PAT---A1001. A+B Format (20)-优快云博客

本文介绍了一种格式化输出两个整数之和的方法，确保输出符合标准格式要求，即每三位数字用逗号分隔。文章提供了一个示例程序，详细展示了如何实现这一功能。

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**题目要求：**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

解题思路：我们可以将输入的两个值相加后的结果转化为一个字符数组，然后在适当的位置进行输出“，”即可。注意如果结果为负时可先打印负号，然后将结果当作正数讨论较为方便

参考代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main()
{
    int a,b;
    cin >>a>>b;
    int sum = a + b;
    char result[20] = {0};
    int i=0;
    //将sum全部转化为正数，这样拆解方便
    if(sum<0)
    {
        cout <<"-";
        sum = -sum;
    }
    //当sum为0时特殊讨论
    if(sum == 0)
        {
            i = 1;
            result[0] = 0+'0';
        }
    //将sum的各个位拆解并转化成字符保存在数组中
    while(sum)
    {
       result[i] = sum%10+'0';
       sum = sum/10;
       i++;
    }
    for(int j = i-1;j>=0;j--)
        {
            cout <<result[j];
            if((j%3==0)&&(j!=0))//当j为3的整数且不是数字结尾，则打印，
                cout <<",";
        }
    return 0;
}