp,0,0,0,t); out.println(val); } public long cal(long dp[][], int mask , int genre , int t ,int T ) { int val = dp.length; if(t>T) { return 0; } if(t==T) { return 1; } if(dp[mask][genre]!=-1) return dp[mask][genre]; dp[mask][genre]=0; int i=1; int count=0; while(i<val) { if((i&mask)==0&&g[count]!=genre) { dp[mask][genre] = ((dp[mask][genre]%1000000007)+ cal(dp,mask|i,g[count],t+time[count],T)%1000000007)%1000000007; } i=2*i; count++; } return dp[mask][genre]; } /////////////////////////////////////////////////////////////////////////////////////////////////////////////////// public static void main(String[] args) { new Thread(null, new realfast(), "", 128 * (1L << 20)).start(); } private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null; private BufferedReader reader; private StringTokenizer tokenizer; private PrintWriter out; @Override public void run() { try { if (ONLINE_JUDGE || !new File("input.txt").exists()) { reader = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } else { reader = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter("output.txt"); } solve(); } catch (IOException e) { throw new RuntimeException(e); } finally { try { reader.close(); } catch (IOException e) { // nothing } out.close(); } } private String readString() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } @SuppressWarnings("unused") private int readInt() throws IOException { return Integer.parseInt(readString()); } @SuppressWarnings("unused") private long readLong() throws IOException { return Long.parseLong(readString()); } @SuppressWarnings("unused") private double readDouble() throws IOException { return Double.parseDouble(readString()); } } class edge implements Comparable<edge>{ int u ; int v ; int val; edge(int u1, int v1 , int val1) { this.u=u1; this.v=v1; this.val=val1; } public int compareTo(edge e) { return this.val-e.val; } } </CODE> <EVALUATION_RUBRIC> - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(1): The execution time is unaffected by the size of the input n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.*; import java.util.ArrayList; import java.util.Arrays; import java.util.List; import java.util.StringTokenizer; public class realfast implements Runnable { private static final int INF = (int) 1e9; int time[]= new int[15]; int g[]= new int [15]; public void solve() throws IOException { int n = readInt(); int t = readInt(); for(int i=0;i<n;i++) { time[i]=readInt(); g[i]=readInt(); } long dp[][]= new long [(int)Math.pow(2,n)][4]; for(int i =0;i<(int)Math.pow(2,n);i++) { for(int j=0;j<=3;j++) dp[i][j]=-1; } long val = cal(dp,0,0,0,t); out.println(val); } public long cal(long dp[][], int mask , int genre , int t ,int T ) { int val = dp.length; if(t>T) { return 0; } if(t==T) { return 1; } if(dp[mask][genre]!=-1) return dp[mask][genre]; dp[mask][genre]=0; int i=1; int count=0; while(i<val) { if((i&mask)==0&&g[count]!=genre) { dp[mask][genre] = ((dp[mask][genre]%1000000007)+ cal(dp,mask|i,g[count],t+time[count],T)%1000000007)%1000000007; } i=2*i; count++; } return dp[mask][genre]; } /////////////////////////////////////////////////////////////////////////////////////////////////////////////////// public static void main(String[] args) { new Thread(null, new realfast(), "", 128 * (1L << 20)).start(); } private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null; private BufferedReader reader; private StringTokenizer tokenizer; private PrintWriter out; @Override public void run() { try { if (ONLINE_JUDGE || !new File("input.txt").exists()) { reader = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } else { reader = new BufferedReader(new FileReader("input.txt")); out = new PrintWriter("output.txt"); } solve(); } catch (IOException e) { throw new RuntimeException(e); } finally { try { reader.close(); } catch (IOException e) { // nothing } out.close(); } } private String readString() throws IOException { while (tokenizer == null || !tokenizer.hasMoreTokens()) { tokenizer = new StringTokenizer(reader.readLine()); } return tokenizer.nextToken(); } @SuppressWarnings("unused") private int readInt() throws IOException { return Integer.parseInt(readString()); } @SuppressWarnings("unused") private long readLong() throws IOException { return Long.parseLong(readString()); } @SuppressWarnings("unused") private double readDouble() throws IOException { return Double.parseDouble(readString()); } } class edge implements Comparable<edge>{ int u ; int v ; int val; edge(int u1, int v1 , int val1) { this.u=u1; this.v=v1; this.val=val1; } public int compareTo(edge e) { return this.val-e.val; } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,160
4,037
2,159
/* ID: andrew42 LANG: JAVA TASK: PROG: */ import java.io.*; import java.util.*; import java.lang.*; public class curling { public static void main(String[] args) throws IOException { Scanner input = new Scanner(System.in); int numD = input.nextInt(); double rad = input.nextInt(); int[] xC = new int[numD]; for (int i = 0; i < numD; i++){ xC[i] = input.nextInt(); } double[] maxY = new double[1001]; for (int i = 0; i < numD; i++){ double h = rad; for (int j = Math.max(1, xC[i]-(int)(2*rad)); j <= Math.min(1000, xC[i]+2*rad); j++){ if (maxY[j] > 0){ h = Math.max(h, Math.sqrt(4*rad*rad-(j-xC[i])*(j-xC[i]))+maxY[j]); } } System.out.print(h + " "); maxY[xC[i]] = h; } } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* ID: andrew42 LANG: JAVA TASK: PROG: */ import java.io.*; import java.util.*; import java.lang.*; public class curling { public static void main(String[] args) throws IOException { Scanner input = new Scanner(System.in); int numD = input.nextInt(); double rad = input.nextInt(); int[] xC = new int[numD]; for (int i = 0; i < numD; i++){ xC[i] = input.nextInt(); } double[] maxY = new double[1001]; for (int i = 0; i < numD; i++){ double h = rad; for (int j = Math.max(1, xC[i]-(int)(2*rad)); j <= Math.min(1000, xC[i]+2*rad); j++){ if (maxY[j] > 0){ h = Math.max(h, Math.sqrt(4*rad*rad-(j-xC[i])*(j-xC[i]))+maxY[j]); } } System.out.print(h + " "); maxY[xC[i]] = h; } } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> /* ID: andrew42 LANG: JAVA TASK: PROG: */ import java.io.*; import java.util.*; import java.lang.*; public class curling { public static void main(String[] args) throws IOException { Scanner input = new Scanner(System.in); int numD = input.nextInt(); double rad = input.nextInt(); int[] xC = new int[numD]; for (int i = 0; i < numD; i++){ xC[i] = input.nextInt(); } double[] maxY = new double[1001]; for (int i = 0; i < numD; i++){ double h = rad; for (int j = Math.max(1, xC[i]-(int)(2*rad)); j <= Math.min(1000, xC[i]+2*rad); j++){ if (maxY[j] > 0){ h = Math.max(h, Math.sqrt(4*rad*rad-(j-xC[i])*(j-xC[i]))+maxY[j]); } } System.out.print(h + " "); maxY[xC[i]] = h; } } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit to any of the given categories or is ambiguous. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
