[Daily] 2020/6/1

B.
题意：给出一个整数n，将从1开始前n个整数换分成两部分，判断两部分各自的和是否为互质数

思路是，首先求出前n个整数和sum，然后让 i 从1到n遍历，判断 i 和 sum 的最大公约数

#include 
#include 
#include 
#include 
using namespace std;

const int xa = 1e5 + 10;
int a[30];
bool cmp(int a, int b) {
	return a < b;
}

int mo(int n) {
	int sum = 0;
	for (int i = 0; i <= n; i++) {
		sum += i;
	}
	return sum;
}

int gcd(int a, int b)
{
	int r;
	while (b > 0)
	{
		r = a % b;
		a = b;
		b = r;
	}
	return a;
}

int main() {
	int n;
	cin >> n;
	int sum = mo(n);
	for (int i = 2; i <= n; i++) {
		if (gcd(sum - i, i) != 1) {
			cout << "Yes" << endl;
			cout << "1 " << i << endl;
			cout << n - 1 << " ";
			for (int j = 1; j <= n; j++) {
				if (j == i)	continue;
				cout << j << " ";
			}
			return 0;
		}
	}
	cout << "No" << endl;
}