开放原子开发者工作坊

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

很简单，数字转字符串即可

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <vector>
#include <sstream>
#include <string>
#include <stack>

using namespace std;

int main()
{
    int a, b,sum;
    string str;
    scanf("%d %d", &a, &b);
    sum = a + b;
    stringstream ss;
    ss << sum;
    ss >> str;

    ///此处可以通过sum/1000，然后转为字符串
    stack<string>res;
    int k;
    string s;
    for (k = str.length(); k > 3;k-=3)
    {    
        s.assign(str.begin() + k - 3, str.begin() + k);
        res.push(s);
        res.push(",");
    }
    s.assign(str.begin(), str.begin() + k);
    if (s == "-")
        res.pop();
    res.push(s);
    while (!res.empty())
    {
        cout << res.top();
        res.pop();
    }
    return 0;
}

更简洁点

#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;
}