c ++ compositions primes and Analyzing

Title Description

Known n integers x1, x2, ..., xn, 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 are integers when 3,7,12,19, available with all combinations thereof and is:

3＋7＋12=22 3＋7＋19＝29 7＋12＋19＝38 3＋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.

Entry

The first line of two integers: n, k (1 <= n <= 20, k <n)

The second line n integers: x1, x2, ..., xn (1 <= xi <= 5000000)

Export

An integer (number satisfying the condition of the program).

Sample input

4 3
3 7 12 19

Sample Output

1

Source Code

#include <iostream>
#include <cmath>
#include <string.h>
#include <stdio.h>
using namespace std;
int n,m,a[100],s[100],sum,ans;
bool isPrime(int);
void dfs(int x,int y)
{
    if(x == m)//如果挑选出来3组数据,就求和,小于3个数,就继续挑选
    {
        sum = 0;
        for(int i = 1;i <= x;i ++)
        {
            printf("a[%d] = %d\n",i,a[i]);
            sum += s[a[i]];//对挑出来的三个数t求和
        }
        cout << "------------" << endl;
        if(isPrime(sum) == 1)
            ans ++;//如果是和是素数,就+1
        return ;
    }
    y ++;
    for(int i = y;i <= n;i ++)//从4个数中挑选三个数字
    {
        a[x + 1] = i;//改变下标形成新的挑选数
        dfs(x + 1,i);//当x + 1等于3 就开始求和  如果小于3就继续挑选
    }
}
bool isPrime(int r)//判断是否是素数
{
    for(int i = 2;i <= sqrt(r);i ++)
        if(r % i == 0)
            return 0;
    return 1;
}
int main()
{
    cin >> n >> m;
    for(int i = 1;i <= n;i ++)
        cin >> s[i];
    dfs(0,0);
    cout << ans << endl;
    return 0;
}