1001. A+B Format (20)-优快云博客

本文介绍了一个简单的C++程序，用于读取两个整数并以标准格式输出它们的和，其中涉及到了负数和不同位数数字的处理。

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


#include <iostream>
#include <stack>
using namespace std;

void PrintNum(int num)
{
    if (num<0)  //处理负数
    {
        cout<<'-';
        num=-num;
    }
    
    if (num<1000)
    {
        cout<<num;  //处理和小于三位数的情况
    }else
    {
        stack<int> Q;
        int remain=0,temp=0;
        while (num>999) //将结果每三位作为一个数字存入栈
        {
            remain=num%1000;
            Q.push(remain);
            num/=1000;
        }
        cout<<num;
        while (!Q.empty())
        {
            temp=Q.top();
            Q.pop();
            if (temp>=100)
            {
                cout<<','<<temp;
            }else if(temp>=10)
            {
                cout<<",0"<<temp;   //该三位数是有效十位数时补一个0
            }else
            {
                cout<<",00"<<temp;  //该三位数是有效个位数时补两个0
            }
            
        }
    }
}

int main(int argc, const char * argv[])
{
    int a=0,b=0,temp=0;
    cin>>a>>b;
    temp=a+b;
    PrintNum(temp);
    return 0;
}