/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
package codejam;

/**
 *
 * @author eblanco
 */
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashSet;
import javax.swing.JFileChooser;

public class RecycledNumbers {

    public static String DatosArchivo[] = new String[10000];

    public static int[] convertStringArraytoIntArray(String[] sarray) throws Exception {
        if (sarray != null) {
            int intarray[] = new int[sarray.length];
            for (int i = 0; i < sarray.length; i++) {
                intarray[i] = Integer.parseInt(sarray[i]);
            }
            return intarray;
        }
        return null;
    }

    public static boolean CargarArchivo(String arch) throws FileNotFoundException, IOException {
        FileInputStream arch1;
        DataInputStream arch2;
        String linea;
        int i = 0;

        if (arch == null) {
            //frame.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE);
            JFileChooser fc = new JFileChooser(System.getProperty("user.dir"));
            int rc = fc.showDialog(null, "Select a File");
            if (rc == JFileChooser.APPROVE_OPTION) {
                arch = fc.getSelectedFile().getAbsolutePath();
            } else {
                System.out.println("No hay nombre de archivo..Yo!");
                return false;
            }
        }
        arch1 = new FileInputStream(arch);
        arch2 = new DataInputStream(arch1);
        do {
            linea = arch2.readLine();
            DatosArchivo[i++] = linea;
            /* if (linea != null) {
            System.out.println(linea);
            }*/
        } while (linea != null);
        arch1.close();
        return true;
    }

    public static boolean GuardaArchivo(String arch, String[] Datos) throws FileNotFoundException, IOException {
        FileOutputStream out1;
        DataOutputStream out2;
        int i;
        out1 = new FileOutputStream(arch);
        out2 = new DataOutputStream(out1);

        for (i = 0; i < Datos.length; i++) {
            if (Datos[i] != null) {
                out2.writeBytes(Datos[i] + "\n");
            }
        }
        out2.close();
        return true;
    }

    public static void echo(Object msg) {
        System.out.println(msg);
    }

    public static void main(String[] args) throws IOException, Exception {
        String[] res = new String[10000], num = new String[2];
        int i, j, k, ilong;
        int ele = 1;
       ArrayList al = new ArrayList();
        String FName = "C-small-attempt0",  istr, irev, linea;
        if (CargarArchivo("C:\\eblanco\\Desa\\CodeJam\\Code Jam\\test\\" + FName + ".in")) {
            for (j = 1; j <= Integer.parseInt(DatosArchivo[0]); j++) {
                linea = DatosArchivo[ele++];
                num = linea.split(" ");
                   al.clear();
                // echo("A: "+num[0]+" B: "+num[1]);
                for (i = Integer.parseInt(num[0]); i <= Integer.parseInt(num[1]); i++) {
                    istr = Integer.toString(i);
                    ilong = istr.length();
                 
                    for (k = 1; k < ilong; k++) {
                        if (ilong > k) {
                            irev = istr.substring(istr.length() - k, istr.length()) + istr.substring(0, istr.length() - k);
                            //echo("caso: " + j + ": isrt: " + istr + " irev: " + irev);
                            if ((Integer.parseInt(irev) > Integer.parseInt(num[0])) && (Integer.parseInt(irev) > Integer.parseInt(istr)) && (Integer.parseInt(irev) <= Integer.parseInt(num[1]))) {
                                al.add("(" + istr + "," + irev + ")");
                            }
                        }
                    }
                }
                HashSet hs = new HashSet();
                hs.addAll(al);
                al.clear();
                al.addAll(hs);
                res[j] = "Case #" + j + ": " + al.size();
                echo(res[j]);
                LeerArchivo.GuardaArchivo("C:\\eblanco\\Desa\\CodeJam\\Code Jam\\test\\" + FName + ".out", res);
            }
        }
    }
}

/*
 * 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-small-attempt0.in";
	private static final String OUTPUT_FILE_NAME = "C-small-attempt0.out";

	/*
	 * Per c Ã¨ una ricerca pseudocubica in un range finito Per ogni m, per ogni
	 * n entrambi compresi tra a e b con m<n esiste una permutazione barrelshift
	 * di m taleche produca n? Questo Ã¨ il succo La difficoltÃ  di c sta in due
	 * punti Prima di tutto nel weeding out dei risultati certamente errati 15
	 * non diventerÃ  mai 42 manco se ti ammazzi E poi nella verifica dei
	 * risultati possibilmente corretti... 11201 potrebbe diventare 20111 Se lo
	 * fa incrementa un contatore E poi stampa i risultati Questi due casi vanno
	 * ottimizzati Successivamente bisogna vedere se c'Ã¨ un barrelshift decimale
	 * "economico" in Java, altrimenti non si puÃ² fare granchÃ© :/ Anche
	 * trattandoli come char [] si perde moto tempo nella stringizzazione e nel
	 * parseint Leggili, magari vengono piÃ¹ facili per te ;)
	 */

	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;
	}

}
