【BZOJ4228】Tibbar的后花园（多项式Exp）

传送门

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题解：

发现就是要求不能有一个点度数超过$2$，且不能存在任何长度为$3$的倍数的环。

那么每个连通分量要么就是链，要么就是长度不为$3$的环。

发现带标号，构造两者的EGF，加起来Exp即可。

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代码：

#include
#define ll long long
#define re register
#define cs const

using std::cerr;
using std::cout;

std::ostream &operator<<(std::ostream &out,cs std::vector<int> &a){
	if(!a.size())out<<"empty ";
	for(int re i:a)out<<i<<" ";
	return out;
}

cs int mod=1004535809;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline int mul(int a,int b){ll r=(ll)a*b;return r>=mod?r%mod:r;}
inline int power(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 Inc(int &a,int b){(a+=b)>=mod&&(a-=mod);}
inline void Dec(int &a,int b){(a-=b)<0&&(a+=mod);}

typedef std::vector<int> Poly;

cs int bit=21,N=1<<bit|1;

int r[N],*w[bit+1];
inline void init_NTT(){
	for(int re i=1;i<=bit;++i)w[i]=new int[(1<<i-1)+1];
	int wn=power(3,mod-1>>bit);
	w[bit][0]=1;
	for(int re i=1;i<(1<<bit-1);++i)w[bit][i]=mul(w[bit][i-1],wn);
	for(int re i=bit-1;i;--i)
	for(int re j=0;j<(1<<i-1);++j)w[i][j]=w[i+1][j<<1];
}
inline void NTT(Poly &A,int len,int typ){
	for(int re i=0;i<len;++i)if(i<r[i])std::swap(A[i],A[r[i]]);
	for(int re i=1,t=1;i<len;i<<=1,++t)
	for(int re j=0;j<len;j+=i<<1)
	for(int re k=0;k<i;++k){
		int x=A[j+k],y=mul(A[j+k+i],w[t][k]);
		A[j+k]=add(x,y),A[j+k+i]=dec(x,y);
	}
	if(typ==-1){
		std::reverse(A.begin()+1,A.begin()+len);
		for(int re i=0,inv=power(len,mod-2);i<len;++i)A[i]=mul(A[i],inv);
	}
}
inline void init_rev(int len){
	for(int re i=0;i<len;++i)r[i]=r[i>>1]>>1|((i&1)?len>>1:0);
}

inline Poly operator*(Poly a,Poly b){
	int deg=a.size()+b.size()-1,l=1;
	while(l<=deg)l<<=1;init_rev(l);
	a.resize(l),NTT(a,l,1);
	b.resize(l),NTT(b,l,1);
	for(int re i=0;i<l;++i)a[i]=mul(a[i],b[i]);
	NTT(a,l,-1),a.resize(deg);
	return a;
}

int fac[N],ifac[N],inv[N];
inline void init_inv(){
	fac[0]=fac[1]=ifac[0]=ifac[1]=inv[0]=inv[1]=1;
	for(int re i=2;i<N;++i){
		fac[i]=mul(fac[i-1],i);
		inv[i]=mul(inv[mod%i],mod-mod/i);
		ifac[i]=mul(ifac[i-1],inv[i]);
	}
}
inline Poly Deriv(Poly a){
	for(int re i=0;i+1<a.size();++i)a[i]=mul(a[i+1],i+1);
	a.pop_back();return a;
}

inline Poly Integ(Poly a){
	a.push_back(0);
	for(int re i=a.size()-1;i;--i)a[i]=mul(a[i-1],inv[i]);
	a[0]=0;return a;
}

inline Poly Inv(cs Poly &a,int lim){
	Poly c,b(1,power(a[0],mod-2));
	for(int re l=4;(l>>2)<lim;l<<=1){
		init_rev(l);
		c.resize(l>>1);for(int re i=0;i<(l>>1);++i)c[i]=i<a.size()?a[i]:0;
		c.resize(l),NTT(c,l,1);
		b.resize(l),NTT(b,l,1);
		for(int re i=0;i<l;++i)b[i]=mul(b[i],dec(2,mul(b[i],c[i])));
		NTT(b,l,-1),b.resize(l>>1);
	}b.resize(lim);
	return b;
}

inline Poly Ln(Poly a,int lim){
	a=Integ(Deriv(a)*Inv(a,lim));
	a.resize(lim);
	return a;
}

inline Poly Exp(cs Poly a){
	Poly c,b(1,1);int lim=a.size();
	for(int re l=2;(l>>1)<lim;l<<=1){
		c=Ln(b,l);
		for(int re i=0;i<l;++i)c[i]=dec(i<a.size()?a[i]:0,c[i]);
		Inc(c[0],1);
		b=b*c;b.resize(l);
	}b.resize(lim);
	return b;
}

Poly f;
int n;
signed main(){
#ifdef zxyoi
//	freopen("tibbar.in","r",stdin);
#endif
	init_NTT();init_inv();
	std::cin>>n;
	f.resize(n+1);
	f[1]=1;f[2]=inv[2];
	for(int re i=3;i<=n;++i){
		f[i]=mul(fac[i],inv[2]);
		if(i%3)Inc(f[i],mul(inv[2],fac[i-1]));
		f[i]=mul(f[i],ifac[i]);
	}
	std::cout<<mul(Exp(f)[n],fac[n])<<"\n";
	return 0;
}