612
2,155
664
import java.util.*; public class Main { public static void main(String[] args) { Main iq = new Main(); Scanner sc = new Scanner(System.in); int n; n = sc.nextInt(); int[] naturalNumbers = new int[n]; for (int i = 0; i < naturalNumbers.length; i++) { naturalNumbers[i] = sc.nextInt(); } System.out.println(iq.diffInEvenness(n, naturalNumbers)); } public int diffInEvenness(int n, int[] naturalNumbers) { int even, odd, lastEvenIndex, lastOddIndex; even = odd = lastEvenIndex = lastOddIndex = 0; for (int i = 0; i < naturalNumbers.length; i++) { if((naturalNumbers[i] % 2) == 0) { even++; lastEvenIndex = i + 1; } else { odd++; lastOddIndex = i + 1; } } return (even > odd ? lastOddIndex : lastEvenIndex); } }
O(n)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; public class Main { public static void main(String[] args) { Main iq = new Main(); Scanner sc = new Scanner(System.in); int n; n = sc.nextInt(); int[] naturalNumbers = new int[n]; for (int i = 0; i < naturalNumbers.length; i++) { naturalNumbers[i] = sc.nextInt(); } System.out.println(iq.diffInEvenness(n, naturalNumbers)); } public int diffInEvenness(int n, int[] naturalNumbers) { int even, odd, lastEvenIndex, lastOddIndex; even = odd = lastEvenIndex = lastOddIndex = 0; for (int i = 0; i < naturalNumbers.length; i++) { if((naturalNumbers[i] % 2) == 0) { even++; lastEvenIndex = i + 1; } else { odd++; lastOddIndex = i + 1; } } return (even > odd ? lastOddIndex : lastEvenIndex); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; public class Main { public static void main(String[] args) { Main iq = new Main(); Scanner sc = new Scanner(System.in); int n; n = sc.nextInt(); int[] naturalNumbers = new int[n]; for (int i = 0; i < naturalNumbers.length; i++) { naturalNumbers[i] = sc.nextInt(); } System.out.println(iq.diffInEvenness(n, naturalNumbers)); } public int diffInEvenness(int n, int[] naturalNumbers) { int even, odd, lastEvenIndex, lastOddIndex; even = odd = lastEvenIndex = lastOddIndex = 0; for (int i = 0; i < naturalNumbers.length; i++) { if((naturalNumbers[i] % 2) == 0) { even++; lastEvenIndex = i + 1; } else { odd++; lastOddIndex = i + 1; } } return (even > odd ? lastOddIndex : lastEvenIndex); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n^2): The running time increases with the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
570
663
3,547
// upsolve with rainboy import java.io.*; import java.util.*; public class CF1187G extends PrintWriter { CF1187G() { super(System.out); } static class Scanner { Scanner(InputStream in) { this.in = in; } InputStream in; int k, l; byte[] bb = new byte[1 << 15]; byte getc() { if (k >= l) { k = 0; try { l = in.read(bb); } catch (IOException e) { l = 0; } if (l <= 0) return -1; } return bb[k++]; } int nextInt() { byte c = 0; while (c <= 32) c = getc(); int a = 0; while (c > 32) { a = a * 10 + c - '0'; c = getc(); } return a; } } Scanner sc = new Scanner(System.in); public static void main(String[] $) { CF1187G o = new CF1187G(); o.main(); o.flush(); } static final int INF = 0x3f3f3f3f; ArrayList[] aa_; int n_, m_; int[] pi, dd, bb; int[] uu, vv, uv, cost; boolean[] iq; int[] cc; void init() { aa_ = new ArrayList[n_]; for (int u = 0; u < n_; u++) aa_[u] = new ArrayList<Integer>(); pi = new int[n_]; dd = new int[n_]; bb = new int[n_]; iq = new boolean[n_]; uu = new int[m_]; vv = new int[m_]; uv = new int[m_]; cost = new int[m_]; cc = new int[m_ * 2]; m_ = 0; } void link(int u, int v, int cap, int cos) { int h = m_++; uu[h] = u; vv[h] = v; uv[h] = u ^ v; cost[h] = cos; cc[h << 1 ^ 0] = cap; aa_[u].add(h << 1 ^ 0); aa_[v].add(h << 1 ^ 1); } boolean dijkstra(int s, int t) { Arrays.fill(pi, INF); pi[s] = 0; TreeSet<Integer> pq = new TreeSet<>((u, v) -> pi[u] != pi[v] ? pi[u] - pi[v] : dd[u] != dd[v] ? dd[u] - dd[v] : u - v); pq.add(s); iq[s] = true; Integer first; while ((first = pq.pollFirst()) != null) { int u = first; iq[u] = false; int d = dd[u] + 1; ArrayList<Integer> adj = aa_[u]; for (int h_ : adj) if (cc[h_] > 0) { int h = h_ >> 1; int p = pi[u] + ((h_ & 1) == 0 ? cost[h] : -cost[h]); int v = u ^ uv[h]; if (pi[v] > p || pi[v] == p && dd[v] > d) { if (iq[v]) { pq.remove(v); iq[v] = false; } pi[v] = p; dd[v] = d; bb[v] = h_; pq.add(v); iq[v] = true; } } } return pi[t] != INF; } void push(int s, int t) { int c = INF; for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; c = Math.min(c, cc[h_]); } for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_] -= c; cc[h_ ^ 1] += c; } } void push1(int s, int t) { for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_]--; cc[h_ ^ 1]++; } } int edmonds_karp(int s, int t) { while (dijkstra(s, t)) push1(s, t); int c = 0; for (int h = 0; h < m_; h++) c += cost[h] * cc[h << 1 ^ 1]; return c; } void main() { int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt(); int c = sc.nextInt(); int d = sc.nextInt(); int[] ii = new int[k]; for (int h = 0; h < k; h++) ii[h] = sc.nextInt() - 1; ArrayList[] aa = new ArrayList[n]; for (int i = 0; i < n; i++) aa[i] = new ArrayList<Integer>(); for (int h = 0; h < m; h++) { int i = sc.nextInt() - 1; int j = sc.nextInt() - 1; aa[i].add(j); aa[j].add(i); } int t = n + k + 1; n_ = n * t + 1; m_ = k + (m * 2 * k + n) * (t - 1); init(); for (int i = 0; i < n; i++) { ArrayList<Integer> adj = aa[i]; for (int s = 0; s < t - 1; s++) { int u = i * t + s; for (int j : adj) { int v = j * t + s + 1; for (int x = 1; x <= k; x++) link(u, v, 1, c + (x * 2 - 1) * d); } } } for (int i = 0; i < n; i++) for (int s = 0; s < t - 1; s++) { int u = i * t + s, v = u + 1; link(u, v, k, i == 0 ? 0 : c); } for (int h = 0; h < k; h++) link(n_ - 1, ii[h] * t + 0, 1, 0); println(edmonds_karp(n_ - 1, 0 * t + t - 1)); } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> // upsolve with rainboy import java.io.*; import java.util.