map初步（由ABBC--->A2BC）

1.题目：

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.

InputThe 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.
OutputFor each test case, output the encoded string in a line.
Sample Input

2
ABC
ABBCCC

Sample Output

ABC
A2B3C


2.代码：

//本题使用G++编译 
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--){
        int sum=1;
        string a;//定义字符数组 
        cin>>a;//输入字符数组内容 
        for(int i=0;i<a.length();i++){
            if(a[i]==a[i+1])
            sum++;//若相邻两个字符相同则sum+1 
            else//若相邻字符不相同则进行else 
            {
                if(sum!=1)//若sum！=1，即相邻字符相同则输出sum再输出字符 
            cout<<sum<<a[i];
            else//若相邻字符不相同则直接输出字符 
            cout<<a[i];
            sum=1;
            }
        }
        cout<<endl;
}
}

3.总结：

应用了map算法！！！