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#include <bits/stdc++.h>
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using namespace std;
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const int N = 30;
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int a[N];
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//是不是质数
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bool isPrime(int n) {
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for (int i = 2; i <= n / i; i++)
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if (n % i == 0) return false;
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return true;
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}
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int main() {
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int n, k, ans = 0;
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cin >> n >> k;
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for (int i = 0; i < n; i++) scanf("%d", &a[i]);
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int U = 1 << n; //U-1即为全集 ,比如 1<<5 就是 2的5次方,就是32,U=32。而U-1=31,就是表示 1 1 1 1 1
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for (int S = 0; S < U; S++) { //枚举所有子集[0,U)
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//找到k元子集
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if (__builtin_popcount(S) == k) {
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int sum = 0;
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//是哪些数存在于子集中呢?
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for (int i = 0; i < n; i++) {
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int bit = (S >> i) & 1;
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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
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}
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//判断子集元素和是不是素数
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if (isPrime(sum)) ans++;
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}
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}
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cout << ans << endl;
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return 0;
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}
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