BestCoder#49

Untitled

Accepts: 504

Submissions: 1542

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

There is an integer aaa and nnn integers b1,…,bnb_1, \ldots, b_nb​1​​,…,b​n​​. After selecting some numbers from b1,…,bnb_1, \ldots, b_nb​1​​,…,b​n​​ in any order, say c1,…,crc_1, \ldots, c_rc​1​​,…,c​r​​, we want to make sure that a mod c1 mod c2 mod… mod cr=0a \ mod \ c_1 \ mod \ c_2 \ mod \ldots \ mod \ c_r = 0a mod c​1​​ mod c​2​​ mod… mod c​r​​=0 (i.e., aaa will become the remainder divided by cic_ic​i​​ each time, and at the end, we want aaa to become 000). Please determine the minimum value of rrr. If the goal cannot be achieved, print −1-1−1 instead.

Input

The first line contains one integer T≤5T \leq 5T≤5, which represents the number of testcases.

For each testcase, there are two lines:

1. The first line contains two integers nnn and aaa (1≤n≤20,1≤a≤1061 \leq n \leq 20, 1 \leq a \leq 10^61≤n≤20,1≤a≤10​6​​).

2. The second line contains nnn integers b1,…,bnb_1, \ldots, b_nb​1​​,…,b​n​​ (∀1≤i≤n,1≤bi≤106\forall 1\leq i \leq n, 1 \leq b_i \leq 10^6∀1≤i≤n,1≤b​i​​≤10​6​​).

Output

Print TTT answers in TTT lines.

Sample Input

2
2 9
2 7
2 9
6 7

Sample Output

2
-1

运用二进制的思想，去枚举每种可能的情况，然后算下最小的。当然，先把比 a 大的数筛掉

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
      int T;
      cin >> T;
      int n, a, b[], i, cnt;
      while(T--) {
            int Min = ;
            bool tag = false;
            cin >>n >> a;
            for(int i = ; i < n; ++i)
                  cin >> b[i];
            sort(b, b+n);
            for(i = n-; i >= ; --i) {
                  if(a >= b[i])
                        break;
            }
            if(i == ) {
                  if(a % b[] == )
                        cout <<  << endl;
                  else
                        cout << - << endl;
            } else {
                  for(int j = ; j < ( << (i+)); ++j) {
                        int tmp = a;
                        cnt = ;
                        for(int k = i; k >=  ; --k) {
                              if(( << k) & j) {
                                    tmp %= b[k];
                                    cnt++;
                                    //cout << j << " " <<tmp << " " << b[k] << " "<<cnt << endl;;
                              }
                        }
                        if(tmp == ) {
                              Min = min(Min, cnt);
                              tag = true;
                        }
                  }
                  if(tag)
                        cout << Min << endl;
                  else
                        cout << - << endl;
            }
      }
      return ;
}