2655: calc

dp方程易得，可證明答案總爲多項式，拉格朗日插值；

#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
const int N=1007;
int mod,n,A;
void upd(int& x,int y){
    x+=y;if(x>=mod)x-=mod;
}
int qpow(int a,int b){int aa=a,res=1;
    for(;b;b>>=1){if(b&1)res=1ll*res*aa%mod;
        aa=1ll*aa*aa%mod;
    }return res;
}
int f[N];
int main(){
    scanf("%d%d%d",&A,&n,&mod);A%=mod;
    int bin=1;
    rep(i,0,2*n+1)f[i]=1;
    rep(i,1,n){bin=1ll*i*bin%mod;
        rep2(j,2*n+1,1)f[j]=1ll*f[j-1]*j%mod;f[0]=0;
        rep(j,1,2*n+1)upd(f[j],f[j-1]);
    }int ans=0;
    rep(i,1,2*n+1){int res=1,a=1;
        rep(j,1,2*n+1)if(j!=i){
            a=1ll*a*(i-j+mod)%mod;
            res=1ll*res*(A-j+mod)%mod;
        }a=qpow(a,mod-2);
        a=1ll*a*f[i]%mod;
        upd(ans,1ll*a*res%mod);
    }
    printf("%d\n",1ll*ans*bin%mod);
}//0->1 +2 n->n+1 +2