骚气的组合题

以二进制为本的骚气操作

    n个数一共2的n次方种组合方式，以循环次数的二进制来控制选与不选，某位为1则选择某位的数字，最后看sum是否为0，是则ans+1；

# include <iostream>
# include <algorithm>
# include <cstdio>
# include <cstdlib>
# include <cmath>
using namespace std;

int main(){
	int n, ans = 0;
	scanf("%d", &n);
	int a[100];
	for(int i = 1; i <= n; i++){
		scanf("%d", &a[i]);
	}
	int k = pow(2, n);
	for(int i = 1; i <= k; i++){
		int temp = i, sum = 0, lab = 1;
		while(temp > 0){
			if(temp & 1){
				sum += a[lab];			
			}
			lab ++;
			temp >>= 1;
		}
		if(sum == 0) ans++;
	}
	
	printf("%d
", ans);
	return 0;
}