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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

22 92 72 96 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;}