1 题目

1001 A+B Format （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(逗号) (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 参考代码：

#include<cstdio>
#include<cstring>
int ans[10];
int main(int argc, char const *argv[])
{
	int a,b;
	scanf("%d %d", &a, &b);
	int sum =0;
	sum = a + b;
	if(sum < 0){
		printf("-");
		sum = - sum;
	}

	int len = 0;
	// if(sum == 0) ans[len++] = 0;
	// while(sum != 0){
	// 	ans[len++] = sum % 10;
	// 	sum /= 10;
	// }
	do{
		ans[len++] = sum % 10;
		sum /= 10; 
	}while(sum);

	for (int i = len - 1; i >= 0; --i)
	{
		printf("%d", ans[i]);
		if(i % 3 == 0 && i != 0) printf(",");
	}
	printf("\n");
	return 0;
}

参考代码2:

#include<cstdio>
#include<cstring>
int main(int argc, char const *argv[])
{
	int a, b;
	scanf("%d%d", &a, &b);
	int sum = a + b;
	
	if(sum < 0){
		printf("-");
		sum = - sum;
	}

	if(sum >= 1000000) printf("%d,%03d,%03d", sum / 1000000, sum % 1000000 / 1000, sum % 1000);
	else if(sum >= 1000) printf("%d,%03d", sum / 1000, sum % 1000);
	else {
		printf("%d", sum);
	}
	printf("\n");
	return 0;
}