Encoding编码 18

Encoding编码
Problem Description
Given a stringcontaining 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.

由于仅包含“A”的字符串 - 'Z'，我们可以用下面的方法对其进行编码：

包含k个相同的字符1.各子串应当被编码为“的kX”，其中“X”是在该子串的唯一的字符。

2.如果子串的长度为1，'1'应该被忽略。
Input
The first linecontains an integer N (1 <= N <= 100) which indicates the number of testcases. The next N lines contain N strings. Each string consists of only 'A' -'Z' and the length is less than 10000.
第一行包含一个整数N（1<= N<=100）表示测试用例的数量。接下来的N行包含N个字符串。每个字符串仅由'A' - 'Z'和长度小于10000。

Output
For each testcase, output the encoded string in a line.
对于每个测试用例，输出线路编码字符串

Sample Input
2
ABC
ABBCCC


Sample Output
ABC
A2B3C

代码如下：

#include <stdio.h>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

//	const int MAXN=10005;
	char a[10005];
int main()
{
    int i,temp;
    int T;
    scanf("%d",&T);//两组测试数据 
    while(T--)
    {
        scanf("%s",&a);//输入的是字符串 
        i=0;
        while(a[i]!='\0')//这是判断字符串结束的标志 
        {
            temp=i;
            while(a[temp+1]==a[i])
            {
                temp++;
            }    
            if(temp>i)
			printf("%d",temp-i+1);
            printf("%c",a[i]);
            i=temp;
            i++;
        } 
        printf("\n");   
    } 
    return 0;   
}