*; public class CF1187G extends PrintWriter { CF1187G() { super(System.out); } static class Scanner { Scanner(InputStream in) { this.in = in; } InputStream in; int k, l; byte[] bb = new byte[1 << 15]; byte getc() { if (k >= l) { k = 0; try { l = in.read(bb); } catch (IOException e) { l = 0; } if (l <= 0) return -1; } return bb[k++]; } int nextInt() { byte c = 0; while (c <= 32) c = getc(); int a = 0; while (c > 32) { a = a * 10 + c - '0'; c = getc(); } return a; } } Scanner sc = new Scanner(System.in); public static void main(String[] $) { CF1187G o = new CF1187G(); o.main(); o.flush(); } static final int INF = 0x3f3f3f3f; ArrayList[] aa_; int n_, m_; int[] pi, dd, bb; int[] uu, vv, uv, cost; boolean[] iq; int[] cc; void init() { aa_ = new ArrayList[n_]; for (int u = 0; u < n_; u++) aa_[u] = new ArrayList<Integer>(); pi = new int[n_]; dd = new int[n_]; bb = new int[n_]; iq = new boolean[n_]; uu = new int[m_]; vv = new int[m_]; uv = new int[m_]; cost = new int[m_]; cc = new int[m_ * 2]; m_ = 0; } void link(int u, int v, int cap, int cos) { int h = m_++; uu[h] = u; vv[h] = v; uv[h] = u ^ v; cost[h] = cos; cc[h << 1 ^ 0] = cap; aa_[u].add(h << 1 ^ 0); aa_[v].add(h << 1 ^ 1); } boolean dijkstra(int s, int t) { Arrays.fill(pi, INF); pi[s] = 0; TreeSet<Integer> pq = new TreeSet<>((u, v) -> pi[u] != pi[v] ? pi[u] - pi[v] : dd[u] != dd[v] ? dd[u] - dd[v] : u - v); pq.add(s); iq[s] = true; Integer first; while ((first = pq.pollFirst()) != null) { int u = first; iq[u] = false; int d = dd[u] + 1; ArrayList<Integer> adj = aa_[u]; for (int h_ : adj) if (cc[h_] > 0) { int h = h_ >> 1; int p = pi[u] + ((h_ & 1) == 0 ? cost[h] : -cost[h]); int v = u ^ uv[h]; if (pi[v] > p || pi[v] == p && dd[v] > d) { if (iq[v]) { pq.remove(v); iq[v] = false; } pi[v] = p; dd[v] = d; bb[v] = h_; pq.add(v); iq[v] = true; } } } return pi[t] != INF; } void push(int s, int t) { int c = INF; for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; c = Math.min(c, cc[h_]); } for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_] -= c; cc[h_ ^ 1] += c; } } void push1(int s, int t) { for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_]--; cc[h_ ^ 1]++; } } int edmonds_karp(int s, int t) { while (dijkstra(s, t)) push1(s, t); int c = 0; for (int h = 0; h < m_; h++) c += cost[h] * cc[h << 1 ^ 1]; return c; } void main() { int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt(); int c = sc.nextInt(); int d = sc.nextInt(); int[] ii = new int[k]; for (int h = 0; h < k; h++) ii[h] = sc.nextInt() - 1; ArrayList[] aa = new ArrayList[n]; for (int i = 0; i < n; i++) aa[i] = new ArrayList<Integer>(); for (int h = 0; h < m; h++) { int i = sc.nextInt() - 1; int j = sc.nextInt() - 1; aa[i].add(j); aa[j].add(i); } int t = n + k + 1; n_ = n * t + 1; m_ = k + (m * 2 * k + n) * (t - 1); init(); for (int i = 0; i < n; i++) { ArrayList<Integer> adj = aa[i]; for (int s = 0; s < t - 1; s++) { int u = i * t + s; for (int j : adj) { int v = j * t + s + 1; for (int x = 1; x <= k; x++) link(u, v, 1, c + (x * 2 - 1) * d); } } } for (int i = 0; i < n; i++) for (int s = 0; s < t - 1; s++) { int u = i * t + s, v = u + 1; link(u, v, k, i == 0 ? 0 : c); } for (int h = 0; h < k; h++) link(n_ - 1, ii[h] * t + 0, 1, 0); println(edmonds_karp(n_ - 1, 0 * t + t - 1)); } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> // upsolve with rainboy import java.io.*; import java.util.*; public class CF1187G extends PrintWriter { CF1187G() { super(System.out); } static class Scanner { Scanner(InputStream in) { this.in = in; } InputStream in; int k, l; byte[] bb = new byte[1 << 15]; byte getc() { if (k >= l) { k = 0; try { l = in.read(bb); } catch (IOException e) { l = 0; } if (l <= 0) return -1; } return bb[k++]; } int nextInt() { byte c = 0; while (c <= 32) c = getc(); int a = 0; while (c > 32) { a = a * 10 + c - '0'; c = getc(); } return a; } } Scanner sc = new Scanner(System.in); public static void main(String[] $) { CF1187G o = new CF1187G(); o.main(); o.flush(); } static final int INF = 0x3f3f3f3f; ArrayList[] aa_; int n_, m_; int[] pi, dd, bb; int[] uu, vv, uv, cost; boolean[] iq; int[] cc; void init() { aa_ = new ArrayList[n_]; for (int u = 0; u < n_; u++) aa_[u] = new ArrayList<Integer>(); pi = new int[n_]; dd = new int[n_]; bb = new int[n_]; iq = new boolean[n_]; uu = new int[m_]; vv = new int[m_]; uv = new int[m_]; cost = new int[m_]; cc = new int[m_ * 2]; m_ = 0; } void link(int u, int v, int cap, int cos) { int h = m_++; uu[h] = u; vv[h] = v; uv[h] = u ^ v; cost[h] = cos; cc[h << 1 ^ 0] = cap; aa_[u].add(h << 1 ^ 0); aa_[v].add(h << 1 ^ 1); } boolean dijkstra(int s, int t) { Arrays.fill(pi, INF); pi[s] = 0; TreeSet<Integer> pq = new TreeSet<>((u, v) -> pi[u] != pi[v] ? pi[u] - pi[v] : dd[u] != dd[v] ? dd[u] - dd[v] : u - v); pq.add(s); iq[s] = true; Integer first; while ((first = pq.pollFirst()) != null) { int u = first; iq[u] = false; int d = dd[u] + 1; ArrayList<Integer> adj = aa_[u]; for (int h_ : adj) if (cc[h_] > 0) { int h = h_ >> 1; int p = pi[u] + ((h_ & 1) == 0 ? cost[h] : -cost[h]); int v = u ^ uv[h]; if (pi[v] > p || pi[v] == p && dd[v] > d) { if (iq[v]) { pq.remove(v); iq[v] = false; } pi[v] = p; dd[v] = d; bb[v] = h_; pq.add(v); iq[v] = true; } } } return pi[t] != INF; } void push(int s, int t) { int c = INF; for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; c = Math.min(c, cc[h_]); } for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_] -= c; cc[h_ ^ 1] += c; } } void push1(int s, int t) { for (int u = t, h_, h; u != s; u ^= uv[h]) { h = (h_ = bb[u]) >> 1; cc[h_]--; cc[h_ ^ 1]++; } } int edmonds_karp(int s, int t) { while (dijkstra(s, t)) push1(s, t); int c = 0; for (int h = 0; h < m_; h++) c += cost[h] * cc[h << 1 ^ 1]; return c; } void main() { int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt(); int c = sc.nextInt(); int d = sc.nextInt(); int[] ii = new int[k]; for (int h = 0; h < k; h++) ii[h] = sc.nextInt() - 1; ArrayList[] aa = new ArrayList[n]; for (int i = 0; i < n; i++) aa[i] = new ArrayList<Integer>(); for (int h = 0; h < m; h++) { int i = sc.nextInt() - 1; int j = sc.nextInt() - 1; aa[i].add(j); aa[j].add(i); } int t = n + k + 1; n_ = n * t + 1; m_ = k + (m * 2 * k + n) * (t - 1); init(); for (int i = 0; i < n; i++) { ArrayList<Integer> adj = aa[i]; for (int s = 0; s < t - 1; s++) { int u = i * t + s; for (int j : adj) { int v = j * t + s + 1; for (int x = 1; x <= k; x++) link(u, v, 1, c + (x * 2 - 1) * d); } } } for (int i = 0; i < n; i++) for (int s = 0; s < t - 1; s++) { int u = i * t + s, v = u + 1; link(u, v, k, i == 0 ? 0 : c); } for (int h = 0; h < k; h++) link(n_ - 1, ii[h] * t + 0, 1, 0); println(edmonds_karp(n_ - 1, 0 * t + t - 1)); } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The running time increases with the square of the input size n. - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,911
3,540
4,314
