encoding（hdoj1020）

最新推荐文章于 2023-05-07 20:27:34 发布

转载最新推荐文章于 2023-05-07 20:27:34 发布 · 119 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/a1225234/p/4522123.html

本文介绍了一种将只包含'A'-'Z'字符的字符串进行编码的方法，通过计数相同字符连续出现的数量，并使用特定格式表示，从而实现压缩存储。

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

ABC

ABBCCC

Sample Output

ABC

A2B3C

 1 #include<stdio.h>
 2 #include<string.h>
 3 int main()
 4 {
 5     char a[10001];
 6     int n;
 7     scanf("%d",&n);
 8     getchar();
 9     while(n--)
10     { 
11         memset(a,0,sizeof(a));
12         gets(a);
13         int len=strlen(a),i,count=1;
14         for(i=0;i<len-1;i++)
15         {
16             if(a[i]==a[i+1])
17             {
18                 count++;
19                 if(i==len-2)
20                     printf("%d%c",count,a[i]);        
21             }
22             else
23             {
24                 if(i==len-2)
25                     {
26                         if(count==1)
27                             printf("%c%c",a[i],a[i+1]);
28                         else
29                             printf("%d%c%c",count,a[i],a[i+1]);
30                     }
31                 else
32                     {
33                         if(count==1)
34                             printf("%c",a[i]);
35                         else
36                             printf("%d%c",count,a[i]);
37                     }
38                     count=1;
39             }
40         }
41         printf("\n");
42     }
43 }

转载于:https://www.cnblogs.com/a1225234/p/4522123.html