1001. A+B Format (20)_acm1001:输入输出实践的a + b(i)-优快云博客

本文介绍了一个简单的程序设计问题：如何将两个整数相加的结果以标准格式输出，即每三位数字用逗号分隔。文章详细展示了输入、输出的要求，并提供了一段完整的C++代码实现。

1001. A+B Format (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
     int n, m;
     scanf("%d%d", &n, &m);
     int sum = n + m;

     int a[10];
     memset(a, 0, sizeof(a));
     int flag = 0;
     int cnt = 0;
     if(sum < 0)
        flag = 1;
    sum = fabs(sum);
    if(sum == 0)
        printf("%d\n",sum);
     while(sum)
     {
        a[cnt++] = sum%10;
        sum/=10;

     }
     if(flag == 1)
        printf("-");
     for(int i = cnt - 1; i >= 0; i--)
     {

        printf("%d",a[i]);
        if(i == 6 || i == 3)
            printf(",");

     }

}