码迷,mamicode.com
首页 > 其他好文 > 详细

HDU 5339 Untitled(暴搜)

时间:2015-08-07 18:58:43      阅读:92      评论:0      收藏:0      [点我收藏+]

标签:

 

                 Untitled



 

Problem Description
There is an integer $a$ and $n$ integers $b_1, \ldots, b_n$. After selecting some numbers from $b_1, \ldots, b_n$ in any order, say $c_1, \ldots, c_r$, we want to make sure that $a \ mod \ c_1 \ mod \ c_2 \ mod \ldots \ mod \ c_r = 0$ (i.e., $a$ will become the remainder divided by $c_i$ 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 $T \leq 5$, which represents the number of testcases.

For each testcase, there are two lines:

1. The first line contains two integers $n$ and $a$ ($1 \leq n \leq 20, 1 \leq a \leq 10^6$).

2. The second line contains $n$ integers $b_1, \ldots, b_n$ ($\forall 1\leq i \leq n, 1 \leq b_i \leq 10^6$).
 

 

Output
Print $T$ answers in $T$ lines.
 

 

Sample Input
2
2 9
2 7
2 9
6 7
 
 

 

Sample Output
2
-1
 
 
 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 int ans,m,n;
 6 int b[21];
 7 
 8 bool cmp(int a,int b)
 9 {
10     return a>b;
11 }
12 
13 void dfs(int total,int cur,int num)
14 {
15     if(total==0)
16     {
17         ans=min(ans,num);
18         return;
19     }
20     if(cur==m)
21     return;
22     dfs(total%b[cur],cur+1,num+1);
23     dfs(total,cur+1,num);
24 }
25 
26 int main()
27 {
28     //freopen("in.txt","r",stdin);
29     int i,t;
30     scanf("%d",&t);
31     while(t--)
32     {
33         ans=21;
34         scanf("%d%d",&m,&n);
35         for(i=0;i<m;i++)
36         scanf("%d",&b[i]);
37         sort(b,b+m,cmp);
38         dfs(n,0,0);
39         if(ans==21)
40         printf("-1\n");
41         else
42         printf("%d\n",ans);
43     }
44     return 0;
45 }

 

HDU 5339 Untitled(暴搜)

标签:

原文地址:http://www.cnblogs.com/homura/p/4711105.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!