[C/C++] 1001 A+B Format (20)（20 分）

1001 A+B Format (20)（20 分）

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>
int main()
{	
	int a,b;
	scanf("%d%d",&a,&b);
	int sum = a+b;
	if(sum < 0) {
		printf("-");
		sum = -sum;
	} 
	
	int ans[20];
	int i = 0;
	if(sum == 0) ans[i++] = 0;	//sum为0时候特殊处理，否则就用do while 
	while(sum != 0) {
		ans[i++] = sum % 10;
		sum /= 10;
	}
	for(int j = i-1;j>=0; j--) {
		printf("%d",ans[j]);
		if(j % 3 == 0&&j!=0) printf(",");
	}
	
	return 0;
 } 

为什么要把负数变成正数？if not,就会这样-9-9-9,-9-9-1
本来我的想法是造一个计数器，每三个数放进一个新数组，之后再放一个逗号。晴神的太简单了，要记得好好利用数组下标，晴神就是根据数组下标判断是否直接输出“，”：

正序输出：

#include<stdio.h>
#include<string.h>

int main()
{
	int a, b;
	scanf("%d %d",&a, &b);
	int sum = a+b;
	if(sum < 0) {
		printf("-");
		sum = -sum;
	}
	char ans[10];
	
	
	sprintf(ans,"%d",sum);	
	int len = strlen(ans);
	for( int i=0; i<len; i++){
		printf("%c",ans[i]);
		if( (i+1)%3==len%3 && i!=len-1) printf(",");
	}
	
	return 0;
}

if( (i+1)%3==len%3 && i!=len-1) printf(",");