传送门

题目

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
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
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

分析

我可能用的方法比较麻烦，但是简单粗暴，比较好想，就是先将结果转化成字符串，然后从末尾压栈，压满三个数就压进一个逗号，需要注意的是，如果首位数字是第三位，则不压入逗号。

源代码

//C/C++实现
#include <iostream>
#include <sstream>
#include <stack>

using namespace std;

int main(){
    int a, b, sum;
    scanf("%d %d", &a, &b);
    sum = a + b;
    stringstream ss;
    string result;
    ss << sum;
    ss >> result;
    stack<char> s;
    if(sum < 0){ // minus
        int count = 0;
        for(int i = result.size() - 1; i >= 1; --i){
            ++count;
            s.push(result[i]);
            if(count == 3){
                count = 0;
                if(i != 1){
                    s.push(',');
                }
            }
        }
        s.push('-');
    }
    else{
        int count = 0;
        for(int i = result.size() - 1; i >= 0; --i){
            ++count;
            s.push(result[i]);
            if(count == 3){
                count = 0;
                if(i != 0){
                    s.push(',');
                }
            }
        }
    }
    while(!s.empty()){
        printf("%c", s.top());
        s.pop();
    }
    printf("\n");
    return 0;
}