hdu1020（Encoding）

水题，相同字母是挨着的，且不会出现空格。

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19863 Accepted Submission(s): 8560

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
#include<iostream>using namespace std;int main(){    char s[10000];    int t;    while(cin>>t&&t)    {                    while(t--)                    {                              cin>>s;                              int count=1;                              for(int i=0;i<strlen(s)-1;i++)                              {                                      if(s[i+1]==s[i]) {count++;                                                       if(i==strlen(s)-2)                                                        cout<<count<<s[i]<<endl;}                                      else {                                           if(i==strlen(s)-2) {if(count>1) cout<<count<<s[i]<<s[i+1]<<endl;                                                               else  cout<<s[i]<<s[i+1]<<endl;}                                           else {                                           if(count==1) cout<<s[i];                                           else cout<<count<<s[i];                                           count=1;                                           continue;                                               }                                           }                              }                    }    }    return 0;}