1097B - Petr and a Combination Lock

位运算：

用在\([0,2^n]\)的每一个数来枚举所有情况，如果\(i >> j \& 1 == 1\)，那么第j个数就要加上，否则要减去

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 20;
int n;
int a[N];
int main () {
	cin >> n;
	for (int i = 0;i < n;i++) cin >> a[i];
	bool flag = false;
	for (int k = 0;k < 1 << n;k++) {
		int degree = 0;
		for (int i = 0;i < n;i++) {
			if (k >> i & 1) degree += a[i];
			else degree -= a[i];
		}
		if (degree%360 == 0) {
			flag = true;
			break;
		}
	}
	if (flag) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}

标签：false,degree,Combination,int,Lock,flag,cin,1097B,include
来源： https://www.cnblogs.com/incra/p/16388118.html