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 + 1)
    {
        int temp = sum;
        int length = 0;
        for (int j = 1; j <=n; j++)
        {
            if (temp == 0)
                break;
            if (c[j] == 0)
            {
                temp %= a[j];
                length++;
            }
        }
        if (temp==0)
        min = Min(min, length);
        return 0;
    }
       //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] = 0;//用来标记哪些被选了
        dfs(m + 1);
        c[m] = 1;//标记未被选
        dfs(m + 1);
}
int cmp(const void*a, const void*b)
{
    return *(int *)b - *(int*)a;
}
int main()
{
    int num;
    scanf("%d", &num);
    while (num--)
    {
        min= 1000000;
        scanf("%d%d", &n, &sum);
        for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
        qsort(&a[1], n, sizeof(a[1]), cmp);
        dfs(1);
        if (min!=1000000)
        printf("%d\n", min);
        else printf("-1\n");
    }
}