hdu 1020 Encoding【字符串处理】

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本文探讨了将字母串转换为紧凑形式的方法，通过识别连续重复字符并使用计数前缀来实现字符串压缩，适用于数据存储与传输优化。

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35512 Accepted Submission(s): 15708

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

用了一个 do while 循环....

注意格式，然后循环加特判就可以了，另外注意不能排序，比如

AABBAA 输出的是 2A2B2B

#include<stdio.h>
int main()
{
	int t;
	//freopen("shuju.txt","r",stdin);
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		char next,cur=0;int cnt=1;
		cur=getchar();
		do
		{
			next=getchar();
			if(next==cur)
			{
				++cnt;
			}
			else
			{
				if(cnt>1)
				{
					printf("%d",cnt);
				}
				printf("%c",cur);
				cur=next;cnt=1;
			}
		}while(next!='\n');
		printf("\n");
	}
	return 0; 
}