Problems C: [entry] recursive determination compositions primes +

1 title

Problems C: [recursive entry] composition + Analyzing prime
time limit: 1 Sec memory limit: 128 MB
submission: 2125 solution: 615
[Submit] [state] [discussion boards] [proposition al: External Import]
Title Description
known in the n integers b1, b2, ..., bn

And an integer k (k <n).

Optionally adding integers k from n integers in a range of available respectively.

For example, when n = 4, k = 3,4 integers, respectively, is 3,7,12,19, available all combinations of thereof and is:
　　　　3 + 7 + 12 = 223 + 7 + 19 = 297 + 12 + 19 = 383 + 12 + 19 = 34.
　　Now, we ask you to calculate the total number of species and is a prime number.

For example the embodiment, only one, and is a prime number: 3 + 7 + 19 = 29.

Input
of the first rows of two integers: n, k (1 <= n <= 20, k <n)
a second row of n integers: x1, x2, ..., xn (1 <= xi <= 5000000)

Output
an integer (number satisfying the condition of the program).

样例输入
4 3
3 7 12 19
样例输出
1

2 reference code

#include <cstdio>
#include <cmath>
#include <vector>

using std::vector;

typedef long long LL;
const int MAXN = 25;
int N,K;
int Count = 0;
LL  A[MAXN], res[MAXN];
bool hashTable[MAXN] = {false};

bool isPrime(LL x){
    if(x <= 1) return false;
    LL sqr = (LL) sqrt(1.0 * x);
    for (LL i = 2; i <= sqr; ++i)
    {
        if(x % i == 0) return false;
    }
    return true;
}

void BFS(int index, int r){
    if(r == K + 1){
        int sum = 0;
        for (int i = 1; i != K + 1; ++i) {
            sum += res[i];
        }
        if(isPrime(sum) == true){
            Count++;
         //   for (int j = 1; j != K + 1; ++j) {
         //       printf("%lld ", res[j]);
         //   }
         //   printf("\n");
        }

        return;
    }

    for (int i = index; i <= N; ++i)
    {
        if(hashTable[i] == false){
            res[r] = A[i];
            hashTable[i] = true;
            BFS(i, r + 1);
            hashTable[i] = false;
        }
    }
}

int main(int argc, char const *argv[])
{
    scanf("%d%d", &N, &K);
    for (int i = 1; i != N + 1; ++i)
    {
        scanf("%lld", &A[i]);
    }
    BFS(1,1);
    printf("%d\n", Count);
    return 0;
}

• Reference code 2

#include <cstdio>
#include <cmath>
#include <vector>

using std::vector;

typedef long long LL;
const int MAXN = 25;
int N,K;
int Count = 0;
LL  A[MAXN];
int sum = 0;

bool isPrime(LL x){
    if(x <= 1) return false;
    LL sqr = (LL) sqrt(1.0 * x);
    for (LL i = 2; i <= sqr; ++i)
    {
        if(x % i == 0) return false;
    }
    return true;
}

void BFS(int index, int r){
    if(r == K + 1){
        if(isPrime(sum) == true){
            Count++;
        }
        return;
    }

    for (int i = index; i <= N; ++i)
    {
            sum += A[i];
            BFS(i + 1, r + 1);
            sum -= A[i];
    }
}

int main(int argc, char const *argv[])
{
    scanf("%d%d", &N, &K);
    for (int i = 1; i != N + 1; ++i)
    {
        scanf("%lld", &A[i]);
    }
    BFS(1,1);
    printf("%d\n", Count);
    return 0;
}