//package contest10D; import java.math.*; import java.util.*; import static java.math.BigInteger.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class Main { static void debug(Object... os) { System.err.println(deepToString(os)); } public static void main(String[] args) { new Main().run(); } void run() { Scanner sc = new Scanner(System.in); int n=sc.nextInt(),m=sc.nextInt(); boolean[][] bs=new boolean[n][n]; for (int i = 0; i < m; i++) { int s=sc.nextInt()-1,t=sc.nextInt()-1; bs[s][t] = bs[t][s] = true; } long res = 0; for(int i=0;i<n;i++){ res += calc(bs,n-1-i); } System.out.println(res/2); } long calc(boolean[][] bs,int n){//start is n. long[][] dp=new long[1<<n][n]; // set,last -> number of path n -> last. for(int i=0;i<n;i++){ if(bs[i][n]) dp[1<<i][i] ++; } for(int i=1;i<1<<n;i++){ for(int j=0;j<n;j++)if(((i>>j)&1)==1) for(int k=0;k<n;k++)if(((i>>k)&1)==0 && bs[j][k]){//add dp[i|(1<<k)][k] += dp[i][j]; } } long res=0; for(int i=0;i<1<<n;i++)for(int j=0;j<n;j++)if(Integer.bitCount(i)>=2&&bs[j][n])res+=dp[i][j]; return res; } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package contest10D; import java.math.*; import java.util.*; import static java.math.BigInteger.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class Main { static void debug(Object... os) { System.err.println(deepToString(os)); } public static void main(String[] args) { new Main().run(); } void run() { Scanner sc = new Scanner(System.in); int n=sc.nextInt(),m=sc.nextInt(); boolean[][] bs=new boolean[n][n]; for (int i = 0; i < m; i++) { int s=sc.nextInt()-1,t=sc.nextInt()-1; bs[s][t] = bs[t][s] = true; } long res = 0; for(int i=0;i<n;i++){ res += calc(bs,n-1-i); } System.out.println(res/2); } long calc(boolean[][] bs,int n){//start is n. long[][] dp=new long[1<<n][n]; // set,last -> number of path n -> last. for(int i=0;i<n;i++){ if(bs[i][n]) dp[1<<i][i] ++; } for(int i=1;i<1<<n;i++){ for(int j=0;j<n;j++)if(((i>>j)&1)==1) for(int k=0;k<n;k++)if(((i>>k)&1)==0 && bs[j][k]){//add dp[i|(1<<k)][k] += dp[i][j]; } } long res=0; for(int i=0;i<1<<n;i++)for(int j=0;j<n;j++)if(Integer.bitCount(i)>=2&&bs[j][n])res+=dp[i][j]; return res; } } </CODE> <EVALUATION_RUBRIC> - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package contest10D; import java.math.*; import java.util.*; import static java.math.BigInteger.*; import static java.lang.Math.*; import static java.util.Arrays.*; import static java.util.Collections.*; public class Main { static void debug(Object... os) { System.err.println(deepToString(os)); } public static void main(String[] args) { new Main().run(); } void run() { Scanner sc = new Scanner(System.in); int n=sc.nextInt(),m=sc.nextInt(); boolean[][] bs=new boolean[n][n]; for (int i = 0; i < m; i++) { int s=sc.nextInt()-1,t=sc.nextInt()-1; bs[s][t] = bs[t][s] = true; } long res = 0; for(int i=0;i<n;i++){ res += calc(bs,n-1-i); } System.out.println(res/2); } long calc(boolean[][] bs,int n){//start is n. long[][] dp=new long[1<<n][n]; // set,last -> number of path n -> last. for(int i=0;i<n;i++){ if(bs[i][n]) dp[1<<i][i] ++; } for(int i=1;i<1<<n;i++){ for(int j=0;j<n;j++)if(((i>>j)&1)==1) for(int k=0;k<n;k++)if(((i>>k)&1)==0 && bs[j][k]){//add dp[i|(1<<k)][k] += dp[i][j]; } } long res=0; for(int i=0;i<1<<n;i++)for(int j=0;j<n;j++)if(Integer.bitCount(i)>=2&&bs[j][n])res+=dp[i][j]; return res; } } </CODE> <EVALUATION_RUBRIC> - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
736
4,303
2,549
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.HashSet; import java.util.List; import java.util.StringTokenizer; public class D { int[][] fast(int n, int m){ int[][] ans = new int[2][n * m]; int c = 0; for (int left = 1, right = m; left < right; left++, right--) { for (int l = 1, r = n; l <= n && r >= 1; l++, r--) { ans[0][c] = l; ans[1][c++] = left; ans[0][c] = r; ans[1][c++] = right; } } if (m % 2 == 1) { int x = m/2 + 1; for(int l = 1, r = n;l < r;l++, r--){ ans[0][c] = l; ans[1][c++] = x; ans[0][c] = r; ans[1][c++] = x; if(n % 2 == 1 && l + 2 == r){ ans[0][c] = l+1; ans[1][c++] = x; } } } if(n == 1 && m % 2 == 1){ ans[0][c] = 1; ans[1][c] = m/2 + 1; } return ans; } void stress(){ for(int i = 3;i<=5;i++){ for(int j = 2;j<=5;j++){ int[][] ans = new int[2][]; try{ ans = fast(i, j); }catch(Exception e){ out.println("ошибка"); out.print(i + " " + j); return; } boolean[][] check = new boolean[i][j]; for(int c = 0;c<ans[0].length;c++){ int x = ans[0][c] - 1; int y = ans[1][c] - 1; check[x][y] = true; } for(int c = 0;c<i;c++){ for(int q = 0;q<j;q++){ if(!check[c][q]){ out.println(i + " " + j); out.println("точки"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } HashSet<String> set = new HashSet<>(); for(int c = 1;c<ans[0].length;c++){ int x = ans[0][c] - ans[0][c- 1]; int y = ans[1][c] - ans[1][c - 1]; set.add(x + " " + y); } if(set.size() < i * j - 1){ out.println(i + " " + j); out.println("вектора"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } } void normal(){ int n =readInt(); int m = readInt(); int[][] ans = fast(n, m); for(int i = 0;i<ans[0].length;i++){ out.println(ans[0][i] + " " + ans[1][i]); } } boolean stress = false; void solve(){ if(stress) stress(); else normal(); } public static void main(String[] args) { new D().run(); } void run(){ init(); solve(); out.close(); } BufferedReader in; PrintWriter out; StringTokenizer tok = new StringTokenizer(""); void init(){ in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } String readLine(){ try{ return in.readLine(); }catch(Exception ex){ throw new RuntimeException(ex); } } String readString(){ while(!tok.hasMoreTokens()){ String nextLine = readLine(); if(nextLine == null) return null; tok = new StringTokenizer(nextLine); } return tok.nextToken(); } int readInt(){ return Integer.parseInt(readString()); } long readLong(){ return Long.parseLong(readString()); } double readDouble(){ return Double.parseDouble(readString()); } }
O(n^2)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.HashSet; import java.util.List; import java.util.StringTokenizer; public class D { int[][] fast(int n, int m){ int[][] ans = new int[2][n * m]; int c = 0; for (int left = 1, right = m; left < right; left++, right--) { for (int l = 1, r = n; l <= n && r >= 1; l++, r--) { ans[0][c] = l; ans[1][c++] = left; ans[0][c] = r; ans[1][c++] = right; } } if (m % 2 == 1) { int x = m/2 + 1; for(int l = 1, r = n;l < r;l++, r--){ ans[0][c] = l; ans[1][c++] = x; ans[0][c] = r; ans[1][c++] = x; if(n % 2 == 1 && l + 2 == r){ ans[0][c] = l+1; ans[1][c++] = x; } } } if(n == 1 && m % 2 == 1){ ans[0][c] = 1; ans[1][c] = m/2 + 1; } return ans; } void stress(){ for(int i = 3;i<=5;i++){ for(int j = 2;j<=5;j++){ int[][] ans = new int[2][]; try{ ans = fast(i, j); }catch(Exception e){ out.println("ошибка"); out.print(i + " " + j); return; } boolean[][] check = new boolean[i][j]; for(int c = 0;c<ans[0].length;c++){ int x = ans[0][c] - 1; int y = ans[1][c] - 1; check[x][y] = true; } for(int c = 0;c<i;c++){ for(int q = 0;q<j;q++){ if(!check[c][q]){ out.println(i + " " + j); out.println("точки"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } HashSet<String> set = new HashSet<>(); for(int c = 1;c<ans[0].length;c++){ int x = ans[0][c] - ans[0][c- 1]; int y = ans[1][c] - ans[1][c - 1]; set.add(x + " " + y); } if(set.size() < i * j - 1){ out.println(i + " " + j); out.println("вектора"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } } void normal(){ int n =readInt(); int m = readInt(); int[][] ans = fast(n, m); for(int i = 0;i<ans[0].length;i++){ out.println(ans[0][i] + " " + ans[1][i]); } } boolean stress = false; void solve(){ if(stress) stress(); else normal(); } public static void main(String[] args) { new D().run(); } void run(){ init(); solve(); out.close(); } BufferedReader in; PrintWriter out; StringTokenizer tok = new StringTokenizer(""); void init(){ in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } String readLine(){ try{ return in.readLine(); }catch(Exception ex){ throw new RuntimeException(ex); } } String readString(){ while(!tok.hasMoreTokens()){ String nextLine = readLine(); if(nextLine == null) return null; tok = new StringTokenizer(nextLine); } return tok.nextToken(); } int readInt(){ return Integer.parseInt(readString()); } long readLong(){ return Long.parseLong(readString()); } double readDouble(){ return Double.parseDouble(readString()); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n): The running time grows linearly with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.HashSet; import java.util.List; import java.util.StringTokenizer; public class D { int[][] fast(int n, int m){ int[][] ans = new int[2][n * m]; int c = 0; for (int left = 1, right = m; left < right; left++, right--) { for (int l = 1, r = n; l <= n && r >= 1; l++, r--) { ans[0][c] = l; ans[1][c++] = left; ans[0][c] = r; ans[1][c++] = right; } } if (m % 2 == 1) { int x = m/2 + 1; for(int l = 1, r = n;l < r;l++, r--){ ans[0][c] = l; ans[1][c++] = x; ans[0][c] = r; ans[1][c++] = x; if(n % 2 == 1 && l + 2 == r){ ans[0][c] = l+1; ans[1][c++] = x; } } } if(n == 1 && m % 2 == 1){ ans[0][c] = 1; ans[1][c] = m/2 + 1; } return ans; } void stress(){ for(int i = 3;i<=5;i++){ for(int j = 2;j<=5;j++){ int[][] ans = new int[2][]; try{ ans = fast(i, j); }catch(Exception e){ out.println("ошибка"); out.print(i + " " + j); return; } boolean[][] check = new boolean[i][j]; for(int c = 0;c<ans[0].length;c++){ int x = ans[0][c] - 1; int y = ans[1][c] - 1; check[x][y] = true; } for(int c = 0;c<i;c++){ for(int q = 0;q<j;q++){ if(!check[c][q]){ out.println(i + " " + j); out.println("точки"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } HashSet<String> set = new HashSet<>(); for(int c = 1;c<ans[0].length;c++){ int x = ans[0][c] - ans[0][c- 1]; int y = ans[1][c] - ans[1][c - 1]; set.add(x + " " + y); } if(set.size() < i * j - 1){ out.println(i + " " + j); out.println("вектора"); for(int w = 0;w<ans[0].length;w++){ out.println(ans[0][w] + " " + ans[1][w]); } return; } } } } void normal(){ int n =readInt(); int m = readInt(); int[][] ans = fast(n, m); for(int i = 0;i<ans[0].length;i++){ out.println(ans[0][i] + " " + ans[1][i]); } } boolean stress = false; void solve(){ if(stress) stress(); else normal(); } public static void main(String[] args) { new D().run(); } void run(){ init(); solve(); out.close(); } BufferedReader in; PrintWriter out; StringTokenizer tok = new StringTokenizer(""); void init(){ in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } String readLine(){ try{ return in.readLine(); }catch(Exception ex){ throw new RuntimeException(ex); } } String readString(){ while(!tok.hasMoreTokens()){ String nextLine = readLine(); if(nextLine == null) return null; tok = new StringTokenizer(nextLine); } return tok.nextToken(); } int readInt(){ return Integer.parseInt(readString()); } long readLong(){ return Long.parseLong(readString()); } double readDouble(){ return Double.parseDouble(readString()); } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit to any of the given categories or is ambiguous. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(1): The execution time is unaffected by the size of the input n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,346
2,543
4,233
import java.util.*; public class Main { public static class Pair implements Comparable<Pair> { int k, x; public Pair(int k) { this.k = k; } public void update(int x) { this.x = Math.max(this.x, x); } public int compareTo(Pair other) { if (x != other.x) { return other.x - x; } return k - other.k; } } public static int sum(int[] arr) { int sum = 0; for (int x : arr) { sum += x; } return sum; } public static int[] join(int[] a, int[] b) { int n = a.length; int[] best = new int[n]; int sum = 0; for (int shift = 0; shift < n; shift++) { int[] curr = new int[n]; for (int i = 0; i < n; i++) { curr[i] = Math.max(a[i], b[(i + shift) % n]); } int now = sum(curr); if (now > sum) { sum = now; best = curr; } } return best; } public static int n; public static int[] pow; public static int[][] dp, real; public static void calc(int mask) { int[] best = new int[n]; int sum = 0; for (int i = 0; i < n; i++) { if ((mask & pow[i]) != 0) { int to = mask ^ pow[i]; int[] init = new int[n]; for (int j = 0; j < n; j++) { init[j] = real[j][i]; } int[] curr = join(dp[to], init); int s = sum(curr); if (s > sum) { sum = s; best = curr; } } } dp[mask] = best; } public static void main(String[] args) { pow = new int[15]; pow[0] = 1; for (int i = 1; i < pow.length; i++) { pow[i] = pow[i - 1] * 2; } Scanner in = new Scanner(System.in); int t = in.nextInt(); for (int i = 0; i < t; i++) { n = in.nextInt(); int m = in.nextInt(); int[][] arr = new int[n][m]; for (int j = 0; j < n; j++) { for (int k = 0; k < m; k++) { arr[j][k] = in.nextInt(); } } Pair[] best = new Pair[m]; for (int j = 0; j < m; j++) { best[j] = new Pair(j); for (int k = 0; k < n; k++) { best[j].update(arr[k][j]); } } Arrays.sort(best); real = new int[n][n]; for (int j = 0; j < n; j++) { for (int k = 0; k < Math.min(n, m); k++) { real[j][k] = arr[j][best[k].k]; } } dp = new int[1 << n][]; Stack<Integer>[] min = new Stack[n + 1]; for (int j = 0; j <= n; j++) { min[j] = new Stack<>(); } for (int j = 0; j < dp.length; j++) { int cnt = 0; for (int k = 0; k < n; k++) { if ((j & pow[k]) != 0) { cnt++; } } min[cnt].add(j); } for (int j = 0; j < min.length; j++) { for (int x : min[j]) { if (j == 0) { dp[x] = new int[n]; } else { calc(x); } } } System.out.println(sum(dp[dp.length - 1])); } } }
non-polynomial
