洛谷 P5408 【模板】第一类斯特林数·行

首先，有

\[x^{\overline n}=\sum_k\begin{bmatrix}{n\\ k}\end{bmatrix}x^{k}\\
\]

那么我们只需要求出\(x^{\overline n}\)即可，考虑倍增

\[x^{\overline 2n}=x^{\overline n}(x+n)^{\overline n}
\]

假设我们现在已经求出了\(x^{\overline n}\)，考虑如何求出\((x+n)^{\overline n}\)

开始颓柿子

\[\begin{aligned}
f(x+n)
&=\sum_{i}f_i(x+n)^i\\
&=\sum_{j}x^j\sum_{i}f_i{i\choose j}n^{i-j}\\
&=\sum_{j}{x^j\over j!}\sum_{i}f_ii!{n^{i-j}\over (i-j)!}\\
\end{aligned}
\]

直接卷就可以了，再把它和原来的多项式卷积来即可

//quming

#include<bits/stdc++.h>

#define R register

#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)

#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)

#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)

template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}

template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}

using namespace std;

const int P=167772161;

inline void upd(R int &x,R int y){(x+=y)>=P?x-=P:0;}

inline int inc(R int x,R int y){return x+y>=P?x+y-P:x+y;}

inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}

inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}

int ksm(R int x,R int y){

    R int res=1;

    for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;

    return res;

}

const int N=(1<<19)+5;

int fac[N],ifac[N],lg[N],r[25][N],rt[2][N],inv[25];

int lim,d;

inline void swap(R int &x,R int &y){R int t=x;x=y,y=t;}

inline int C(R int n,R int m){return m>n?0:1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}

void init(){

    fac[0]=ifac[0]=1;fp(i,1,262144)fac[i]=mul(fac[i-1],i);

    ifac[262144]=ksm(fac[262144],P-2);fd(i,262143,1)ifac[i]=mul(ifac[i+1],i+1);

    fp(d,1,19){

        fp(i,1,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1));

        inv[d]=ksm(1<<d,P-2),lg[1<<d]=d;

    }

    for(R int t=(P-1)>>1,i=1,x,y;i<=262144;i<<=1,t>>=1){

        x=ksm(3,t),y=ksm(55924054,t),rt[0][i]=rt[1][i]=1;

        fp(k,1,i-1)

            rt[1][i+k]=mul(rt[1][i+k-1],x),

            rt[0][i+k]=mul(rt[0][i+k-1],y);

    }

}

void NTT(int *A,int ty){

    fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]);

    R int t;

    for(R int mid=1;mid<lim;mid<<=1)

        for(R int j=0;j<lim;j+=(mid<<1))

            fp(k,0,mid-1)

                A[j+k+mid]=dec(A[j+k],t=mul(rt[ty][mid+k],A[j+k+mid])),

                A[j+k]=inc(A[j+k],t);

    if(!ty){

        t=inv[d];

        fp(i,0,lim-1)A[i]=mul(A[i],t);

    }

}

int f[N],n;

void solve(int *b,int len){

	if(!len)return b[0]=1,void();

	solve(b,len>>1);

	lim=1,d=0;while(lim<=len)lim<<=1,++d;

	int dm=(len>>1);

	static int A[N],B[N];

	for(R int i=0,c=1;i<=dm;++i,c=mul(c,dm))A[i]=mul(c,ifac[i]);

	fp(i,0,dm)B[dm-i]=mul(b[i],fac[i]);

	fp(i,dm+1,lim-1)A[i]=B[i]=0;

	NTT(A,1),NTT(B,1);

	fp(i,0,lim-1)A[i]=mul(A[i],B[i]);

	NTT(A,0);

	reverse(A,A+dm+1);

	fp(i,0,dm)A[i]=mul(A[i],ifac[i]);fp(i,dm+1,lim-1)A[i]=0;

	fp(i,0,dm)B[i]=b[i];fp(i,dm+1,lim-1)B[i]=0;

	NTT(A,1),NTT(B,1);

	fp(i,0,lim-1)A[i]=mul(A[i],B[i]);

	NTT(A,0);

	fp(i,0,len)b[i]=A[i];

	if(len&1){

		fd(i,len,1)b[i]=inc(mul(b[i],len-1),b[i-1]);

		b[0]=mul(b[0],len-1);

	}

}

int main(){

//	freopen("testdata.in","r",stdin);

    scanf("%d",&n);

    init();

    solve(f,n);

    fp(i,0,n)printf("%d ",f[i]);

    return 0;

}

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洛谷 P5408 【模板】第一类斯特林数·行

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