[SRM686]CyclesNumber

题意：求$n$个数的所有排列形成的轮换个数的$m$次方之和

我以前只知道这是GDKOI的题，今天在ckw博客上发现它是TC题...原题真是哪里都有...

就是求$\sum\limits_{i=1}^n{n\brack i}i^m$

一个引理${n+1\brack m+1}=\sum\limits_k{n\brack k}\binom km$，可以直接用生成函数暴力证

$\begin{aligned}{n+1\brack m+1}=(n+1)![z^{n+1}]\frac{\ln^{m+1}\left(\frac1{1-z}\right)}{(m+1)!}=\frac{n!}{m!}[z^n]\frac1{1-z}\ln^m\left(\frac1{1-z}\right)\end{aligned}$

$\begin{aligned}\sum\limits_k{n\brack k}\binom km&=n![z^n]\sum\limits_k\frac{\ln^k\left(\frac1{1-z}\right)}{k!}\binom km\\&=\frac{n!}{m!}[z^n]\sum\limits_k\frac{\ln^k\left(\frac1{1-z}\right)}{(k-m)!}\\&=\frac{n!}{m!}[z^n]\ln^m\left(\frac1{1-z}\right)\frac1{1-z}\end{aligned}$

然后推式子

$\begin{aligned}\sum\limits_{i=1}^n{n\brack i}i^m&=\sum\limits_{i=1}^n{n\brack i}\sum\limits_{j=1}^m{m\brace j}i^{\underline j}\\&=\sum\limits_{j=1}^m{m\brace j}j!\sum\limits_{i=1}^n{n\brack i}\binom ij\\&=\sum\limits_{j=1}^mj!{m\brace j}{n+1\brack j+1}\end{aligned}$

$O(nm)$预处理第一类斯特林数，$O(m^2)$预处理第二类斯特林数，就可以$O(m)$回答一个询问了

#include<stdio.h>
#include<vector>
using namespace std;
typedef long long ll;
const int mod=1000000007;
int mul(int a,int b){return a*(ll)b%mod;}
int s1[100010][310],s2[310][310],fac[100010];
void pre(int n,int k){
	int i,j;
	fac[0]=1;
	for(i=1;i<=n;i++)fac[i]=mul(fac[i-1],i);
	s1[0][0]=1;
	for(i=1;i<=n+1;i++){
		for(j=1;j<=k+1&&j<=i;j++)s1[i][j]=(mul(s1[i-1][j],i-1)+s1[i-1][j-1])%mod;
	}
	s2[0][0]=1;
	for(i=1;i<=k;i++){
		for(j=1;j<=i;j++)s2[i][j]=(mul(s2[i-1][j],j)+s2[i-1][j-1])%mod;
	}
}
int solve(int n,int k){
	if(k==0)return fac[n];
	int j,f,s;
	f=1;
	s=0;
	for(j=1;j<=k;j++){
		f=mul(f,j);
		(s+=mul(f,mul(s2[k][j],s1[n+1][j+1])))%=mod;
	}
	return s;
}
class CyclesNumber{
	public:
		vector<int>getExpectation(vector<int>n,vector<int>m){
			pre(100000,300);
			vector<int>res;
			for(int i=0;i<(int)n.size();i++)res.push_back(solve(n[i],m[i]));
			return res;
		}
};
/*
int main(){
	vector<int>a,b;
	CyclesNumber c;
	int n,m;
	scanf("%d%d",&n,&m);
	a.push_back(n);
	b.push_back(m);
	printf("%d",*c.getExpectation(a,b).begin());
}
*/