正题

题目链接:https://www.luogu.com.cn/problem/CF932E

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题目大意

给出$n,k$，求$$\sum _{i=1}^{n}C(n,i)\ast i^k$$

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解题思路

上式子的话，大体是先拆开$i^k$变成$$\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}\left(\genfrac{}{}{0ex}{}{i}{j}\right)j!$$
$$\Rightarrow \sum _{i=1}^n\sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}\left(\genfrac{}{}{0ex}{}{i}{j}\right)\left(\genfrac{}{}{0ex}{}{n}{i}\right)j!$$
拆开$C$然后化解一下
$$\Rightarrow \sum _{i=1}^n\sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}\ast \frac{1}{(i-j)!}\ast \frac{n!}{(n-i)!}$$
$$\Rightarrow \sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}\frac{n!}{(n-j)!}\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n-j}{n-i}\right)$$
$$\sum _{j=1}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}P(n,j)\ast 2^{n-j}$$

理解一下的话就是最开的式子的意义是：在$n$种颜色中选择若干种然后再去涂$k$个物品，求方案种数。转换一下的话我们可以把$k$个物品分成若干个集合，每个物品集合颜色相同，然后再$n$个颜色中排列出这些颜色，剩下的都是可选可不选的。

然后递推处理第二类斯特林数就好了，时间复杂度$O(k^2)$

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=5100,XJQ=1e9+7;
ll n,k,S[N][N],ans;
ll power(ll x,ll b){ll ans=1;if(b<0)return power(power(x,-b),XJQ-2);while(b){if(b&1)ans=ans*x%XJQ;x=x*x%XJQ;b>>=1;}return ans;
}
ll P(ll n,ll m){ll tmp=1;if(n<m)return 0;for(ll i=n-m+1;i<=n;i++)tmp=tmp*i%XJQ;return tmp;
}
int main()
{scanf("%lld%lld",&n,&k);S[0][0]=1;for(ll i=1;i<=k;i++)for(ll j=1;j<=i;j++)S[i][j]=(S[i-1][j-1]+j*S[i-1][j]%XJQ)%XJQ;for(ll i=0;i<=k;i++)(ans+=S[k][i]*P(n,i)%XJQ*power(2,n-i)%XJQ)%=XJQ;if(!k)ans--;printf("%lld",ans);return 0;
}