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; public class Main { public static class Pair implements Comparable<Pair> { int k, x; public Pair(int k) { this.k = k; } public void update(int x) { this.x = Math.max(this.x, x); } public int compareTo(Pair other) { if (x != other.x) { return other.x - x; } return k - other.k; } } public static int sum(int[] arr) { int sum = 0; for (int x : arr) { sum += x; } return sum; } public static int[] join(int[] a, int[] b) { int n = a.length; int[] best = new int[n]; int sum = 0; for (int shift = 0; shift < n; shift++) { int[] curr = new int[n]; for (int i = 0; i < n; i++) { curr[i] = Math.max(a[i], b[(i + shift) % n]); } int now = sum(curr); if (now > sum) { sum = now; best = curr; } } return best; } public static int n; public static int[] pow; public static int[][] dp, real; public static void calc(int mask) { int[] best = new int[n]; int sum = 0; for (int i = 0; i < n; i++) { if ((mask & pow[i]) != 0) { int to = mask ^ pow[i]; int[] init = new int[n]; for (int j = 0; j < n; j++) { init[j] = real[j][i]; } int[] curr = join(dp[to], init); int s = sum(curr); if (s > sum) { sum = s; best = curr; } } } dp[mask] = best; } public static void main(String[] args) { pow = new int[15]; pow[0] = 1; for (int i = 1; i < pow.length; i++) { pow[i] = pow[i - 1] * 2; } Scanner in = new Scanner(System.in); int t = in.nextInt(); for (int i = 0; i < t; i++) { n = in.nextInt(); int m = in.nextInt(); int[][] arr = new int[n][m]; for (int j = 0; j < n; j++) { for (int k = 0; k < m; k++) { arr[j][k] = in.nextInt(); } } Pair[] best = new Pair[m]; for (int j = 0; j < m; j++) { best[j] = new Pair(j); for (int k = 0; k < n; k++) { best[j].update(arr[k][j]); } } Arrays.sort(best); real = new int[n][n]; for (int j = 0; j < n; j++) { for (int k = 0; k < Math.min(n, m); k++) { real[j][k] = arr[j][best[k].k]; } } dp = new int[1 << n][]; Stack<Integer>[] min = new Stack[n + 1]; for (int j = 0; j <= n; j++) { min[j] = new Stack<>(); } for (int j = 0; j < dp.length; j++) { int cnt = 0; for (int k = 0; k < n; k++) { if ((j & pow[k]) != 0) { cnt++; } } min[cnt].add(j); } for (int j = 0; j < min.length; j++) { for (int x : min[j]) { if (j == 0) { dp[x] = new int[n]; } else { calc(x); } } } System.out.println(sum(dp[dp.length - 1])); } } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.*; public class Main { public static class Pair implements Comparable<Pair> { int k, x; public Pair(int k) { this.k = k; } public void update(int x) { this.x = Math.max(this.x, x); } public int compareTo(Pair other) { if (x != other.x) { return other.x - x; } return k - other.k; } } public static int sum(int[] arr) { int sum = 0; for (int x : arr) { sum += x; } return sum; } public static int[] join(int[] a, int[] b) { int n = a.length; int[] best = new int[n]; int sum = 0; for (int shift = 0; shift < n; shift++) { int[] curr = new int[n]; for (int i = 0; i < n; i++) { curr[i] = Math.max(a[i], b[(i + shift) % n]); } int now = sum(curr); if (now > sum) { sum = now; best = curr; } } return best; } public static int n; public static int[] pow; public static int[][] dp, real; public static void calc(int mask) { int[] best = new int[n]; int sum = 0; for (int i = 0; i < n; i++) { if ((mask & pow[i]) != 0) { int to = mask ^ pow[i]; int[] init = new int[n]; for (int j = 0; j < n; j++) { init[j] = real[j][i]; } int[] curr = join(dp[to], init); int s = sum(curr); if (s > sum) { sum = s; best = curr; } } } dp[mask] = best; } public static void main(String[] args) { pow = new int[15]; pow[0] = 1; for (int i = 1; i < pow.length; i++) { pow[i] = pow[i - 1] * 2; } Scanner in = new Scanner(System.in); int t = in.nextInt(); for (int i = 0; i < t; i++) { n = in.nextInt(); int m = in.nextInt(); int[][] arr = new int[n][m]; for (int j = 0; j < n; j++) { for (int k = 0; k < m; k++) { arr[j][k] = in.nextInt(); } } Pair[] best = new Pair[m]; for (int j = 0; j < m; j++) { best[j] = new Pair(j); for (int k = 0; k < n; k++) { best[j].update(arr[k][j]); } } Arrays.sort(best); real = new int[n][n]; for (int j = 0; j < n; j++) { for (int k = 0; k < Math.min(n, m); k++) { real[j][k] = arr[j][best[k].k]; } } dp = new int[1 << n][]; Stack<Integer>[] min = new Stack[n + 1]; for (int j = 0; j <= n; j++) { min[j] = new Stack<>(); } for (int j = 0; j < dp.length; j++) { int cnt = 0; for (int k = 0; k < n; k++) { if ((j & pow[k]) != 0) { cnt++; } } min[cnt].add(j); } for (int j = 0; j < min.length; j++) { for (int x : min[j]) { if (j == 0) { dp[x] = new int[n]; } else { calc(x); } } } System.out.println(sum(dp[dp.length - 1])); } } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - O(n): The running time grows linearly with the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(1): The running time does not change regardless of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,270
4,222
3,487
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), md = sc.nextInt(); int k = (n + 1) / 2;int ans = 0; int[][] dp = new int[k + 1][n + 1];dp[0][0] = 1; for (int h = 1; h <= k; h++) for (int l = h; l <= n - h + 1; l++) dp[h][l] = (int) ((dp[h][l - 1] * 2L + dp[h - 1][l - 1]) * h % md); for (int h = 1; h <= k; h++) ans = (ans + dp[h][n - h + 1]) % md; System.out.println(ans); } }
O(n^3)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), md = sc.nextInt(); int k = (n + 1) / 2;int ans = 0; int[][] dp = new int[k + 1][n + 1];dp[0][0] = 1; for (int h = 1; h <= k; h++) for (int l = h; l <= n - h + 1; l++) dp[h][l] = (int) ((dp[h][l - 1] * 2L + dp[h - 1][l - 1]) * h % md); for (int h = 1; h <= k; h++) ans = (ans + dp[h][n - h + 1]) % md; System.out.println(ans); } } </CODE> <EVALUATION_RUBRIC> - O(n): The running time grows linearly with the input size n. - O(n^3): The running time increases with the cube of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(1): The running time does not change regardless of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(), md = sc.nextInt(); int k = (n + 1) / 2;int ans = 0; int[][] dp = new int[k + 1][n + 1];dp[0][0] = 1; for (int h = 1; h <= k; h++) for (int l = h; l <= n - h + 1; l++) dp[h][l] = (int) ((dp[h][l - 1] * 2L + dp[h - 1][l - 1]) * h % md); for (int h = 1; h <= k; h++) ans = (ans + dp[h][n - h + 1]) % md; System.out.println(ans); } } </CODE> <EVALUATION_RUBRIC> - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(1): The time complexity is constant to the input size n. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
536
3,481
3,287
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.StringTokenizer; public class Solver { StringTokenizer st; BufferedReader in; PrintWriter out; public static void main(String[] args) throws NumberFormatException, IOException { Solver solver = new Solver(); solver.open(); solver.solve(); solver.close(); } public void open() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } public String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws NumberFormatException, IOException { return Double.parseDouble(nextToken()); } public void solve() throws NumberFormatException, IOException { int n = nextInt(); out.println(0+" "+0+" "+n); } public void close() { out.flush(); out.close(); } }
O(1)
