Bzoj3456:城市規劃:NTT

題目鏈接：城市規劃

設g[i]表示i個點的連通圖的個數，h[i]表示i個點的圖有多少個

有

容斥一下，假設有j個點組成了合法的圖，剩下的i-j個點不合法，那麼新圖不合法

有

組合數拆掉，等式除以(i-1)!，移項合併後可得

這是卷積

令

那麼

多項式求逆即可

#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=500001;
const int G=3;
const int mod=1004535809;
int n,h[maxn],g[maxn],h2[maxn],tmp[maxn],jc[maxn];
int rev[maxn],dig[maxn],tmp2[maxn],fac[maxn];

int quick_power(int x,ll y){
	int ret=1;
	while (y){
		if (y&1) ret=1ll*ret*x%mod;
		x=1ll*x*x%mod; y>>=1;
	}return ret;
}

void NTT(int a[],int flag,int N){
	for (int i=0,j=0;i<N;++i){
		if (i>j) swap(a[i],a[j]);
		for (int t=N>>1;(j^=t)<t;t>>=1);
	}
	for (int i=2;i<=N;i<<=1){
		int wn=quick_power(G,flag==1?(mod-1)/i:(mod-1-(mod-1)/i));
		int t=i>>1;
		for (int k=0;k<N;k+=i){
			int w=1;
			for (int j=k;j<k+i/2;++j){
				int x=a[j],y=1ll*a[j+i/2]*w%mod;
				a[j]=(x+y)%mod; 
				a[j+t]=x-y; if (a[j+t]<0) a[j+t]+=mod;
				w=1ll*w*wn%mod;
			}
		}
	}
	if (flag==-1){
		int inv=quick_power(N,mod-2);
		for (int i=0;i<N;++i) a[i]=1LL*a[i]*inv%mod;
	}
}

void polynomial_inv(int *a,int *b,int n){
	if(n==1){b[0]=quick_power(a[0],mod-2);return;}
	polynomial_inv(a,b,(n+1)>>1);
	int N=1; for(;N<=n;N<<=1); N<<=1;
	for (int i=0;i<=n;++i) tmp[i]=a[i];
	for (int i=n+1;i<N;++i) tmp[i]=0;
	NTT(tmp,1,N); NTT(b,1,N);
	for (int i=0;i<N;++i) tmp[i]=1ll*b[i]*(2ll-1ll*tmp[i]*b[i]%mod+mod)%mod;
	NTT(tmp,-1,N);
	for (int i=0;i<=n;++i) b[i]=tmp[i];
	for (int i=n+1;i<N;++i) b[i]=0;
}

void polynomial_mul(int *a,int *b,int n){
	int N=1; for(;N<=n;N<<=1); N<<=1;
	for (int i=n+1;i<N;++i) a[i]=b[i]=0;
	NTT(a,1,N); NTT(b,1,N);
	for (int i=0;i<N;++i) a[i]=1ll*a[i]*b[i]%mod;
	NTT(a,-1,N);
}

int main(){
	scanf("%d",&n); h[0]=1;
	for (int i=1;i<=n;++i) h[i]=quick_power(2,1ll*i*(i-1)/2);
	jc[0]=1; for(int i=1;i<=n;++i)jc[i]=1ll*jc[i-1]*i%mod;
	for (int i=0;i<=n;++i) fac[i]=quick_power(jc[i],mod-2);
	for (int i=1;i<=n;++i) g[i]=1ll*h[i]*fac[i-1]%mod;
	for (int i=0;i<=n;++i) h[i]=1ll*h[i]*fac[i]%mod;
	polynomial_inv(h,h2,n+1);
    polynomial_mul(g,h2,n);
    printf("%d",(1ll*g[n]*jc[n-1]%mod+mod)%mod);
}