P1036 [NOIP2002 Popularization Group] Selection Number

[NOIP2002 Popularization Group] Selection Number

Question description

Wachi $n$ 个整数 $x_1,x_2,\cdots ,x_n$, hereafter $1$ integer $k$（$k<n$).从 $n$ Integer selection $Adding\; k$ integers can produce a series of sums. For example, when $n=4$，$k=3$，$4$ Integer division $3,7,12,When\; 19$, all combinations and their sum can be obtained:

$3+7+12=22$

$3+7+19=29$

$7+12+19=38$

$3+12+19=34$

Now, you are asked to calculate how many kinds of sums are prime numbers.

For example, in the above example, there is only one sum that is a prime number: $3+7+19=29$。

Input format

Two spaces separated integers in the first line $n,k$($1\le n\le 20$，$k<n$）。

Second line $n$ integers, respectively $x_1,x_2,\cdots ,x_n$（$1\le x_i\le 5\times 10^6$）。

Output format

Output an integer representing the number of categories.

Example #1

Sample input #1

4 3
3 7 12 19

Sample output #1

1

hint

[Source of the question]

NOIP 2002 Popularization Group Question 2

solution

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int n,k,ans,a[30],b[30];
bool zs(int x)
{
    
    
	if (x==1)
		return 1;
	for (int i=2;i*i<=x;i++)
		if (x%i==0)
			return 0;
	return 1;
}
int main()
{
    
    
	scanf ("%d%d",&n,&k);
	for (int i=1;i<=n;i++)
		scanf ("%d",&a[i]);
	while (b[0]==0)
	{
    
    
		int kk,sum=0,shu=0;
		for (kk=n;b[kk];kk--);
		b[kk]=1;
		for (int j=kk+1;j<=n;j++)
			b[j]=0;
		for (int i=1;i<=n;i++)
			if (b[i])
			{
    
    
				sum++;
				shu+=a[i];
			}
		if (sum==k)
			if (zs(shu))
				ans++;
	}
	printf ("%d",ans);
	return 0;
}