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.StringTokenizer; public class Solver { StringTokenizer st; BufferedReader in; PrintWriter out; public static void main(String[] args) throws NumberFormatException, IOException { Solver solver = new Solver(); solver.open(); solver.solve(); solver.close(); } public void open() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } public String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws NumberFormatException, IOException { return Double.parseDouble(nextToken()); } public void solve() throws NumberFormatException, IOException { int n = nextInt(); out.println(0+" "+0+" "+n); } public void close() { out.flush(); out.close(); } } </CODE> <EVALUATION_RUBRIC> - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - O(n^2): The execution time ascends in proportion to the square of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.StringTokenizer; public class Solver { StringTokenizer st; BufferedReader in; PrintWriter out; public static void main(String[] args) throws NumberFormatException, IOException { Solver solver = new Solver(); solver.open(); solver.solve(); solver.close(); } public void open() throws IOException { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } public String nextToken() throws IOException { while (st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); } return st.nextToken(); } public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(nextToken()); } public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(nextToken()); } public double nextDouble() throws NumberFormatException, IOException { return Double.parseDouble(nextToken()); } public void solve() throws NumberFormatException, IOException { int n = nextInt(); out.println(0+" "+0+" "+n); } public void close() { out.flush(); out.close(); } } </CODE> <EVALUATION_RUBRIC> - O(1): The running time does not change regardless of the input size n. - O(n): The running time grows linearly with the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. - O(n^2): The running time increases with the square of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - O(n^3): The running time increases with the cube of the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
608
3,281
1,803
import java.util.Arrays; import java.util.Scanner; public class Main2 { public static void main(String args[]){ Scanner input = new Scanner(System.in); int n = input.nextInt(); int m = input.nextInt(); int k = input.nextInt(); int[] num = new int[n]; for(int i = 0 ; i < n ; i++){ num[i] = input.nextInt(); } System.out.println(str(n,m,k,num)); } static int str(int n,int m,int k,int[] num){ Arrays.sort(num); int total = k; int count = 0; while(k < m){ if(count == num.length)return -1; k += num[num.length-count-1]-1; count++; } return count; } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Main2 { public static void main(String args[]){ Scanner input = new Scanner(System.in); int n = input.nextInt(); int m = input.nextInt(); int k = input.nextInt(); int[] num = new int[n]; for(int i = 0 ; i < n ; i++){ num[i] = input.nextInt(); } System.out.println(str(n,m,k,num)); } static int str(int n,int m,int k,int[] num){ Arrays.sort(num); int total = k; int count = 0; while(k < m){ if(count == num.length)return -1; k += num[num.length-count-1]-1; count++; } return count; } } </CODE> <EVALUATION_RUBRIC> - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - O(n^2): The time complexity grows proportionally to the square of the input size. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(1): The time complexity is constant to the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Determine the worst-case time complexity category of a given Java code based on input size n. </TASK> <CODE> import java.util.Arrays; import java.util.Scanner; public class Main2 { public static void main(String args[]){ Scanner input = new Scanner(System.in); int n = input.nextInt(); int m = input.nextInt(); int k = input.nextInt(); int[] num = new int[n]; for(int i = 0 ; i < n ; i++){ num[i] = input.nextInt(); } System.out.println(str(n,m,k,num)); } static int str(int n,int m,int k,int[] num){ Arrays.sort(num); int total = k; int count = 0; while(k < m){ if(count == num.length)return -1; k += num[num.length-count-1]-1; count++; } return count; } } </CODE> <EVALUATION_RUBRIC> - O(n): The time complexity increases proportionally to the input size n in a linear manner. - O(nlog(n)): The time complexity combines linear and logarithmic growth factors. - non-polynomial: The time complexity is not polynomial. It grows very fast, often exponentially. - others: The time complexity does not fit any of the above categories or cannot be clearly classified. - O(1): The time complexity is constant to the input size n. - O(log(n)): The time complexity increases logarithmically in relation to the input size. - O(n^3): The time complexity scales proportionally to the cube of the input size. - O(n^2): The time complexity grows proportionally to the square of the input size. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
503
1,799
2,021
//package codeforces.br22; import java.io.*; import java.util.ArrayList; import java.util.Collections; import java.util.List; /** * User: Kandy * Date: 29.06.2010 * Time: 21:07:24 */ public class ProblemA { public void solve() { boolean oj = true; try { Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("A.in"); Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("A.out"); BufferedReader br = new BufferedReader(reader); StreamTokenizer st = new StreamTokenizer(reader); PrintWriter out = new PrintWriter(writer); MyTokenizer tok = new MyTokenizer(br.readLine()); int n = (int)tok.getNum(); tok =new MyTokenizer(br.readLine()); List<Integer> a = new ArrayList<Integer>(); for(int i=0;i<n;i++) { int r = (int)tok.getNum(); if (!a.contains(r)) { a.add(r); } } Collections.sort(a); if (a.size() < 2) { out.printf("NO"); } else { out.printf("%d", a.get(1)); } br.close(); out.close(); reader.close(); writer.close(); } catch (Exception ex) { ex.printStackTrace(); } finally { } } public static void main(String[] args) { ProblemA f = new ProblemA(); f.solve(); } private class MyTokenizer { private String s; private int cur; public MyTokenizer(String s) { this.s = s; cur = 0; } public void skip() { while (cur < s.length() && (s.charAt(cur) == ' ' || s.charAt(cur) == '\n')) { cur++; } } public double getNum() { skip(); String snum = ""; while (cur < s.length() && (s.charAt(cur) >= '0' && s.charAt(cur) <= '9' || s.charAt(cur) == '.' || s.charAt(cur) == '-')) { snum += s.charAt(cur); cur++; } return Double.valueOf(snum); } public String getString() { skip(); String s2 = ""; while (cur < s.length() && (((s.charAt(cur) >= 'a' && s.charAt(cur) <= 'z')) || ((s.charAt(cur) >= 'A' && s.charAt(cur) <= 'Z')))) { s2 += s.charAt(cur); cur++; } return s2; } public char getCurrentChar() throws Exception { if (cur < s.length()) return s.charAt(cur); else throw new Exception("Current character out of string length"); } public void moveNextChar() { if (cur < s.length()) cur++; } public boolean isFinished() { return cur >= s.length(); } } }
