Untitled
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 840 Accepted Submission(s): 472
Problem Description
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
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
T answers in T lines.
Sample Input
2 2 9 2 7 2 9 6 7
Sample Output
2 -1
Source
BestCoder Round #49 ($)
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解析：直接暴力搜索就好了。
第一次提交时，惊喜的发现居然排在第二位，exe.memory为1416kb，然后不断测试，显示删掉多余变量，然后又直接把对输入数据的排序给去掉，就排第一了，好开森。。。。。
代码：
<span style="font-size:14px;">#include<cstdio>
#include<algorithm>
using namespace std;
int a[30],n,m,ans;
void dfs(int step,int x)
{
if(step>=ans)return;
for(int i=1;i<=n;i++)
{
if(x%a[i]==0){ans=step;return;}
if(x>a[i])dfs(step+1,x%a[i]);
}
}
int main()
{
int i,j,t;
while(scanf("%d",&t)==1)
for(i=1;i<=t;i++)
{
scanf("%d%d",&n,&m);
for(j=1;j<=n;j++)scanf("%d",&a[j]);
ans=30,dfs(1,m);
if(ans==30)printf("%d\n",-1);
else printf("%d\n",ans);
}
return 0;
}</span>
