CodeForce Round#49 untitled (Hdu 5339)

Untitled

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 947    Accepted Submission(s): 538

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

 

Source

BestCoder Round #49 ($)

 

本次题目就是个纯暴力搜索的一道题目，唯一一点要注意的就是序列应该先降序排列，为什么呢？(因为如果升序排列的话i<j，Ci小于Cj,Ci先进行取余运算，那么Cj就没有意义了)

除此之外就是简单的深搜，附上本人的代码

 #include<stdio.h>
 #include<stdlib.h>
 #define maxn 30
 int min, sum,n;
 int a[maxn], b[maxn],c[maxn];
 int Min(int a, int b)
 {
     return a> b ? b:a;
 }
 int dfs(int m)
 {
     if (m == n + )
     {
         int temp = sum;
         int length = ;
         for (int j = ; j <=n; j++)
         {
             if (temp == )
                 break;
             if (c[j] == )
             {
                 temp %= a[j];
                 length++;
             }
         }
         if (temp==)
         min = Min(min, length);
         return ;
     }
       //这里利用选择为0，未选为1，方便代码调试时按照二进制运算来看，比较直观
         c[m] = ;//用来标记哪些被选了
         dfs(m + );
         c[m] = ;//标记未被选
         dfs(m + );
 }
 int cmp(const void*a, const void*b)
 {
     return *(int *)b - *(int*)a;
 }
 int main()
 {
     int num;
     scanf("%d", &num);
     while (num--)
     {
         min= ;
         scanf("%d%d", &n, &sum);
         for (int i = ; i <= n; i++)
         scanf("%d", &a[i]);
         qsort(&a[], n, sizeof(a[]), cmp);
         dfs();
         if (min!=)
         printf("%d\n", min);
         else printf("-1\n");
     }
 }