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.

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.

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

2
ABC
ABBCCC

ABC
A2B3C

ZHANG Zheng

JGShining   |   We have carefully selected several similar problems for you:  1019 1021 1062 1032 1018 

很简单的水题，我不会告诉你我是无聊才来切水题的！

题目的主要意思就是：让你求一个字符串里面连续的不同字符的个数,注意是连续的字符！

所以AABCCA的结果是2AB2CA而不是3AB2C!就只有这一个坑点了，其他没有了......

另外注意如果字符是1前面不输出1......

import java.io.*;
import java.util.*;

public class Main
{

	public static void main(String[] args)
	{
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		int a[] = new int[10005];
		int n = input.nextInt();
		input.nextLine();
		for (int i = 0; i < n; i++)
		{
			int count = 1;
			String str = input.nextLine();
			char c[] = str.toCharArray();
			for (int j = 1; j < c.length; j++)
			{
				if (c[j] == c[j - 1])
				{
					++count;
				}
				else
				{
					if (count == 1)
					{
						System.out.print(c[j - 1]);
					} 
					else
					{
						System.out.print(count + "" + c[j - 1]);
						count = 1;
					}
				}
				if (j == c.length - 1)
				{
					if (count == 1)
					{
						System.out.print(c[j]);
					} 
					else
					{
						System.out.print(count + "" + c[j]);
						count = 1;
					}
				}
			}
			System.out.println();
		}
	}

}