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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 570    Accepted Submission(s): 291

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 ($)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5339

题目大意：重排列bi，问a对重排列的数不断取模最快能为0的取模次数

题目分析：其实是水题，首先对数字小的取完余后再对数字大的取余等于没取，所以先对大数字取余，从大到小排序，因为n很小，DFS随意搜，也可以用状态压缩做，把每个数字选或不选的状态二进制压缩

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1 << 22;
int const INF = 0x3fffffff;
int b[25], sta[MAX];

int lowbit(int x)
{
    return x & (-x);
}

bool cmp(int a, int b)
{
    return a > b;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        memset(sta, 0, sizeof(sta));
        int n, a;
        scanf("%d %d", &n, &a);
        for(int i = 1; i <= n; i++)
            scanf("%d", &b[i]);
        sort(b + 1, b + n + 1, cmp);
        for(int i = 1; i <= n; i++)
            sta[1 << (i - 1)] = b[i];
        int cnt, ans = INF;
        for(int i = 1; i < (1 << n); i++)
        {
            int tmp = a;
            cnt = 0;
            for(int j = i; j > 0; j -= lowbit(j))
            {
                tmp %= sta[lowbit(j)];
                cnt ++;
            }
            if(tmp == 0)
                ans = min(ans, cnt);
        }
        if(ans == INF)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }
}

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