PAT甲级——A+B Format

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

思路：

我想的这个思路有点傻瓜，就是将输入的两个数加起来，记录ans是正数还是负数，取ans绝对值然后从个位数开始将每一位存入vector中。当ans 小于1000的时候直接输出ans就行。当ans>1000的时候，需要用vector的长度除以3得到余数，然后用长度减去3*余数，这样就可以知道是否存在不足3个元素的group（也就是最前面是否要单独输出1到2个数）。后面依次倒着输出vector中的单个元素并且补上逗号即可。代码可能有些冗余，但是如果将int换成long long那么可以应用到更大数的加法，结果也是对的，不限于题目所限定的数据范围。

代码：

#include 
#include
#include
using namespace std;

int main()
{
	int a, b, ans = 0;
	vectornum;
	bool flag = true;
	scanf_s("%d %d", &a, &b);
	ans = a + b;
	if (ans < 0)
		flag = false;
	ans = abs(ans);
	int temp = 0, dot = 0, frontNum = 0,number=ans;
	while (number != 0)
	{
		temp = number % 10;
		number = number / 10;
		num.push_back(temp);
	}
	dot = num.size() / 3;
	//if(dot>=1)
	frontNum = num.size() - dot * 3;
	int len = num.size() - 1;
	if (flag == false)
		printf_s("-");
	//if (ans < 1000)
	//	printf_s("%ld", ans);
	if (ans < 1000)
		printf_s("%d", ans);
	else {
		if (frontNum != 0) {
			while (frontNum != 0)
			{
				printf_s("%d", num[len]);
				frontNum--;
				len--;
			}
			printf_s(",");
		}
		//else printf_s("%ld", ans);

		for (int i = 0; i < dot; i++)
		{
			for (int j = 0; j < 3; j++)
			{
				printf_s("%d", num[len]);
				len--;
			}
			if (i < dot - 1)
				printf_s(",");
		}
	}
	
	
	return 0;


}