A1001A+Bforamt

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 −. 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

Sample Output:

-999,991

思路：

判断sum正负，若为负输出-，然后将sum的绝对值存到字符串c中；遍历c，每三个元素且不是最后一个元素时输出逗号。

问题：

部分测试用例答案错误

解决：

从高位开始每三位加一个，这样最低一组可能不足三个，改成，从低位开始每三位一组。给出的样例太有迷惑性了…

 

#include <iostream>
#include <string>
using namespace std;
int main() {
    int a, b;
    cin >> a >> b;
    int sum = a + b;
    if (sum < 0)cout << "-";
    string c = to_string(abs(sum));
    int len = c.length();
    for (int i = 0; i < len; i++) {
        cout << c [i];
        if ((i+1) % 3 == len%3 &&i!=len-1)
            cout << ",";
    }
    return 0;
}