/*
 * 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 com.google.code.codejam._2012.qualification;

import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

/**
 * 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?
 * 
 * The RecycledNumbers is the second problem.
 * 
 * <p>
 * Apr 14, 2012 9:25:48 PST AM
 * </p>
 * 
 * @author Marcello de Sales (marcello.desales@gmail.com)
 * 
 */
public class RecycledNumbers extends CommandLineClient {

  /**
   * 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
   * 
   * @param args the input arguments from the stdin.
   */
  public static void main(String[] args) {
    try {
      new RecycledNumbers().execute(args);
      System.exit(0);

    } catch (Throwable t) {
      t.printStackTrace();
      System.exit(1);
    }
  }

  @Override
  public void process(Scanner input, PrintWriter output) throws Exception {
    int numberOfTestCases = Integer.valueOf(input.nextLine());

    StringBuilder builder = null;

    for (int testIndex = 0; testIndex < numberOfTestCases; testIndex++) {
      int numberOfPairs = 0;

      output.printf("Case #%d: ", testIndex + 1);

      // get the value of A and B
      String[] aAndB = input.nextLine().split(" ");

      // if the number is only one digit, then leave as it is zero.
      if (aAndB[0].length() == 1 || aAndB[1].length() == 1) {
        output.printf("%d\n", numberOfPairs);
        continue;
      }

      // the inputs A and B
      int a = Integer.valueOf(aAndB[0]);
      int b = Integer.valueOf(aAndB[1]);

      // caching the current results to avoid counting the same value twice.
      Set<String> resultsSoFar = new HashSet<>();

      // Generating all the numbers between a and b, inclusive [a, b]
      String[] ranges = new String[b - a];
      for (int i = 0, rangeValue = a; i < ranges.length; i++) {
        ranges[i] = String.valueOf(rangeValue++);
      }

      String n = "";
      String m = "";

      for (int i = 0; i < ranges.length; i++) {
        n = ranges[i];
        m = ranges[i];

        int length = String.valueOf(ranges[i]).length();
        for (int j = 0; j < length; j++) {
          // rotate number m
          String stringValue = String.valueOf(m);
          builder = new StringBuilder();
          builder.append(stringValue.charAt(stringValue.length() - 1));
          builder.append(stringValue.subSequence(0, stringValue.length() - 1));

          m = builder.toString();

          // a recycle pair is one such that A â¤ n < m â¤ B
          if (a <= Integer.valueOf(n) && Integer.valueOf(n) < Integer.valueOf(m)
              && Integer.valueOf(m) <= b) {
            if (!resultsSoFar.contains(n + m)) {
              resultsSoFar.add(n+m);
              numberOfPairs++;
            }
            //output.printf("%s, %s count: %d \n", n, m, numberOfPairs); // could see the repeated value!!!
          }
        }
      }
      output.printf("%d\n", numberOfPairs);
    }
  }
}
