字符串编码(Encoding) - ACM

题目描述
     Given a string containing only "A"-"Z", we could encode it using the following method:
We use 3 characters to represent a sub-string if there are some consecutive repeating
characters in one line, which "0" is as the mark, and then the repeat number , and the character itself.

输入
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 80.

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

样例输入
2
ABBC
BBCCC

样例输出
A02BC
02B03C

　　关键:
　　　　The first line contains an integer N (1≤N ≤100) which indicates the number of test cases.
　　The next N lines contain N strings.

#include <stdio.h>

// 处理每一行数据
void solve(char* str)
{
    char ch;
    int count;

    while(*str){
        //每个字符至少存在一个的
        count = 1;
        //遍历与当前第1个字符相同的字符并计数
        for(ch=*str++;*str&&*str==ch;str++){
            count++;
        }
        if(count==1){//--------仅一次,输出原字符
            printf("%c",ch);
        }else{//---------------最多3个,不足用零补足
            printf("%02d%c",count,ch);
        }
    }
    printf("\n");
}

int main(void)
{
    //1<=N<=100,字符串长度<80
    char astr[100][80];
    int N,i;
    scanf("%d\n",&N);
    //按照题目要求,先读入所有待处理的数据
    //不能读入一个就处理一个
    for(i=0;i<N;i++){
        gets(astr[i]);
    }
    //一次性处理所有输入数据
    for(i=0; i<N; i++){
        solve(astr[i]);
    }
    return 0;
}

 女孩不哭 @ 2013-05-08 22:24:30 @ http://www.cnblogs.com/nbsofer