Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −106≤a,b≤106. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

结尾无空行

Sample Output:

-999,991

结尾无空行

题目大意

从后往前，每三位中间都要加上一个逗号。如果前面部分少于四位就不用。

比如1000，就是1，000。

10900就是10，900。 

题目分析

因为懒得动脑所以动脑想了一个不用动脑的解决方法！！

首先因为怕麻烦所以先把－号给处理掉了。

然后倒序遍历数字，把遍历到的元素头插到output里去， 每隔3位插入一个，

然后输出output。

AC代码

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    int a, b, cnt = 0;
    cin >> a >> b;
    a += b;
    string str = to_string(a), output = "";
    int flag = 0;
    if(str[0] == '-'){
            flag = 1;
            str = str.substr(1, str.size() - 1);
        }
    for (int i = str.size() - 1; i >= 0;i--){
        if (cnt % 3 == 0&&(cnt>=3)) output += ',';
        if(str[i]>='0'&&str[i]<='9') cnt++;
        output += str[i];
    }
    if(flag)output += '-';
    reverse(output.begin(), output.end());
    cout << output;
}

总结

开始刷甲级了！