题解 P5396 【【模板】第二类斯特林数·列】
Great_Influence
2019-05-19 16:24:26
我们根据第二类斯特林数的递推式来找方法。
$$\begin{Bmatrix}n\\ m\end{Bmatrix}=\begin{Bmatrix}n-1\\m-1\end{Bmatrix}+m\begin{Bmatrix}n-1\\m\end{Bmatrix}$$
我们设 $S_n(x)$ 表示第二类斯特林数第 $n$ 列的生成函数，即
$$S_m(x)=\sum_{i=0}^\infty \begin{Bmatrix}i\\ m\end{Bmatrix}x^i$$
那么，有:
$$S_m(x)=xS_{m-1}(x)+mxS_m$$
$$(1-mx)S_m(x)=xS_{m-1}(x)$$
$$S_m(x)=\frac{x}{1-mx}S_{m-1}(x)$$
$$S_m(x)=\frac{x^m}{\prod_{i=1}^m (1-ix)}$$
我们通过分治乘法算出下面，然后求逆后位移即可。
复杂度 $O(n\log^2 n)$ 。
代码:
```cpp
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cassert>
#include<iostream>
#define Rep(i,a,b) for(register int i=(a);i<=(b);++i)
#define Repe(i,a,b) for(register int i=(a);i>=(b);--i)
#define rep(i,a,b) for(register int i=(a);i<(b);++i)
#define pb push_back
#define mx(a,b) (a>b?a:b)
#define mn(a,b) (a<b?a:b)
typedef unsigned long long uint64;
typedef unsigned int uint32;
typedef long long ll;
using namespace std;
namespace IO
{
const uint32 Buffsize=1<<15,Output=1<<24;
static char Ch[Buffsize],*S=Ch,*T=Ch;
inline char getc()
{
return((S==T)&&(T=(S=Ch)+fread(Ch,1,Buffsize,stdin),S==T)?0:*S++);
}
static char Out[Output],*nowps=Out;
inline void flush(){fwrite(Out,1,nowps-Out,stdout);nowps=Out;}
template<typename T>inline void read(T&x)
{
x=0;static char ch;T f=1;
for(ch=getc();!isdigit(ch);ch=getc())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getc())x=x*10+(ch^48);
x*=f;
}
template<typename T>inline void write(T x,char ch='\n')
{
if(!x)*nowps++='0';
if(x<0)*nowps++='-',x=-x;
static uint32 sta[111],tp;
for(tp=0;x;x/=10)sta[++tp]=x%10;
for(;tp;*nowps++=sta[tp--]^48);
*nowps++=ch;
}
}
using namespace IO;
void file()
{
#ifndef ONLINE_JUDGE
FILE*DSD=freopen("water.in","r",stdin);
FILE*CSC=freopen("water.out","w",stdout);
#endif
}
const int MAXN=1<<19;
namespace poly
{
const int mod=167772161,gen=3;
static int g[23][MAXN];
inline int power(int u,int v)
{
register int sm=1;
for(;v;v>>=1,u=(uint64)u*u%mod)if(v&1)
sm=(uint64)sm*u%mod;
return sm;
}
inline void predone()
{
Rep(i,1,21)
{
g[i][0]=1,g[i][1]=power(gen,(mod-1)>>i);
Rep(j,2,4e5)g[i][j]=(ll)g[i][j-1]*g[i][1]%mod;
}
}
static int Len,rev[MAXN];
inline void calrev()
{
int II=log(Len)/log(2)-1;
Rep(i,1,Len-1)rev[i]=rev[i>>1]>>1|(i&1)<<II;
}
inline int ad(int u,int v){return(u+=v)>=mod?u-mod:u;}
inline void NTT(int*F,int typ)
{
Rep(i,1,Len-1)if(i<rev[i])swap(F[i],F[rev[i]]);
for(register int i=2,ii=1,t=1;i<=Len;i<<=1,ii<<=1,++t)
for(register int j=0;j<Len;j+=i)rep(k,0,ii)
{
register int tt=(uint64)*(F+j+k+ii)*g[t][k]%mod;
