題目描述 ：

連結：https://ac.nowcoder.com/acm/contest/5672/G
來源：牛客網

• (Vji,Vj+Ni)(V^i_j, V^i_{j+N})(Vji​,Vj+Ni​), for all integer j such that 1≤j≤N1 \le j \le N1≤j≤N;
• (Vji,Vj+1i)(V^i_j, V^i_{j+1})(Vji​,Vj+1i​), for all integer j such that 1≤j≤N−11 \le j \le N-11≤j≤N−1 or N+1≤j≤2N−1N + 1 \leq j \le 2N-1N+1≤j≤2N−1.

• (Vi+N1,V1i+1)(V^1_{i+N}, V^{i+1}_1)(Vi+N1​,V1i+1​), for all integer i such that 1≤i≤N−11 \le i \le N - 11≤i≤N−1;
• (Vi1,V1+Ni)(V^1_{i}, V^{i}_{1+N})(Vi1​,V1+Ni​), for all integer i such that 2≤i≤N2 \le i \le N2≤i≤N.

輸入描述:

The input contains only two integers N, M (1≤N≤3000,N∗N∗2<M≤230)(1 \le N \le 3000, N*N*2 < M \le 2^{30})(1≤N≤3000,N∗N∗2<M≤230), and M is a prime.

輸出描述:

Output one integer indicating the answer.

分析

#include<bits/stdc++.h>
using namespace std;
long long a[18000010],b[18000010];
int dp[3010][3010],g[3010][3010],w[3010][3010],ans[3010],n,m,mod;
int get(int x,int y)
{
    return a[x]*b[y]%mod*b[x-y]%mod;
}
int main()
{
    int x,y,i,j;
    scanf("%d%d",&n,&mod);
    m=n*n*2;
    a[0]=1;
    a[1]=1;
    b[0]=1;
    b[1]=1;
    ans[0]=1;
    for(i=2;i<=m;i++)
    {
        a[i]=a[i-1]*i%mod;
        b[i]=mod-(mod/i)*b[mod%i]%mod;
    }
    for(i=1;i<=m;i++) b[i]=b[i-1]*b[i]%mod;
     
     
    for(i=0;i<=n;i++)
    for(j=i;j<=n;j++)
    {
        if(i==0) w[i][j]=1;
        else
        {
            w[i][j]=w[i-1][j];
            if(i<j) w[i][j]+=w[i][j-1];
            if(w[i][j]>=mod) w[i][j]-=mod;
        }
    }
     
     
    for(j=2;j<=n;j++)
    {
        if(j>=3) y=dp[0][j-1];
        else y=ans[0];
        x=(j-1)*(n*2+1)-n*2;
        (dp[0][j]+=1ll*w[n][n]*y%mod*get(x+n*2,n*2)%mod)%=mod;
    }
     
    for(i=1;i<n;i++)
    {
        for(j=0;j<=n;j++)
        {
            x=2*i-1+(i-1)*(n*2);
            g[i][j]=1ll*w[j][n]*ans[i-1]%mod*get(n+j+x,x)%mod;
            if(j!=0) (g[i][j]+=g[i][j-1])%=mod;
        }
         
         
        ans[i]=dp[i-1][i+1]+g[i][n];
        if(ans[i]>=mod) ans[i]-=mod;
         
         
        for(j=2;j<=n;j++)
        {
            dp[i][j]=dp[i-1][j];
            if(j-1>i+1) y=dp[i][j-1];
            else y=ans[i];
            x=i+(j-1)*(n*2+1)-n*2;
            (dp[i][j]+=1ll*w[n][n]*y%mod*get(x+n*2,n*2)%mod)%=mod;
        }
    }
    printf("%d\n",ans[n-1]);
    return 0;
}