/*
 * Problem C. Recycled Numbers
 * 
 * Do you ever become frustrated with television because you keep seeing the
 * same things, recycled over and over again? Well I personally don't care about
 * television, but I do sometimes feel that way about numbers.
 * 
 * Let's say a pair of distinct positive integers (n, m) is recycled if you can
 * obtain m by moving some digits from the back of n to the front without
 * changing their order. For example, (12345, 34512) is a recycled pair since
 * you can obtain 34512 by moving 345 from the end of 12345 to the front. Note
 * that n and m must have the same number of digits in order to be a recycled
 * pair. Neither n nor m can have leading zeros.
 * 
 * Given integers A and B with the same number of digits and no leading zeros,
 * how many distinct recycled pairs (n, m) are there with A â¤ n < m â¤ B?
 * 
 * Input
 * 
 * The first line of the input gives the number of test cases, T. T test cases
 * follow. Each test case consists of a single line containing the integers A
 * and B.
 * 
 * Output
 * 
 * For each test case, output one line containing "Case #x: y", where x is the
 * case number (starting from 1), and y is the number of recycled pairs (n, m)
 * with A â¤ n < m â¤ B.
 * 
 * Limits
 * 
 * 1 â¤ T â¤ 50.
 * 
 * A and B have the same number of digits.

Small dataset

1 â¤ A â¤ B â¤ 1000.

Large dataset

1 â¤ A â¤ B â¤ 2000000.

Sample


Input 
 	
Output 
 
4
1 9
10 40
100 500
1111 2222
Case #1: 0
Case #2: 3
Case #3: 156
Case #4: 287
 */

package google.codejam.y2012.qualifications;

import google.codejam.commons.Utils;

import java.io.IOException;
import java.util.StringTokenizer;

/**
 * @author Marco Lackovic <marco.lackovic@gmail.com>
 * @version 1.0, Apr 14, 2012
 */
public class RecycledNumbers {

	private static final String INPUT_FILE_NAME = "C-large.in";
	private static final String OUTPUT_FILE_NAME = "C-large.out";

	public static void main(String[] args) throws IOException {
		String[] input = Utils.readFromFile(INPUT_FILE_NAME);
		String[] output = new String[input.length];
		int x = 0;
		for (String line : input) {
			StringTokenizer st = new StringTokenizer(line);
			int a = Integer.valueOf(st.nextToken());
			int b = Integer.valueOf(st.nextToken());
			System.out.println("Computing case #" + (x + 1));
			output[x++] = "Case #" + x + ": " + recycledNumbers(a, b);
		}
		Utils.writeToFile(output, OUTPUT_FILE_NAME);
	}

	private static int recycledNumbers(int a, int b) {
		int count = 0;
		for (int n = a; n < b; n++) {
			count += getNumGreaterRotations(n, b);
		}
		return count;
	}

	private static int getNumGreaterRotations(int n, int b) {
		int count = 0;
		char[] chars = Integer.toString(n).toCharArray();
		for (int i = 0; i < chars.length; i++) {
			chars = rotateRight(chars);
			if (chars[0] != '0') {
				int m = Integer.valueOf(new String(chars));
				if (n == m) {
					break;
				} else if (n < m && m <= b) {
					count++;
				}
			}
		}
		return count;
	}

	private static char[] rotateRight(char[] chars) {
		char last = chars[chars.length - 1];
		for (int i = chars.length - 2; i >= 0; i--) {
			chars[i + 1] = chars[i];
		}
		chars[0] = last;
		return chars;
	}

}

package round03;

import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.IOException;
import java.util.HashSet;
import java.util.Scanner;

/**
 * Problem

Do you ever become frustrated with television because you keep seeing the same things, recycled over and over again? Well I personally don't care about television, but I do sometimes feel that way about numbers.

Let's say a pair of distinct positive integers (n, m) is recycled if you can obtain m by moving some digits from the back of n to the front without changing their order. For example, (12345, 34512) is a recycled pair since you can obtain 34512 by moving 345 from the end of 12345 to the front. Note that n and m must have the same number of digits in order to be a recycled pair. Neither n nor m can have leading zeros.

Given integers A and B with the same number of digits and no leading zeros, how many distinct recycled pairs (n, m) are there with A ¡Â n < m ¡Â B?
Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of a single line containing the integers A and B.
Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1), and y is the number of recycled pairs (n, m) with A ¡Â n < m ¡Â B.
Limits

1 ¡Â T ¡Â 50.
A and B have the same number of digits.
Small dataset

1 ¡Â A ¡Â B ¡Â 1000.
Large dataset

1 ¡Â A ¡Â B ¡Â 2000000.
Sample


input

4
1 9
10 40
100 500
1111 2222

output 

Case #1: 0
Case #2: 3
Case #3: 156
Case #4: 287


 * @author chmin
 *
 */
public class RecycledNumbers {
	
	public static class RNumber {
		String n;
		int cntShift = 0;
		public RNumber( int n){
			init(n);
		}
		
		public void init( int  n ){
			this.n = "" + n;
			cntShift = 0;
		}
		
		public int next() {
			int pos = n.length()-1;
			cntShift ++;
			
			while ( n.charAt(pos) == '0'){
				pos --;
				cntShift ++;
			}
			
			if ( pos > 0 ) {
				String v = n.substring(pos , n.length()) + n.substring(0, pos);
				n = v;
			}
			
			return Integer.parseInt(n);
		}
		
		public boolean hasNext() {
			return cntShift < n.length();
		}
	}
	public void solve() throws IOException{
		Scanner scanner = new Scanner (getClass().getResourceAsStream("R3-large.in"));
		int T = scanner.nextInt();
		RNumber rnum = new RNumber(0);
		HashSet<String> set = new HashSet<>();
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < T ; i++) {
			int A = scanner.nextInt();
			int B = scanner.nextInt();
			set.clear();
			for( int n = A ; n <= B ; n++){
				rnum.init(n);
				while ( rnum.hasNext() ){
					int m = rnum.next();
					if ( n < m  && m <= B) {
						set.add(n + " " + m);
					}
				}
			}
			sb.append("Case #" + (i+1) + ": " + set.size() + System.getProperty("line.separator"));
		}
		
		BufferedWriter bw = new BufferedWriter(new FileWriter("R3-large.out"));
		bw.write(sb.toString());
		bw.close();
	}
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		
		try {
			new RecycledNumbers().solve();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}

	}

}
