【洛谷 P5409】 【模板】—第一类斯特林数·列(多项式Ln+多项式快速幂)
考虑这样一个展开
有
所以
又由于
所以
对左边做一个再做个快速幂就可以了
可以用能处理的快速幂解决
复杂度
#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<<" ";
}
}