*(F+j+k+ii)=ad(*(F+j+k),mod-tt);
*(F+j+k)=ad(*(F+j+k),tt);
}
if(typ==-1)
{
reverse(F+1,F+Len);
register uint64 invn=power(Len,mod-2);
rep(i,0,Len)*(F+i)=invn**(F+i)%mod;
}
}
static int X[MAXN],Y[MAXN],Iv[MAXN];
inline void mul(int *a,int *b,int *c,int lenl,int lenr)
{
if((ll)lenl*lenr<=100)
{
Rep(i,0,lenl+lenr)X[i]=0;
Rep(i,0,lenl)Rep(j,0,lenr)
X[i+j]=(X[i+j]+(ll)a[i]*b[j])%mod;
Rep(i,0,lenl+lenr)c[i]=X[i];
return;
}
for(Len=2;Len<=lenl+lenr;Len<<=1);
calrev();
Rep(i,0,lenl)X[i]=a[i];
Rep(i,0,lenr)Y[i]=b[i];
rep(i,lenl+1,Len)X[i]=0;
rep(i,lenr+1,Len)Y[i]=0;
NTT(X,1),NTT(Y,1);
rep(i,0,Len)X[i]=(ll)X[i]*Y[i]%mod;
NTT(X,-1);
Rep(i,0,lenl+lenr)c[i]=X[i];
rep(i,lenl+lenr+1,Len)c[i]=0;
}
inline void Inv(int*F,int*G,int ln)
{
Iv[0]=power(F[0],mod-2);
for(register int Ln=2;Ln>>1<=ln;Ln<<=1)
{
rep(i,ln+1,Ln)F[i]=0;
rep(i,0,Ln)X[i]=F[i],Y[i]=0;
rep(i,0,(Ln>>1))Y[i]=Iv[i];
Len=Ln,calrev();
NTT(X,1),NTT(Y,1);
rep(i,0,Ln)X[i]=(uint64)X[i]*Y[i]%mod;
NTT(X,-1);
rep(i,0,(Ln>>1))X[i]=0;
NTT(X,1);
rep(i,0,Ln)X[i]=(uint64)X[i]*Y[i]%mod;
NTT(X,-1);
rep(i,(Ln>>1),Ln)Iv[i]=mod-X[i];
}
Rep(i,0,ln)G[i]=Iv[i];
}
}
using poly::mul;
using poly::power;
using poly::Len;
using poly::calrev;
using poly::NTT;
using poly::mod;
using poly::predone;
using poly::Inv;
using poly::ad;
static int n,k,F[MAXN];
static int st[27][2][MAXN];
void getmul(int l,int r,int lev,int dr)
{
if(l==r){st[lev][dr][0]=1,st[lev][dr][1]=mod-l;return;}
int md=(l+r)>>1;
getmul(l,md,lev+1,0),getmul(md+1,r,lev+1,1);
mul(st[lev+1][0],st[lev+1][1],st[lev][dr],md-l+1,r-md);
}
int main()
{
file();
predone();
read(n),read(k);
if(k>n)Rep(i,0,n)write(0,' ');
else
{
Rep(i,0,k-1)write(0,' ');
getmul(1,k,0,0);
Inv(st[0][0],F,n-k);
Rep(i,0,n-k)write(F[i],' ');
}
flush();
return 0;
}
```
还有另一个式子:
$$\sum _{n=k}^{\infty }\left\{{n \atop k}\right\}{\frac {x^{n}}{n!}}={\frac {(e^{x}-1)^{k}}{k!}}$$
(从维基百科上扣的)
我们可以通过 $\ln$ 和 $\exp$ 来做到 $O(n\log n)$ 。不过一般的 $\exp$ 跑得都比分治乘法慢。
代码:
```cpp
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cassert>
#include<iostream>
#define Rep(i,a,b) for(register int i=(a);i<=(b);++i)
#define Repe(i,a,b) for(register int i=(a);i>=(b);--i)
#define rep(i,a,b) for(register int i=(a);i<(b);++i)
#define pb push_back
#define mx(a,b) (a>b?a:b)
#define mn(a,b) (a<b?a:b)
typedef unsigned long long uint64;
typedef unsigned int uint32;
typedef long long ll;
using namespace std;
namespace IO
{
const uint32 Buffsize=1<<15,Output=1<<24;
static char Ch[Buffsize],*S=Ch,*T=Ch;
inline char getc()
{
return((S==T)&&(T=(S=Ch)+fread(Ch,1,Buffsize,stdin),S==T)?0:*S++);
