【LuoguP5850】calc加强版

【LuoguP5850】calc加强版

by AmanoKumiko

Description

对于\(n=1,2,3 ... m\)，求出所有合法序列的权值

其中，合法的定义为每个元素都为\([1,k]\)中的整数，同时每个元素都不相同

同时，权值为所有元素的乘积

两个序列不同当且仅当他们任意一位不一样。

对\(998244353\)取模

Input

一行两个数\(k,m\)

Output

共\(m\)行表示答案

Sample Input

13 8

Sample Output

91
7462
546546
35387352
3869654
396558319
363789591
879373476

Data Constraint

\(1\le m\le k\le998244353\)

Solution

不难发现对于一个\(n\)，要求的就是\(n![x^n]\prod_{i=1}^k(ix+1)\)

由于\(k\)很大，所以不能分治\(NTT\)

但是可以运用积化和的\(trick\)

令\(F(x)=\prod_{i=1}^k(ix+1)\)

那么\(ln\ F(x)=\sum_{i=1}^kln(ix+1)\)

根据\(-ln(1-x)=\sum \frac{x^i}{i}\)

原式变为\(\sum_{i=1}^kln(1-(-ix))=\sum_{i=1}^k-\sum_{j=1}^{+∞}\frac{(-ix)^j}{j}\)

即\(-\sum_{i=1}^{+∞}\frac{(-1)^ix^i}{i}\sum_{j=1}^kj^i\)

那么现在需要的就是快速预处理自然数幂和

考虑其\(EGF\)

即\(\sum_{i=1}^{k}\sum_{j=1}^{+∞}\frac{(ix)^j}{j!}=\sum_{i=1}^{k}e^{ix}=\frac{e^{(k+1)x}\ \ -e^x}{e^x-1}\)

由于分母常数项为\(0\)，所以上下要同时除以\(x\)

Code

#include<bits/stdc++.h>
using namespace std;
#define F(i,a,b) for(int i=a;i<=b;i++)
#define Fd(i,a,b) for(int i=a;i>=b;i--)
#define N 2000010
#define mo 998244353

vector<int>Is;

int rev[N],G1[N],G2[N],fac[N],ifac[N],inv[N];

int mod(int x){return x>=mo?x-mo:x;}

int mi(int x,int y){
	if(y==1)return x;
	return y%2?1ll*x*mi(1ll*x*x%mo,y/2)%mo:mi(1ll*x*x%mo,y/2);
}

void init(){
	Is.push_back(1);
	fac[0]=ifac[0]=1;
	F(i,1,N-10)fac[i]=1ll*fac[i-1]*i%mo,inv[i]=(i==1?1:1ll*mo/i*mod(mo-1ll*inv[mo%i]%mo)%mo);
	ifac[N-10]=mi(fac[N-10],mo-2);
	Fd(i,N-11,1)ifac[i]=1ll*ifac[i+1]*(i+1)%mo;
	for(int l=1;l<=N-10;l<<=1)G1[l]=mi(3,(mo-1)/(l*2)),G2[l]=mi(G1[l],mo-2);
}

