【洛谷P4233】— 射命丸文的笔记（竞赛图+多项式求逆）

考虑强连通竞赛图哈密顿回路总数很好求
为 $(n-1)!2^{{nchoose 2}-n}$

现在问题是求强连通竞赛图数
设 $f_i$ 为 $i$ 个点的图数
$f_n=2^{{nchoose 2}}-sum_{i=1}^{n-1}f_i{nchoose i}2^{{n-ichoose 2}}$
即枚举拓扑序最小的强连通竞赛子图，剩下部分的随便连

发现这个可以多项式求逆

注意特判 $n=1,2$

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#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 ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
template<class T>inline void chemx(T &a,T b){a<b?a=b:0;}
template<class T>inline void chemn(T &a,T b){a>b?a=b:0;}
cs int mod=998244353,G=3;
inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);}
inline void Add(int &a,int b){a+=b-mod;a+=a>>31&mod;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline void Dec(int &a,int b){a-=b;a+=a>>31&mod;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);};
cs int C=18;
poly w[C+1];
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];
}
cs int N=100005;
int fac[N],ifac[N],inv[N];
inline void init_inv(){
	fac[0]=ifac[0]=inv[0]=inv[1]=1;
	for(int i=1;i<N;i++)fac[i]=mul(fac[i-1],i);
	ifac[N-1]=Inv(fac[N-1]);
	for(int i=N-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	for(int i=2;i<N;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
}
int rev[(1<<C)|5];
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 mid=1,l=1,a0,a1;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,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
	}
}
inline poly operator *(poly a,poly b){
	int deg=a.size()-b.size()+1,lim=1;
	if(deg<=32){
		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,Inv(a[0])),c;
	for(int lim=4;lim<(deg<<2);lim<<=1){
		c=a,c.resize(lim>>1);
		init_rev(lim);
		b.resize(lim),ntt(b,lim,1);
		c.resize(lim),ntt(c,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;
}
poly f,g;
int n;
int main(){
	init_w();
	init_inv();
	n=read();
	g.resize(n+1),f.resize(n+1);
	for(int i=0;i<=n;i++)g[i]=mul(ksm(2,(1ll*i*(i-1)/2)%(mod-1)),ifac[i]);
	g=Inv(g,n+1);
	for(int i=0;i<=n;i++)f[i]=dec(0,g[i]);
	Add(f[0],1);
	for(int i=0;i<=n;i++)Mul(f[i],fac[i]);
	if(n>=1)cout<<1<<'
';
	if(n>=2)cout<<-1<<'
';
	for(int i=3;i<=n;i++)
	cout<<mul(mul(fac[i-1],ksm(2,(1ll*i*(i-1)/2-i)%(mod-1))),Inv(f[i]))<<'
';
}