O(nlog(n))
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Identify the category of worst-case time complexity for the Java program, considering how algorithm performance grows with input n. </TASK> <CODE> //package codeforces.br22; import java.io.*; import java.util.ArrayList; import java.util.Collections; import java.util.List; /** * User: Kandy * Date: 29.06.2010 * Time: 21:07:24 */ public class ProblemA { public void solve() { boolean oj = true; try { Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("A.in"); Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("A.out"); BufferedReader br = new BufferedReader(reader); StreamTokenizer st = new StreamTokenizer(reader); PrintWriter out = new PrintWriter(writer); MyTokenizer tok = new MyTokenizer(br.readLine()); int n = (int)tok.getNum(); tok =new MyTokenizer(br.readLine()); List<Integer> a = new ArrayList<Integer>(); for(int i=0;i<n;i++) { int r = (int)tok.getNum(); if (!a.contains(r)) { a.add(r); } } Collections.sort(a); if (a.size() < 2) { out.printf("NO"); } else { out.printf("%d", a.get(1)); } br.close(); out.close(); reader.close(); writer.close(); } catch (Exception ex) { ex.printStackTrace(); } finally { } } public static void main(String[] args) { ProblemA f = new ProblemA(); f.solve(); } private class MyTokenizer { private String s; private int cur; public MyTokenizer(String s) { this.s = s; cur = 0; } public void skip() { while (cur < s.length() && (s.charAt(cur) == ' ' || s.charAt(cur) == '\n')) { cur++; } } public double getNum() { skip(); String snum = ""; while (cur < s.length() && (s.charAt(cur) >= '0' && s.charAt(cur) <= '9' || s.charAt(cur) == '.' || s.charAt(cur) == '-')) { snum += s.charAt(cur); cur++; } return Double.valueOf(snum); } public String getString() { skip(); String s2 = ""; while (cur < s.length() && (((s.charAt(cur) >= 'a' && s.charAt(cur) <= 'z')) || ((s.charAt(cur) >= 'A' && s.charAt(cur) <= 'Z')))) { s2 += s.charAt(cur); cur++; } return s2; } public char getCurrentChar() throws Exception { if (cur < s.length()) return s.charAt(cur); else throw new Exception("Current character out of string length"); } public void moveNextChar() { if (cur < s.length()) cur++; } public boolean isFinished() { return cur >= s.length(); } } } </CODE> <EVALUATION_RUBRIC> - O(n^2): The execution time ascends in proportion to the square of the input size n. - O(1): The execution time is unaffected by the size of the input n. - O(n): The execution time ascends in a one-to-one ratio with the input size n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The execution time ascends in proportion to the cube of the input size n. - O(nlog(n)): The execution time ascends in non-polynomial way with input size n, typically exponentially. - O(log(n)): The execution time ascends in proportion to the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
Evaluate the code based on the given task, code, and evaluation rubric. Provide a fair and detailed assessment following the rubric. The XML tags are defined as follows: - <TASK>: Describes what the responses are supposed to accomplish. - <CODE>: The code provided to be evaluated. - <EVALUATION_RUBRIC>: Evaluation rubric to select which label is appropriate. - <OUTPUT_FORMAT>: Specifies the required format for your final answer. <TASK> Classify the following Java code's worst-case time complexity according to its relationship to the input size n. </TASK> <CODE> //package codeforces.br22; import java.io.*; import java.util.ArrayList; import java.util.Collections; import java.util.List; /** * User: Kandy * Date: 29.06.2010 * Time: 21:07:24 */ public class ProblemA { public void solve() { boolean oj = true; try { Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("A.in"); Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("A.out"); BufferedReader br = new BufferedReader(reader); StreamTokenizer st = new StreamTokenizer(reader); PrintWriter out = new PrintWriter(writer); MyTokenizer tok = new MyTokenizer(br.readLine()); int n = (int)tok.getNum(); tok =new MyTokenizer(br.readLine()); List<Integer> a = new ArrayList<Integer>(); for(int i=0;i<n;i++) { int r = (int)tok.getNum(); if (!a.contains(r)) { a.add(r); } } Collections.sort(a); if (a.size() < 2) { out.printf("NO"); } else { out.printf("%d", a.get(1)); } br.close(); out.close(); reader.close(); writer.close(); } catch (Exception ex) { ex.printStackTrace(); } finally { } } public static void main(String[] args) { ProblemA f = new ProblemA(); f.solve(); } private class MyTokenizer { private String s; private int cur; public MyTokenizer(String s) { this.s = s; cur = 0; } public void skip() { while (cur < s.length() && (s.charAt(cur) == ' ' || s.charAt(cur) == '\n')) { cur++; } } public double getNum() { skip(); String snum = ""; while (cur < s.length() && (s.charAt(cur) >= '0' && s.charAt(cur) <= '9' || s.charAt(cur) == '.' || s.charAt(cur) == '-')) { snum += s.charAt(cur); cur++; } return Double.valueOf(snum); } public String getString() { skip(); String s2 = ""; while (cur < s.length() && (((s.charAt(cur) >= 'a' && s.charAt(cur) <= 'z')) || ((s.charAt(cur) >= 'A' && s.charAt(cur) <= 'Z')))) { s2 += s.charAt(cur); cur++; } return s2; } public char getCurrentChar() throws Exception { if (cur < s.length()) return s.charAt(cur); else throw new Exception("Current character out of string length"); } public void moveNextChar() { if (cur < s.length()) cur++; } public boolean isFinished() { return cur >= s.length(); } } } </CODE> <EVALUATION_RUBRIC> - O(n): The running time grows linearly with the input size n. - O(1): The running time does not change regardless of the input size n. - O(nlog(n)): The running time increases with the product of n and logarithm of n. - non-polynomial: The running time increases non-polynomially with input size n, typically exponentially. - O(n^3): The running time increases with the cube of the input size n. - others: The code does not clearly correspond to the listed classes or has an unclassified time complexity. - O(n^2): The running time increases with the square of the input size n. - O(log(n)): The running time increases with the logarithm of the input size n. </EVALUATION_RUBRIC> <OUTPUT_FORMAT> Return a JSON response in the following format: { "explanation": "Explanation of why the response received a particular time complexity", "time_complexity": "Time complexity assigned to the response based on the rubric" } </OUTPUT_FORMAT>
1,002
2,017