(PAT Advanced)1001.A+B Format（字符串处理） C++

原题：https://pintia.cn/problem-sets/994805342720868352/problems

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

题目分析：
风格一：
先计算A+B的和，然后利用to_string转成字符串；除了第一位是负号的情况，只要当前下标满足(len-1-i)%3==0即，剩下的位数是3的倍数的时候，且i不是最后一位，就在逐位输出后加上一个逗号。

#include <iostream>
#include <string>

using namespace std;
int main() {
	int a, b;
	cin >> a >> b;
	int res = a + b;
	string ans = to_string(res);
	for (int i = 0; i < ans.length(); i++) {
		cout << ans[i];
		if (ans[i] == '-') continue;// 负数不能考虑第一符号位
		//if ((i + 1) % 3 == ans.length() % 3 && i != ans.length() - 1)
		if ((ans.length() - 1 - i) % 3 == 0 && i != ans.length() - 1)
			cout << ",";
	}
	cout << endl;
	system("pause");
	return 0;
}