题目
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);
}