import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

public class Q3M {
	public static void main(String[] args) throws Exception {
		compute();
	}

	private static void compute() throws Exception {
		BufferedReader br = new BufferedReader(new FileReader(new File(
				"C:\\work\\Q3\\C-small-attempt0.in")));
		String line = null;
		int i = Integer.parseInt(br.readLine());
		List l = new ArrayList();
		for (int j = 0; j < i; j++) {
			line = br.readLine();
			String[] nums = line.split(" ");
			l.add(calculate(nums));
		}
		writeOutput(l);
	}

	private static int calculate(String[] nums) {
		int min = Integer.parseInt(nums[0]);
		int max = Integer.parseInt(nums[1]);
		int count = 0;
		List l = new ArrayList();
		for (int i = min; i <= max; i++) {
			for (int times = 1; times < countDigits(i); times++) {
				int res = shiftNum(i, times);
				if (res <= max && i < res) {
					if ((!l.contains((i + ":" + res)))) {
						l.add(i + ":" + res);
						l.add(res + ":" + i);
						count++;
					}
				}
			}
		}
		return count;
	}

	private static boolean checkZeros(int temp, int res) {
		if (temp % 10 == 0 || res % 10 == 0)
			return false;
		return true;
	}

	private static int shiftNum(int n, int times) {
		int base = (int) Math.pow(10, times);
		int temp2 = n / base;
		int placeHolder = (int) Math.pow((double) 10,
				(double) countDigits(temp2));
		int res = placeHolder * (n % base) + temp2;
		if (countDigits(res) == countDigits(n)) {
			return res;
		} else {
			return 2000001;
		}
	}

	public static int countDigits(int x) {
		if (x < 10)
			return 1;
		else {
			return 1 + countDigits(x / 10);
		}
	}

	private static void writeOutput(List l) throws Exception {
		StringBuffer b = new StringBuffer();
		int i = 1;
		BufferedWriter br = new BufferedWriter(new FileWriter(new File(
				"C:\\work\\Q3\\ans.txt")));
		for (Iterator iterator = l.iterator(); iterator.hasNext();) {
			br.write("Case #" + i++ + ": " + iterator.next());
			br.newLine();
		}
		br.close();
	}

}
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-small.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-small.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();
		}

	}

}
