传送门

斯特林数学习笔记

考虑这样一个展开

$$\begin{gathered} (x+1{)}^n=\sum _{i=0}^{\infty }{x}^i(\genfrac{}{}{0ex}{}{n}{i}) \\ =\sum _{i=0}^{\infty }\frac{x^i}{i!}{n}^{\underline{i}} \end{gathered}$$

有
$x^{\underline{n}}={\sum }_{i=0}^n(-1{)}^{n-i}s(n,i)x^i$

所以$$\begin{gathered} (x+1{)}^n=\sum _{i=0}^{\infty }\frac{x^i}{i!}\sum _{j=0}^i(-1{)}^{i-j}s(i,j)n^j \\ =\sum _{j=0}^{\infty }{n}^j\sum _{i=j}^{\infty }(-1{)}^{i-j}s(i,j)\frac{x^i}{i!} \\ =\sum _{i=0}^{\infty }{n}^i\sum _{j=i}^{\infty }(-1{)}^{j-i}s(j,i)\frac{x^j}{j!} \end{gathered}$$

又由于$$(x+1{)}^n=e^{n\ast Ln(x+1)}=\sum _{i=0}^{\infty }{n}^i\frac{Ln(x+1{)}^i}{i!}$$

所以$$\sum _{i=0}^{\infty }{n}^i\frac{Ln(x+1{)}^i}{i!}=\sum _{i=0}^{\infty }{n}^i\sum _{j=i}^{\infty }(-1{)}^{j-i}s(j,i)\frac{x^j}{j!}$$

$$\frac{Ln(x+1{)}^k}{k!}=\sum _{j=k}^{\infty }(-1{)}^{j-k}s(j,k)\frac{z^j}{j!}$$

对左边做一个$Ln$再做个快速幂就可以了
可以用能处理$a_0≠1$的快速幂解决

复杂度$O(nlogn)$

#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){
    
	if(a.size()<b.size())a.resize(b.size());
	for(int i=0;i<b.size();i++)Add(a[i],b[i]);
	return a;
}
inline poly operator -(poly a,poly b){
    
	if(a.size()<b.size())a.resize(b.size());
	for(int i=0;i<b.size();i++)Dec(a[i],b[i]);
	return a;
}
inline poly operator *(poly a,int b){
    
	for(int i=0;i<a.size();i++)Mul(a[i],b);
	return a;
}
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 Inv(poly a,int deg){
    
	poly b(1,ksm(a[0],mod-2)),c;
	for(int lim=4;lim<(deg<<2);lim<<=1){
    
		c=a,c.resize(lim>>1);
		init_rev(lim);
		c.resize(lim),ntt(c,lim,1);
		b.resize(lim),ntt(b,lim,1);
		for(int i=0;i<lim;i++)Mul(b[i],dec(2,mul(b[i],c[i])));
		ntt(b,lim,-1),b.resize(lim>>1);
	}b.resize(deg);return b;
}
inline poly deriv(poly a){
    
	for(int i=0;i<a.size()-1;i++)a[i]=mul(a[i+1],i+1);
	a.pop_back();return a;
}
inline poly integ(poly a){
    
	a.pb(0);
	for(int i=a.size()-1;i;i--)a[i]=mul(a[i-1],inv[i]);
	a[0]=0;return a;
}
inline poly Ln(poly a,int deg){
    
	a=integ(Inv(a,deg)*deriv(a)),a.resize(deg);return a;
}
inline poly ksm(poly a,int b,int deg){
    
	poly res(1,1);
	for(;b;b>>=1){
    
		if(b&1){
    
			res=res*a;if(res.size()>deg)res.resize(deg);
		}
		a=a*a;
		if(a.size()>deg)a.resize(deg);
	}
	res.resize(deg);
	return res;
}
poly f;
int n,k;
int main(){
    
	n=read(),k=read();
	if(n<k){
    for(int i=0;i<=n;i++)cout<<0<<" ";return 0;}
	init_w(),init();
	f.pb(1),f.pb(1);
	f=Ln(f,n+1),f=ksm(f,k,n+1);
	f=f*ifac[k];
	for(int i=0;i<k;i++)cout<<0<<" ";
	for(int i=k;i<=n;i++){
    
		int res=f[i];
		if((i-k)&1)Mul(res,mod-fac[i]);
		else Mul(res,fac[i]);
		cout<<res<<" ";
	}
}