A1001 A+B Format（字符串处理）

问题

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

• Sample Output:

-999,991

思考

将求和问题用字符串进行表示和处理是一种常规做法

即用 to_String(int) 进行转换

这一道题的关键在于 何时插入‘，’ 即从前往后输出 而又重后往前数数

((i + 1) % 3 == len % 3 && i != len - 1)

用取余比对的方法来解决这个问题 即通过总长度得出余数 并且保证最后一位不会多输出

代码示例

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

(解题时参考了柳婼大大的代码 有兴趣可以去看一下 -> 柳婼のBlog)