A1001 A+B Format (20 point(s))

a+b和按要求输出

1. 原文

原题

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 −${10}^6$≤a,b≤${10}^6$. 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

2. 解析

a,b两数相加的和按格式输出

求和结果的位数

do{
	num[idx++]=sum%10;
	sum/=10;
}while(sum!=0);

求得的数组是从个位到最高位排列的199999

从个位开始，cnt=1，循环遍历，每次cnt++表示前进一位，当遍历满三位时ans+=’,’ —>199,

⚠️注意处理最后一位，9前面不加 199,999

最后字符串反向输出 999,991

3. AC代码

#include<cstdio>
int num[20]={};
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
    int c = a+b;
    if (c<0)
    {
    	printf("-");
    	c = -c;
    }
    int idx = 0;
    do{
    	num[idx++]=c%10;
    	c/=10;
    }while(c!=0);
    for (int i = idx-1; i >= 0; i--)
    {
    	printf("%d",num[i]);
    	if (i%3==0&&i>0)
    	{
    		printf(",");
    	}
    }
    printf("\n");

	return 0;
}