1001. A+B Format

最新推荐文章于 2024-10-14 16:58:09 发布

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本文详细解析了一道PAT题目，要求按照标准格式输出两个整数的和，包括负数和不同位数情况的处理。通过C语言实现，展示了如何根据数字大小分组并用逗号分隔。

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：这道题就是按照给定的格式输出2个数的和，由于a,b都是<=1000000的，所以可以直接进行判断，在编写中要注意0的情况

code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    long int a,b;               //定义变量a,b和sum
    scanf("%ld %ld",&a,&b);     //输入a,b
    long sum = a + b;                //将a,b之和赋值给sum
    if(sum<0){                  //判断是否为负数
        sum = -1*sum;         //变为正的
        printf("-");            //输出负号
    }
    if(sum>=1000000){            //当大于1000000时
        printf("%ld,%03ld,%03ld",sum/1000000,sum/1000%1000,sum%1000);
    }
    else if(sum>=1000){               //当大于1000时
        printf("%ld,%03ld",sum/1000,sum%1000);
    }
    else{                       //小于1000时
        printf("%ld",sum);
    }
    
    return 0;
}

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今天又开始啦，做完数据结构的题后继续做做PAT的啦，继续努力啦。。。^_^