Encoding

最新推荐文章于 2016-10-26 23:35:22 发布

runnerzyc

最新推荐文章于 2016-10-26 23:35:22 发布

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分类专栏： oj入门 acm水题

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本文介绍了一种用于压缩字符串的简单编码算法，该算法通过将连续重复的字符替换为字符与其出现次数来实现字符串的压缩。文章提供了具体的输入输出示例，并附带了完整的 C 语言实现代码。

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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

代码：

#include<stdio.h>
#include<string.h>
int main()
{
	int n,i,sum,p;
	char a[10010];
	scanf("%d",&n);
	while(~scanf("%s",a))
	{ 
	  p=strlen(a);
      for(i=0,sum=1;i<p;i++)	
	  {
	  	if(a[i+1]==a[i])
	  	 sum++;
	  	else { 
	  	if(sum==1)
	  		printf("%c",a[i]);
	  	else if(a[i+1]!=a[i])
	  	{
	  		printf("%d%c",sum,a[i]);
	  	 sum=1;
	    }
	  	}
		  
	  }	printf("\n");
	}
	return 0;
}