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hdu 1020 Encoding

时间:2015-09-12 10:51:23      阅读:164      评论:0      收藏:0      [点我收藏+]

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Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33473    Accepted Submission(s): 14895


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
 

 

Author
ZHANG Zheng
 

 

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很水的题,题意看输入输出就应该能懂。
 
题意:统计一个连续出现的字符个数。
 
附上代码:
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,i,t,s;
 8     char a[10010];
 9     scanf("%d",&n);
10     while(n--)
11     {
12         scanf("%s",a);
13         s=strlen(a);
14         t=1;
15         for(i=1; i<=s; i++)
16         {
17             if(a[i-1]==a[i])
18                 t++;
19             else
20             {
21                 if(t>1)
22                     printf("%d%c",t,a[i-1]);
23                 else
24                     printf("%c",a[i-1]);
25                 t=1;
26             }
27         }
28         printf("\n");
29     }
30     return 0;
31 }

 

hdu 1020 Encoding

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原文地址:http://www.cnblogs.com/pshw/p/4802653.html

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