CF932E Team Work(第二类斯特林数)

题目

CF932E Team Work

前置：斯特林数(Longrightarrow)点这里

做法

[egin{aligned}\ &sumlimits_{i=1}^n C_n^ii^k\ &sumlimits_{i=1}^n C_n^isumlimits_{j=0}^iC_i^jegin{Bmatrix}k\jend{Bmatrix}j!\ &sumlimits_{i=1}^n frac{n!}{(n-i)!}sumlimits_{j=0}^ifrac{egin{Bmatrix}k\jend{Bmatrix}}{(i-j)!}\ &sumlimits_{j=0}^{min(n,k)}egin{Bmatrix}k\jend{Bmatrix}sumlimits_{i=j}^nfrac{n!}{(n-i)!}frac{1}{(i-j)!}\ &sumlimits_{j=0}^{min(n,k)}egin{Bmatrix}k\jend{Bmatrix}sumlimits_{i=j}^nfrac{n!}{(n-j)!}frac{(n-j)!}{(n-i)!(i-j)!}\ &sumlimits_{j=0}^{min(n,k)}egin{Bmatrix}k\jend{Bmatrix}frac{n!}{(n-j)!}sumlimits_{i=j}^nC_{n-j}^{i-j}\ &sumlimits_{j=0}^{min(n,k)}egin{Bmatrix}k\jend{Bmatrix}frac{n!}{(n-j)!}2^{n-j}\ end{aligned}]

至此我们可以通过(O(k^2))处理第二类斯特林数达到(O(n^2))通过此题

Code

更多斯特林数及反演的姿势(Longrightarrow)点这里

#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);
}