C++ :

#include<iostream>
#include<cmath>
using namespace std;
int n,m,i,ans=0,s=0,a[21],f[10001]={0},p[2000];
void get()//建立10000以内的素数表
{ int i,j,k=0;
f[1]=1;
for(i=2;i<=int(sqrt(10000));i++)
if(f[i]==0)
{ p[k++]=i;
for(j=i*i;j<=10000;j+= i)f[j]=1;
}
} void work(int s)
{ int i;
if(s<=10000){if(f[s]==0)ans++;}//直接判断
else{ for(i=0;p[i]*p[i]<=s;i++)if(s%p[i]==0)return;
ans++;
}
} void search(int d,int pre)//d选的个数,pre前一次选的编号
{ int i;
if(d>m)work(s);
else for(i=pre+1;i<=n-(m-d);i++)
{ s+=a[i];
search(d+1,i);
s-=a[i];
}
}
int main()
{ cin>>n>>m;
for(i=1;i<=n;i++)cin>>a[i];
get();
search(1,0);
cout<<ans;
}

Pascal :

var  
 n,k,i,ans:longint;  
 a:array[1..20] of longint;    
function prime(x:longint):boolean;  
var  
 i:longint;  
begin  
 if x=1 then exit(false);  
 for i:=2 to trunc(sqrt(x)) do  
  if x mod i=0 then exit(false);  
exit(true);  
end;    
procedure search(m,depth,sum:longint);  
var  
 i:longint;  
begin  
 if depth>k then  
 begin  
  if prime(sum) then inc(ans);  
  exit;  
 end; 
 for i:=m+1 to n do search(i,depth+1,sum+a[i]);  
end;   
begin  
 readln(n,k);  
 for i:=1 to n do read(a[i]);  
 ans:=0;  
 search(0,1,0);  
 writeln(ans);  
end. 

Java :

import java.util.*;
public class Main{
	static int[] a;
	static int n,k,ret = 0;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		k = sc.nextInt();
		a = new int[n+3];
		for(int i=0;i<n;i++)
			a[i] = sc.nextInt();
		f(0,0,0);
		System.out.println(ret);
		
	}
	static void f(int num,int pos,int sum){
		if(num==k){
			if(isPrime(sum))
				ret++;
				return ;
		}
		if(pos>=n)  return;
		f(num+1,pos+1,sum + a[pos]);
		f(num,pos+1,sum);
	}
	static boolean isPrime(int sum){
		for(int i=2;i<=Math.sqrt(sum);i++)
			if(sum%i==0) return false;
		return true;
	}
}