# Encoding
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 34681    Accepted Submission(s): 15377  

Problem Description

Given a string containing only "A" - "Z", we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, "1" should be ignored.

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only "A" - "Z" and the length is less than 10000.

Output

For each test case, output the encoded string in a line.

Sample Input

2
ABC
ABBCCC

Sample Output

ABC
A2B3C解题思路整理：对字符串进行遍历，如果当前的字符与前面的字符相同的话那么我加1记录其个数，如果不相同的时候，排除只有1个的情况，只有在字符重复个数大于1的时候那么认为是有多个在一起，这是将字符串拼接组成新的串继续下一步操作。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int count = scanner.nextInt();
		int i = 0, countNum = 0, j = 0, j2 = 0;
		String str ;
		String str2 = "";
		char sCopy = 0;
		List<String> list = new ArrayList<String>();
		while(i < count)
		{
			str = scanner.next();
			
			sCopy = 0;
			j = 0;
			str2 = "";
			countNum = 0;
			
			while(j < str.length())
			{
				if(sCopy != str.charAt(j))
				{
					if(countNum > 1)
					{
						str2 = str2 + countNum + sCopy;
					}
					else if(j != 0)
						str2 = str2 + sCopy;
					
					countNum = 0;
					sCopy = str.charAt(j);
				}
				
				if(sCopy == str.charAt(j))
				{
					countNum++;
				}
				j++;
				if(j == str.length())
				{
					if(countNum > 1)
					{
						str2 = str2 + countNum + sCopy;
					}
					else if(j != 0)
						str2 = str2 + sCopy;
					list.add(str2);
				}
			}
			i++;
		}
		for(i = 0; i < count; i++)
		{
			System.out.println(list.get(i));
		}
	}
}