TLE - Time Limit Exceeded

TLE - Time Limit Exceeded

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Given integers N (1 ≤ N ≤ 50) and M (1 ≤ M ≤ 15), compute the number of sequences a₁, ..., a_N such that:

• 0 ≤ a_i < 2^M
• a_i is not divisible by c_i (0 < c_i ≤ 2^M)
• a_i & a_i+1 = 0 (that is, a_i and a_i+1 have no common bits in their binary representation)

Input

The first line contains the number of test cases, T (1 ≤ T ≤ 10). For each test case, the first line contains the integers N and M, and the second line contains the integers c₁, ..., c_N.

Output

For each test case, output a single integer: the number of sequences described above, modulo 1,000,000,000.

Example

Input:
1
2 2
3 2

Output:
1

The only possible sequence is 2, 1.

分析：考虑相邻a[i]&a[i+1]=0；

　　　初始化状态转移为dp[i][j]=dp[i-1][j^((1<<m)-1)]；

　　　dp[i][j]表示第i步为j的方案数；

　　　其次，若j&k=j，则dp[i][j]也应包含dp[i][k]，这个可以推回去与一下；

　　　dp[i][j]=Σdp[i][k],j&k=j，这个即为高维前缀和；

　　　剩下不被整除特判一下即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000000
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
const int N=2e2+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,a[51],dp[51][1<<15];
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        rep(i,1,n)scanf("%d",&a[i]);
        rep(i,0,(1<<m)-1)if(i%a[1]!=0)dp[1][i]++;
        rep(i,2,n)
        {
            rep(j,0,(1<<m)-1)dp[i][j]=dp[i-1][j^((1<<m)-1)];
            rep(j,0,m-1)
            {
                rep(k,0,(1<<m)-1)
                {
                    if((~k)&(1<<j))(dp[i][k]+=dp[i][k^(1<<j)])%=mod;
                }
            }
            for(j=0;j<(1<<m);j+=a[i])dp[i][j]=0;
        }
        ll ret=0;
        rep(i,0,(1<<m)-1)(ret+=dp[n][i])%=mod;
        printf("%lld\n",ret);
    }
    return 0;
}