hdu 5339 Untitled dfs

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

 

 

    dfs题：不过这题很容易超时。要注意剪枝。剪枝地方我会在代码中说明，看代码就可以了。

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,b[30],r,mn;
int v[30];
void dfs(int k,int a)
{
    int i,s;
    for (i=1;i<=n;i++)
    {
        if (!v[i])
        {
            s=a%b[i];
            if (s==0) {r=min(r,k+1);return ;}
            if (s<mn) continue; //剪枝。如果s比这些数中最小值还小，那么s就不可能为0.
            v[i]=1;
            dfs(k+1,s);
            v[i]=0;
        }
    }
    return ;
}
int main()
{
    int t,i,a,c,d;
    scanf("%d",&t);
    while (t--)
    {
        r=30;
        scanf("%d%d",&n,&a);
        c=n;
        i=1;
        mn=1000000;
        while (c--)
        {
            scanf("%d",&d);
            if (d>a) continue; //剪枝。a模比a大的数还是a，就可以不用管。
            if (a%d==0) r=1;  //剪枝。如果a是这些数中一些数的倍数，那么最小值就是一。
            b[i++]=d;
            mn=min(mn,d);
        }
        n=i-1;
        sort(b+1,b+i);
        if (r==1) {printf("1\n");continue;}
        memset(v,0,sizeof(v));
        dfs(0,a);
        if (r==30) printf("-1\n");
        else printf("%d\n",r);
    }
}