【洛谷 P5408】【模板】—第一类斯特林数·行（倍增+NTT）

传送门

斯特林数学习笔记

---

考虑构造生成函数$F_n=\sum_{i=0}^{\infty}s(n,i)x^i$

由第一类斯特林数的递推式可得

$F_n=xF_{n-1}+(n-1)F(n-1)=(x+n-1)F(n-1)=x^{\overline n}$

考虑倍增

如果$n$为奇数，递归求解$n-1$,乘一个$x+n-1$

如果$n$为偶数
有$x^{\overline {2n}}=x^{\overline n}(x+n)^{\overline n}$
假设已经求出了$x^{\overline n}=f1(x)=\sum_{i=0}^{n}a_ix^i$
考虑如何求出
$f2(x)=\sum_{i=0}^n a_i(x+n)^i$

$f2(x)=\sum_{i=0}^{n}a_i\sum_{j=0}^{i}x^jn^{i-j}{i\choose j}$

$\ \ \ \ \ \ \ \ \ \ \ =\sum_{j=0}^{n}\frac{x^j}{j!}\sum_{i=j}^na_ii!\frac{n^{i-j}}{(i-j)!}$

$\ \ \ \ \ \ \ \ \ \ \ =\sum_{j=0}^{n}\frac{x^j}{j!}\sum_{i=0}^{n-j}a_{i+j}(i+j)!\frac{n^{i}}{i!}$

令$A(x)=\sum_{i=0}^n a_ii!x^i,B(x)=\sum_{i=0}^n\frac{n^{n-i}}{(n-i)!}x^i$

把$(A*B)(x)$乘出来左移$n$位即可

复杂度$T(n)=T(\frac n 2)+O(nlogn)=O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define poly vector<int>
#define bg begin
cs int mod=167772161,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=(1<<20)|5,C=20;
poly w[C+1];
int rev[N],fac[N],ifac[N],inv[N];
inline void init(cs int len=N-5){
	fac[0]=ifac[0]=inv[0]=inv[1]=1;
	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
	ifac[len]=ksm(fac[len],mod-2);
	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	for(int i=2;i<=len;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
}
inline void init_w(){
	for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
	int wn=ksm(G,(mod-1)/(1<<C));
	w[C][0]=1;
	for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn);
	for(int i=C-1;i;i--)
	for(int j=0;j<(1<<(i-1));j++)
	w[i][j]=w[i+1][j<<1];
}
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void ntt(poly &f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int a0,a1,l=1,mid=1;mid<lim;mid<<=1,l++)
	for(int i=0;i<lim;i+=(mid<<1))
	for(int j=0;j<mid;j++)
	a0=f[i+j],a1=mul(w[l][j],f[i+j+mid]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	if(kd==-1){
		reverse(f.bg()+1,f.bg()+lim);
		for(int i=0;i<lim;i++)Mul(f[i],inv[lim]);
	}
}
inline poly operator *(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	if(deg<=64){
		poly c(deg,0);
		for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
		Add(c[i+j],mul(a[i],b[j]));
		return c;
	}
	while(lim<deg)lim<<=1;
	init_rev(lim);
	a.resize(lim),ntt(a,lim,1);
	b.resize(lim),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1),a.resize(deg);
	return a;
}
inline poly calc_s(int n){
	poly res;
	if(n==1){res.pb(0),res.pb(1);return res;}
	if(n&1){
		res=calc_s(n-1);
		res.resize(n+1);
		for(int i=n;i;i--)res[i]=add(res[i-1],mul(res[i],n-1));
		return res;
	}
	int mid=n>>1;
	poly a=calc_s(mid),b(mid+1),c(mid+1);
	for(int i=0;i<=mid;i++)c[i]=mul(a[i],fac[i]);
	for(int i=0,p=1;i<=mid;i++,Mul(p,mid))b[i]=mul(p,ifac[i]);
	reverse(b.bg(),b.bg()+mid+1);
	c=c*b;for(int i=0;i<=mid;i++)c[i]=mul(c[i+mid],ifac[i]);
	c.resize(mid+1),res=a*c;return res;
}
int n;
int main(){
	n=read();
	init_w(),init();
	poly ans=calc_s(n);
	for(int i=0;i<=n;i++)cout<<ans[i]<<" ";
}