HDU 1020 Encoding 字符串操作

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43902 Accepted Submission(s): 19397

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

2ABCABBCCC

Sample Output

ABCA2B3C

Author

ZHANG Zheng

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给你一个字符串然后让你改变一下输出格式，切记不能排序啊，排序就是错的。

ac代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;bool cmp(char x,char y){return x < y ? true : false;}int main(){int t,i,j,n,num,cnt[10005];char a[10005],b[10005],tmp;int place;while(~scanf("%d",&t)){while(t--){scanf("%s",a);//printf("%s\n",a);n=strlen(a);//sort(a,a+n,cmp);memset(cnt,0,sizeof(cnt));tmp=a[0];num=1;place=0;for(i=1;i<n;i++){if(a[i]==tmp)num++;else{b[place]=tmp;cnt[place]=num;place++;num=1;tmp=a[i];}}b[place]=tmp;cnt[place]=num;place++;for(i=0;i<place;i++){if(cnt[i]>1)printf("%d",cnt[i]);printf("%c",b[i]);}printf("\n");}}return 0;}

万里长城永不倒，再来一题好不好

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