}
static char Out[Output],*nowps=Out;
inline void flush(){fwrite(Out,1,nowps-Out,stdout);nowps=Out;}
template<typename T>inline void read(T&x)
{
x=0;static char ch;T f=1;
for(ch=getc();!isdigit(ch);ch=getc())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getc())x=x*10+(ch^48);
x*=f;
}
template<typename T>inline void write(T x,char ch='\n')
{
if(!x)*nowps++='0';
if(x<0)*nowps++='-',x=-x;
static uint32 sta[111],tp;
for(tp=0;x;x/=10)sta[++tp]=x%10;
for(;tp;*nowps++=sta[tp--]^48);
*nowps++=ch;
}
}
using namespace IO;
void file()
{
#ifndef ONLINE_JUDGE
FILE*DSD=freopen("water.in","r",stdin);
FILE*CSC=freopen("water.out","w",stdout);
#endif
}
const int MAXN=1<<19;
namespace poly
{
const int mod=167772161,gen=3;
static int g[23][MAXN],iv[MAXN];
inline int power(int u,int v)
{
register int sm=1;
for(;v;v>>=1,u=(uint64)u*u%mod)if(v&1)
sm=(uint64)sm*u%mod;
return sm;
}
inline void predone()
{
Rep(i,1,21)
{
g[i][0]=1,g[i][1]=power(gen,(mod-1)>>i);
Rep(j,2,4e5)g[i][j]=(ll)g[i][j-1]*g[i][1]%mod;
}
iv[1]=1;
Rep(i,2,4e5)iv[i]=mod-(uint64)mod/i*iv[mod%i]%mod;
}
static int Len,rev[MAXN];
inline void calrev()
{
int II=log(Len)/log(2)-1;
Rep(i,1,Len-1)rev[i]=rev[i>>1]>>1|(i&1)<<II;
}
inline int ad(int u,int v){return(u+=v)>=mod?u-mod:u;}
inline void NTT(int*F,int typ)
{
Rep(i,1,Len-1)if(i<rev[i])swap(F[i],F[rev[i]]);
for(register int i=2,ii=1,t=1;i<=Len;i<<=1,ii<<=1,++t)
for(register int j=0;j<Len;j+=i)rep(k,0,ii)
{
register int tt=(uint64)*(F+j+k+ii)*g[t][k]%mod;
*(F+j+k+ii)=ad(*(F+j+k),mod-tt);
*(F+j+k)=ad(*(F+j+k),tt);
}
if(typ==-1)
{
reverse(F+1,F+Len);
register uint64 invn=power(Len,mod-2);
rep(i,0,Len)*(F+i)=invn**(F+i)%mod;
}
}
static int X[MAXN],Y[MAXN],Iv[MAXN];
inline void mul(int *a,int *b,int *c,int lenl,int lenr)
{
if((ll)lenl*lenr<=100)
{
Rep(i,0,lenl+lenr)X[i]=0;
Rep(i,0,lenl)Rep(j,0,lenr)
X[i+j]=(X[i+j]+(ll)a[i]*b[j])%mod;
Rep(i,0,lenl+lenr)c[i]=X[i];
return;
}
for(Len=2;Len<=lenl+lenr;Len<<=1);
calrev();
Rep(i,0,lenl)X[i]=a[i];
Rep(i,0,lenr)Y[i]=b[i];
rep(i,lenl+1,Len)X[i]=0;
rep(i,lenr+1,Len)Y[i]=0;
NTT(X,1),NTT(Y,1);
rep(i,0,Len)X[i]=(ll)X[i]*Y[i]%mod;
NTT(X,-1);
Rep(i,0,lenl+lenr)c[i]=X[i];
rep(i,lenl+lenr+1,Len)c[i]=0;
}
inline void Inv(int*F,int*G,int ln)
{
Iv[0]=power(F[0],mod-2);
for(register int Ln=2;Ln>>1<=ln;Ln<<=1)
{
rep(i,ln+1,Ln)F[i]=0;
rep(i,0,Ln)X[i]=F[i],Y[i]=0;
rep(i,0,(Ln>>1))Y[i]=Iv[i];
Len=Ln,calrev();
NTT(X,1),NTT(Y,1);
rep(i,0,Ln)X[i]=(uint64)X[i]*Y[i]%mod;
NTT(X,-1);
rep(i,0,(Ln>>1))X[i]=0;
NTT(X,1);
rep(i,0,Ln)X[i]=(uint64)X[i]*Y[i]%mod;
