HDU-1020 Encoding

问题描述：

Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:

Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
If the length of the sub-string is 1, ‘1’ should be ignored.

输入说明：

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.

输出说明：

For each test case, output the encoded string in a line.

思路：

理解题意后挺简单的一道题，题意大概就是给你一串全部由A~Z构成的字符串，相邻的相同字母合并，保留一个数字（代表有多少个这个字母）和一个这个字母，举个例子：AAAA 经过转化变为4A 。（不要理解成了整个字符串有多少个这样的字母，都要全部合并，憨憨的我因为这个原因多交了2发wa）。解题就很简单了，直接遍历一遍，如果a[i]=a[i+1]，计数+1，如果a[i]!=a[i+1]，就进行输出（注意要分输出的有没有前缀数字，分两种情况输出）

AC代码：

#include<bits/stdc++.h>
using namespace std;

int main()
{
    
    
	int count,i,n;
	char a[100001];
	cin>>n;
	while(n--)
    {
    
    
		scanf("%s",a);
		count=1;
		for(i=0;i<strlen(a);i++)
		{
    
    
			if(a[i]==a[i+1])
			{
    
    
			    count++;
			}
			else
            {
    
    
				if(count==1)
				cout<<a[i];
				else
				{
    
    
                    cout<<count<<a[i];
                    count=1;
				}
			}
		}
		cout<<endl;
	}
	return 0;
}