1001 A+B Format (20)

1001A+B问题解析

最新推荐文章于 2022-03-31 10:41:10 发布

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本文介绍了一个简单的编程问题——1001A+B，重点在于如何正确地格式化输出结果，确保数字每三位用逗号分隔。通过示例代码详细展示了输入输出的处理方法，并特别注意了负数和零的特殊情况。

1001 A+B Format (20)
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：
1.将输入数字相加，按千位分节。为0时特判输出，负时先输出负号再转为正值。
2.直接根据数据大小判断所需输出的位数，要注意保留三位有效数字，否则像1000和0这样的输入会变成1，0.

#include<cstdio> 
int main(){
    int a,b,c;
    scanf("%d%d",&a,&b);
    c=a+b;
    if(c==0) {
        printf("0");
        return 0;   
    }
    if(c<0){

        printf("-");
        c=-c;
    }
    if(c<1000) printf("%d",c);
    else if(c<1000000){
        printf("%d,%03d",c/1000,c%1000);
    }
    else{
        printf("%d,%03d,%03d",c/1000000,(c%1000000)/1000,c%1000);
    }
    return 0;
}