NTT(X,-1);
rep(i,(Ln>>1),Ln)Iv[i]=mod-X[i];
}
Rep(i,0,ln)G[i]=Iv[i];
}
static int ExX[MAXN],ExY[MAXN],Op[MAXN];
inline void Deriv(int*F,int*G,int ln)
{Rep(i,1,ln)G[i-1]=(uint64)F[i]*i%mod;G[ln]=0;}
inline void Inter(int*F,int*G,int ln)
{Repe(i,ln,1)G[i]=(uint64)F[i-1]*iv[i]%mod;G[0]=0;}
static int LnX[MAXN];
inline void Ln(int*F,int*G,int ln)
{
Deriv(F,LnX,ln),Inv(F,G,ln);
for(Len=2;Len<=ln<<1;Len<<=1);
rep(i,ln+1,Len)LnX[i]=G[i]=0;
calrev();
NTT(LnX,1),NTT(G,1);
rep(i,0,Len)G[i]=(uint64)G[i]*LnX[i]%mod;
NTT(G,-1);
Inter(G,G,ln);
}
void cdq_Exp(int*a,int*F,int l,int r)
{
if(l==r)
{
if(!l)F[l]=1;
else F[l]=(ll)F[l]*iv[l]%mod;
return;
}
int md=(l+r)>>1;cdq_Exp(a,F,l,md);
for(Len=2;Len<=r-l;Len<<=1);
calrev();
memset(X,0,sizeof(int)*Len);
memset(Y,0,sizeof(int)*Len);
memcpy(X,F+l,sizeof(int)*(md-l+1));
memcpy(Y,a,sizeof(int)*(r-l));
NTT(X,1),NTT(Y,1);
rep(i,0,Len)X[i]=(ll)X[i]*Y[i]%mod;
NTT(X,-1);
Rep(i,md+1,r)F[i]=ad(F[i],X[i-l-1]);
cdq_Exp(a,F,md+1,r);
}
inline void Exp(int *F,int *G,int ln)
{
Rep(i,1,ln)Op[i-1]=(ll)F[i]*i%mod,Op[i]=0;
memset(G,0,sizeof(int)*(ln+1));
cdq_Exp(Op,G,0,ln);
/*Op[0]=1;
for(register int Lx=2;Lx>>1<=ln;Lx<<=1)
{
rep(i,Lx>>1,Lx)Op[i]=0;
Ln(Op,ExX,Lx);
rep(i,0,Lx>>1)ExX[i]=ad(F[i+(Lx>>1)],mod-ExX[i+(Lx>>1)]);
rep(i,0,Lx>>1)ExY[i]=Op[i];
rep(i,Lx>>1,Lx)ExX[i]=ExY[i]=0;
Len=Lx;
calrev();
NTT(ExY,1),NTT(ExX,1);
rep(i,0,Len)ExX[i]=(ll)ExX[i]*ExY[i]%mod;
NTT(ExX,-1);
rep(i,0,Lx>>1)Op[i+(Len>>1)]=ExX[i];
}
Rep(i,0,ln)G[i]=Op[i];*/
}
inline void Pow(int*F,int*G,int ln,ll k)
{
int lst=ln+1;
Rep(i,0,ln)if(F[i]){lst=i;break;}
if(ln&&(__int128)lst*k>ln)
{memset(G,0,sizeof(int)*(ln+1));return;}
int md=ln-k*lst,bs=F[lst],iv=power(bs,mod-2);
Rep(i,lst,ln)G[i]=(ll)G[i]*iv%mod;
Ln(G+lst,G,md);
k%=mod;
Rep(i,0,md)G[i]=(ll)G[i]*k%mod;
Exp(G,G,md);
bs=power(bs,k);
Repe(i,md,0)G[i+lst*k]=(ll)G[i]*bs%mod;
memset(G,0,sizeof(int)*(lst*k));
}
}
using poly::mul;
using poly::power;
using poly::Len;
using poly::calrev;
using poly::NTT;
using poly::mod;
using poly::predone;
using poly::Inv;
using poly::Inter;
using poly::Deriv;
using poly::Ln;
using poly::Exp;
using poly::Pow;
using poly::ad;
static int n,k,X[MAXN],fac[MAXN],inv[MAXN];
int main()
{
file();
read(n),read(k);
if(k>n)Rep(i,0,n)write(0,' ');
else
{
predone();
fac[0]=1;
Rep(i,1,n)fac[i]=(ll)fac[i-1]*i%mod;
inv[n]=power(fac[n],mod-2);
Repe(i,n,1)inv[i-1]=(ll)inv[i]*i%mod;
Rep(i,1,n)X[i]=inv[i];
Pow(X,X,n,k);
Rep(i,0,n)X[i]=(ll)X[i]*inv[k]%mod;
Rep(i,0,n)write((ll)X[i]*fac[i]%mod,' ');
}
flush();
return 0;
}
```