Hakata's stock market

Subject:
https://ac.nowcoder.com/acm/problem/15550

$A[0][0]=p$，且
A [ i ] [ j ] = { A [ i ] [ j − 1 ] i = 0 , 0 < j < m max ⁡ 0 ≤ k ≤ j { A [ i − 1 ] [ k ] + ∑ l = k + 1 j A [ i − 1 ] [ l ] j − k } 1 ≤ i , 0 ≤ j < m A[i][j]=\begin{cases} A[i][j-1]\quad i=0,0<j<m\\ \max\limits_{0\le k\le j}\{A[i-1][k]+\sum_{l=k+1}^{j}\frac{A[i-1][l]}{j-k}\}\quad 1\le i,0\le j<m \end{cases} A[i][j]=⎩⎨⎧​A[i][j−1]i=0,0<j<m0≤k≤jmax​{ A[i−1][k]+∑l=k+1j​j−kA[i−1][l]​}1≤i,0≤j<m​
求$A\left[x\right]\left[y\right]$

Idea:
Calculate by hand and find that it is AA0 0 of the$A$ arrayRow$0$ and$0$ columns are$p$，然后
$$A[i][j]=A[i-1][j]+A[i-1][j-1]1\le i,j<m$$
Use combinatorics to calculate the0th 0Row$0$ and0 0The contribution of the number in the$0$ column is sufficient.

#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
const int N=1000009;
int T;
ll p[N],p1[N],m,x,y,P;
ll qpow(ll a, ll b) {
    
    
    ll res=1;
    a%=mod;
    while(b) {
    
    
        if(b&1)
            res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
ll cal(ll a,ll b) {
    
    
    if(a<0||b<0||a-b<0)
        return 0;
    return p[a]*p1[b]%mod*p1[a-b]%mod;
}
int main() {
    
    
    p[0]=1;
    for(int i=1; i<N; i++)
        p[i]=p[i-1]*i%mod;
    p1[N-1]=qpow(p[N-1],mod-2);
    for(int i=N-2; i>=0; i--)
        p1[i]=p1[i+1]*(i+1)%mod;
    while(scanf("%lld%lld%lld%lld",&m,&x,&y,&P)==4) {
    
    
        ll ans=0;
        x--;
        if(!x||!y) {
    
    
            cout<<P<<endl;
            continue;
        }
        ans+=cal(x,y);
        for(int i=1; i<=y; i++)
            ans=(ans+cal(x,y-i))%mod;
        cout<<ans*P%mod<<endl;
    }
    return 0;
}