void BRT(int x){F(i,0,x-1)rev[i]=(rev[i>>1]>>1)|((i&1)?(x>>1):0);}

struct poly{
	vector<int>val;
	void clear(){vector<int>().swap(val);}
	int sz(){return val.size();}
	void rsz(int x){val.resize(x);}
	void shrink(){for(;sz()&&!val.back();val.pop_back());}
	poly modxn(int x){
		if(val.size()<=x)return (poly){val};
		else return (poly){vector<int>(val.begin(),val.begin()+x)};
	}
	poly I(){return (poly){Is};}
	int operator[](int x)const{
		if(x<0||x>=val.size())return 0;
		return val[x];
	}
	void NTT(int x){
		F(i,0,sz()-1)if(rev[i]<i)swap(val[rev[i]],val[i]);
		for(int mid=1;mid<sz();mid<<=1){
			for(int i=0,gn=1;i<sz();i+=(mid<<1),gn=1){
				F(j,0,mid-1){
					int le=val[i+j],ri=1ll*gn*val[i+j+mid]%mo;
					val[i+j]=mod(le+ri);val[i+j+mid]=mod(le-ri+mo);
					gn=1ll*gn*(x==1?G1[mid]:G2[mid])%mo;
				}
			}
		}
		if(x==-1){F(i,0,sz()-1)val[i]=1ll*val[i]*inv[sz()]%mo;}
	}
	void DFT(){NTT(1);}
	void IDFT(){NTT(-1);}
	friend poly operator*(poly x,poly y){
		if(x.sz()<30||y.sz()<30){
			if(x.sz()>y.sz())swap(x,y);
			poly ret;
			ret.rsz(x.sz()+y.sz());
			F(i,0,ret.sz()-1){
				for(int j=0;j<=i&&j<x.sz();j++)
					ret.val[i]=mod(ret.val[i]+1ll*x[j]*y[i-j]%mo);
			}
			return ret;
		}
		int l=1;
		while(l<x.sz()+y.sz()-1)l<<=1;
		x.rsz(l);y.rsz(l);BRT(l);
		x.DFT();y.DFT();
		F(i,0,l-1)x.val[i]=1ll*x[i]*y[i]%mo;
		x.IDFT();
		return x;
	}
	friend poly operator+(poly x,poly y){
		poly ret;
		ret.rsz(max(x.sz(),y.sz()));
		F(i,0,ret.sz()-1)ret.val[i]=mod(x[i]+y[i]);
		return ret;
	}
	friend poly operator-(poly x,poly y){
		poly ret;
		ret.rsz(max(x.sz(),y.sz()));
		F(i,0,ret.sz()-1)ret.val[i]=mod(x[i]-y[i]+mo);
		return ret;
	}
	poly &operator*=(poly x){return (*this)=(*this)*x;}
	poly &operator+=(poly x){return (*this)=(*this)+x;}
	poly &operator-=(poly x){return (*this)=(*this)-x;}
	poly inver(int Len){
		poly f,g;
		f.clear();g.clear();
		g.val.push_back(mi(val[0],mo-2));
		for(int i=2;i<Len*2;i<<=1){
			f.rsz(i<<1);
			g.rsz(i<<1);
			BRT(i<<1);
			F(j,0,i-1)f.val[j]=(j<val.size()?val[j]:0);
			f.DFT();g.DFT();
			F(j,0,(i<<1)-1)g.val[j]=1ll*g[j]*mod(mo+2-1ll*f[j]*g[j]%mo)%mo;
			g.IDFT();
			g.rsz(i);
		}
		return g.modxn(Len);
	}
	poly Ln(int Len){
		poly f,g;
		f.clear();g.clear();
		f.rsz(Len);
		F(i,0,Len-2)f.val[i]=1ll*(i+1>=sz()?0:val[i+1])*(i+1)%mo;
		g=inver(Len);
		g*=f;
		g.modxn(Len);
		Fd(i,Len-1,1)g.val[i]=1ll*g[i-1]*inv[i]%mo;
		g.val[0]=0;
		return g.modxn(Len);
	}
	poly Exp(int Len){
		poly f;
		f.clear();
		f.val.push_back(1);
		for(int i=2;i<Len*2;i<<=1)f=(f*(I()-f.Ln(i)+modxn(i))).modxn(i);
		return f.modxn(Len);
	}
};

int Pk[N],Msum[N];

int main(){
	init();
	int k,m;
	scanf("%d%d",&k,&m);
	poly f,g;
	f.clear();g.clear();
	Pk[0]=1;
	F(i,1,m+1)Pk[i]=1ll*Pk[i-1]*(k+1)%mo;
	F(i,0,m+1)f.val.push_back(1ll*(Pk[i]-1)*ifac[i]%mo),g.val.push_back(ifac[i]);
	F(i,0,m)f.val[i]=f.val[i+1],g.val[i]=g.val[i+1];
	f.val.pop_back();g.val.pop_back();
	g=g.inver(m+1);
	f=(f*g).modxn(m+1);
	F(i,1,m)Msum[i]=1ll*f[i]*fac[i]%mo;
	f.clear();
	f.val.push_back(0);
	F(i,1,m)f.val.push_back(1ll*Msum[i]*((i&1)?1:mo-1)%mo*inv[i]%mo);
	f=f.Exp(m+1);
	F(i,1,m)printf("%d\n",1ll*f[i]*fac[i]%mo);
	return 0;
}