Untitled

Problem Description

There is an integer a and n integers b1,…,bn. After selecting some numbers from b1,…,bn in any order, say c1,…,cr, we want to make sure that a mod c1 mod c2 mod… mod cr=0 (i.e., a will become the remainder divided by ci each time, and at the end, we want a to become 0). Please determine the minimum value of r. If the goal cannot be achieved, print −1 instead.

 

Input

The first line contains one integer T≤5, which represents the number of testcases.

For each testcase, there are two lines:

1. The first line contains two integers n and a (1≤n≤20,1≤a≤106).

2. The second line contains n integers b1,…,bn (∀1≤i≤n,1≤bi≤106).

 

Output

Print T answers in T lines.

 

Sample Input

2
2 9
2 7
2 9
6 7

 

Sample Output

2
-1

题解：暴力做法，不过用的是递归，每次选一个数我可以选择除当前数或者不除当前数，就这两个选择。时间复杂度2的n次方，n<=20，可以过。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int a[30];
int res;

int min(int a,int b)
{
	return a > b ? b : a;
}

bool com(int a,int b) 
{
	return a > b;
}

void dfs(int n,int r,int c,int d)
{
	if(c >= n)
	{
		return;
	}
	if(0 == r % a[c])   //找到一组 
	{
		res = min(d + 1,res);
	}
	dfs(n,r % a[c],c + 1,d + 1); //除当前数 
	dfs(n,r,c + 1,d);        //不除当前数 
}

int main()
{
	int T;
	cin>>T;
	int n,m;
	while(T--)
	{
		res = 100;
		scanf("%d%d",&n,&m);
		for(int i = 0;i < n;i++)
		{
			scanf("%d",a + i);
		}
		sort(a,a + n,com); //必须从最大的开始，从小的开始的话，一旦取余就永远得到余数了，因为余数肯定小于后面的数 
		dfs(n,m,0,0);
		
		if(100 == res)
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n",res);
		}
	}
	
	
	return 0;
}