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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 30;
int a[N];
//是不是质数
bool isPrime(int n) {
for (int i = 2; i <= n / i; i++)
if (n % i == 0) return false;
return true;
}
int main() {
int n, k, ans = 0;
cin >> n >> k;
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
int U = 1 << n; //U-1即为全集 ,比如 1<<5 就是 2的5次方就是32U=32。而U-1=31就是表示 1 1 1 1 1
for (int S = 0; S < U; S++) { //枚举所有子集[0,U)
//找到k元子集
if (__builtin_popcount(S) == k) {
int sum = 0;
//是哪些数存在于子集中呢?
for (int i = 0; i < n; i++) {
int bit = (S >> i) & 1;
if (bit) sum += a[i]; //遍历数字S的每一位如果不是0表示这一位上的数字是存在的需要加进来
}
//判断子集元素和是不是素数
if (isPrime(sum)) ans++;
}
}
cout << ans << endl;
return 0;
}