hdu 1020 简单的字符串处理

来源:互联网 发布:网络爬虫基本原理 编辑:程序博客网 时间:2024/03/29 21:16

题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1020

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23937    Accepted Submission(s): 10515


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

 

代码如下:

 1 #include<iostream> 2 #include<stdio.h> 3 #include<string> 4 #include<string.h> 5 #include<map> 6 #include<math.h> 7 #include<algorithm> 8 #define N 10005 9 using namespace std;10 char str[N];11 int main()12 {13     int n,i,t;14     scanf("%d",&n);15     while(n--)16     {17         scanf("%s",str);18         i=0;19         while(str[i]!='\0')20         {21             t=i;22             while(str[t+1] == str[t])23                 t++;24             if(t>i)25                 printf("%d%c",t-i+1,str[t]);26             else27                 printf("%c",str[t]);28             i=t+1;29         }30         printf("\n");31     }32     return 0 ;33 }
0 0
原创粉丝点击