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

题意：按照字符串的顺序，输出字符的个数和字符。

如果字符出现次数为1次，只要输出原字符。 如果输入为：ABBCCCBBB 输出为：A2B3C3B 而不是：A5B3C

import java.util.Scanner;public class Main {
         public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int t = sc.nextInt();        while (t-- > 0) {            String strs = sc.next();            // System.out.println(strs+"=strs");            boolean isSee[] = new boolean[strs.length()];            for (int i = 1; i < isSee.length; i++) {                isSee[i] = false;            }            int sum = 0;            boolean isLast=false;            for (int i = 0; i < strs.length()-1; i++) {                if(strs.charAt(i)==strs.charAt(i+1)) {                    isSee[i]=true;                    isSee[i+1]=true;                    sum=sum+1;                }else{                    sum=sum+1;                    isSee[i]=false;                }                if(!isSee[i]){                    if(sum==1){                        System.out.print(strs.charAt(i));                    }else{                        System.out.print(""+sum+strs.charAt(i));                    }                }                if(!isSee[i]){                    sum=0;                }            }            if(isSee[strs.length()-1]){                System.out.print(""+(sum+1)+strs.charAt(strs.length()-1));            }else{                System.out.print(strs.charAt(strs.length()-1));            }            System.out.println();        }    }}