CF932E Team Work(第二類斯特林數)

題目

CF932E Team Work

前置：斯特林數\(\Longrightarrow\)點這裏

作法

\[\begin{aligned}\\ &\sum\limits_{i=1}^n C_n^ii^k\\ &\sum\limits_{i=1}^n C_n^i\sum\limits_{j=0}^iC_i^j\begin{Bmatrix}k\\j\end{Bmatrix}j!\\ &\sum\limits_{i=1}^n \frac{n!}{(n-i)!}\sum\limits_{j=0}^i\frac{\begin{Bmatrix}k\\j\end{Bmatrix}}{(i-j)!}\\ &\sum\limits_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\sum\limits_{i=j}^n\frac{n!}{(n-i)!}\frac{1}{(i-j)!}\\ &\sum\limits_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\sum\limits_{i=j}^n\frac{n!}{(n-j)!}\frac{(n-j)!}{(n-i)!(i-j)!}\\ &\sum\limits_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{(n-j)!}\sum\limits_{i=j}^nC_{n-j}^{i-j}\\ &\sum\limits_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{(n-j)!}2^{n-j}\\ \end{aligned}\]
至此咱們能夠經過\(O(k^2)\)處理第二類斯特林數達到\(O(n^2)\)經過此題

Code

更多斯特林數及反演的姿式\(\Longrightarrow\)點這裏

#include<bits/stdc++.h>
typedef int LL;
const LL maxn=5e3+9,mod=1e9+7,inv2=500000004;
inline LL Pow(LL base,LL b){
    LL ret(1);
    while(b){
        if(b&1) ret=1ll*ret*base%mod; base=1ll*base*base%mod; b>>=1;
    }return ret;
}
LL ans[maxn][maxn];
inline void Fir(LL n){
    ans[1][1]=1;
    for(LL i=2;i<=n;++i)
        for(LL j=1;j<=i;++j)
            ans[i][j]=1ll*(ans[i-1][j-1]+1ll*j*ans[i-1][j]%mod)%mod;
}
inline LL Get(LL l,LL r){
    LL ret(1);
    for(LL i=l;i<=r;++i) ret=1ll*ret*i%mod;
    return ret;
}
LL n,k,ret;
int main(){
    scanf("%d%d",&n,&k);
    Fir(k);
    for(LL j=0,val1=1,val2=Pow(2,n);j<=k;++j,val1=1ll*val1*(n-j+1)%mod,val2=1ll*val2*inv2%mod)
        ret=1ll*(ret+1ll*ans[k][j]*val1%mod*val2%mod)%mod;
    printf("%d ",ret);
}