【PAT甲级】1001 A+B Format (20 分)

1001 A+B Format (20 分)

题目描述

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 Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −10⁶≤a,b≤10⁶​​ . The numbers are separated by a space.

Output Specification:

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

思路

由于a，b的范围在−10⁶到10⁶​​ ，因此a+b的范围最多在 ·−2x10⁶到2x10⁶​​之间，因此我们我们可以用至多三个数字储存按照逗号分割的和。

例如：1123312，可以用1，123，312这三个数字储存

代码

注意在这里使用n1,n2,n3的形式来输出结果，当和小于0的时候要先转化为正数在计算，同时还要依次判断首位的数字是否为0来决定是否输出

#include "stdio.h"

int main()
{
    
    
    int a,b;
    scanf("%d %d",&a,&b);
    int c=a+b;
    bool flag=(c>=0);//because remainder operation may have erro for the negative numbers
    //so we need to flag wheather the sum is negative and change the negative to positive
    if(!flag)
      c=-c;
    int n1,n2,n3;//three number to represent the sum
    n3=c%1000;n2=c/1000%1000;n1=c/1000000%1000;
    if(!flag)
        printf("-");
    if(n1!=0)
    {
    
    
        printf("%d,%03d,%03d\n",n1,n2,n3);
    }
    else if(n2!=0)
    {
    
    
        printf("%d,%03d\n",n2,n3);
    }
    else
    {
    
    
        printf("%d\n",n3);
    }
    return 0;
}

git仓库:1001 A+B Format