BC - 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

      第一次打BC的第一道题，比赛结束前没A出来。这道题方法其实和poj1753一样，枚举步数然后在DFS找当前步数下所有元素能否构成符合题意的情况。此题关键在于要对b数组由大到小排序。排序之后只需要依次遍历如1,2,3；1,2,4；1,3,4;因为该题是不断对a取余，如果先对小的数取余那么余数一定比小的还要小，那么这时再对较大的数取余没有任何意义。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#define INF 0x3f3f3f3f

int cmp(int x,int y)
{
    return x>y;
}
using namespace std;
int arr[100];
int vis[100];
int ans,k,z;
int flag;

void fun(int step,int m,int a)
{
    int i;
    if(step==k)
    {
        for (i=1;i<z;i++)
        {
            if(vis[i])
                a %= arr[i];
            if(a==0)
                {flag=1;ans=step;return;}
        }
    }
    for (i=m+1;i<z;i++)
    {
        vis[i]=1;
        fun(step+1,i,a);
        if(flag) return;
        vis[i]=0;
    }
}
int main()
{
    int t,n,a,x,i;
    scanf("%d",&t);
    while (t--)
    {
        flag=0;z=1;
        scanf("%d%d",&n,&a);
        for ( i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(x<=a) arr[z++]=x;
        }
        sort(arr+1,arr+z,cmp);
        for (i=1;i<z;i++)
        {
            memset(vis,0,sizeof(vis));
            k=i;
            fun(0,-1,a);
            if(flag) break;
        }
        if(!flag)
            printf("-1\n");
        else
            printf("%d\n",ans);
    }
}