bzoj#3456. 城市规划

题目描述

Solution

用组合意义推很简单。

$i$个点的简单无向图个数为$2^{\left(\genfrac{}{}{0ex}{}{i}{2}\right)}$个。
则其$EGF$为
$$G(x)=\sum _{i>=0}\frac{2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}}{i!}{x}^i$$
令$i$个点的简单无向连通图个数为$f_i$，则其$EGF$为：
$$F(x)=\sum _{i>=0}\frac{f_i}{i!}{x}^i$$
考虑它的含义，简单无向图就相当于若干个简单无向连通图拼接起来，因此$G(x)=1+\frac{F(x)}{1!}+\frac{F(x{)}^2}{2!}+...$

因此$G(x)=e^{F(x)}$，$F(x)=\ln G(x)$。

因此时间复杂度为$O(nlgn)$

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1004535809;
const int inv2=(mods+1)>>1;
const int G=3;
const int Gi=(mods+1)/G;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int a[MAXN],F[MAXN];
namespace Poly
{int c[MAXN],f[MAXN],b[MAXN],d[MAXN],rev[MAXN],Limit,l;inline int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }inline int quick_pow(int x,int y){int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;}void Number_Theoretic_Transform(int *A,int n,int opt){for (int i=0;i<Limit;i++) if (i<rev[i]) swap(A[i],A[rev[i]]);for (int mid=1;mid<Limit;mid<<=1){int Wn=quick_pow((opt==1)?G:Gi,(mods-1)/(mid<<1));for (int j=0;j<Limit;j+=mid<<1)for (int k=j,w=1;k<j+mid;k++,w=1ll*w*Wn%mods){int x=A[k],y=1ll*A[k+mid]*w%mods;A[k]=upd(x,y),A[k+mid]=upd(x,mods-y);}}if (opt==-1){int invLimit=quick_pow(Limit,mods-2);for (int i=0;i<n;i++) A[i]=1ll*A[i]*invLimit%mods;for (int i=n;i<Limit;i++) A[i]=0;}}void getinv(int *F,int *G,int n) //F->invG {if (n==1) { F[0]=quick_pow(G[0],mods-2); return; }getinv(F,G,(n+1)>>1);Limit=1,l=0;while (Limit<(n<<1)) Limit<<=1,l++;for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));for (int i=0;i<n;i++) c[i]=G[i];for (int i=n;i<Limit;i++) c[i]=0;Number_Theoretic_Transform(c,n,1);Number_Theoretic_Transform(F,n,1);for (int i=0;i<Limit;i++) F[i]=1ll*upd(2,mods-1ll*F[i]*c[i]%mods)*F[i]%mods;Number_Theoretic_Transform(F,n,-1);for (int i=n;i<Limit;i++) F[i]=0;}void getln(int *a,int n) //a->ln a (a[0]=1){memset(f,0,sizeof f);getinv(f,a,n);for (int i=0;i<n;i++) a[i]=1ll*a[i+1]*(i+1)%mods;Limit=1,l=0;while (Limit<(n<<1)) Limit<<=1,l++;for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));Number_Theoretic_Transform(a,n,1);Number_Theoretic_Transform(f,n,1);for (int i=0;i<Limit;i++) f[i]=1ll*f[i]*a[i]%mods;Number_Theoretic_Transform(f,n,-1);for (int i=1;i<n;i++) a[i]=1ll*f[i-1]*quick_pow(i,mods-2)%mods; for (int i=n;i<=Limit;i++) a[i]=0; a[0]=0;}void getexp(int *F,int *a,int n) //F->exp a (a[0]=0){memset(b,0,sizeof b);if (n==1) { F[0]=1; return; }getexp(F,a,(n+1)>>1);for (int i=0;i<n;i++) b[i]=F[i];getln(b,n);Limit=1,l=0;while (Limit<(n<<1)) Limit<<=1,l++;for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));for (int i=0;i<n;i++) b[i]=upd(a[i],mods-b[i]);for (int i=n;i<Limit;i++) b[i]=F[i]=0;b[0]++;Number_Theoretic_Transform(F,n,1);Number_Theoretic_Transform(b,n,1);for (int i=0;i<Limit;i++) F[i]=1ll*F[i]*b[i]%mods;Number_Theoretic_Transform(F,n,-1);for (int i=n;i<Limit;i++) F[i]=0; }void getsqrt(int *F,int *a,int n,int k) //F->sqrt^k a(a[0]=1){memset(d,0,sizeof d);for (int i=0;i<n;i++) d[i]=a[i];getln(d,n);int invk=quick_pow(k,mods-2);for (int i=0;i<n;i++) d[i]=1ll*d[i]*invk%mods;getexp(F,d,n);} 	void getpow(int *F,int *a,int n,int k) //F->pow^k a(a[0]=1){memset(d,0,sizeof d);for (int i=0;i<n;i++) d[i]=a[i];getln(d,n);for (int i=0;i<n;i++) d[i]=1ll*d[i]*k%mods;getexp(F,d,n);}
}
int fac[MAXN],inv[MAXN];
void Init(int n)
{fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;inv[n]=Poly::quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mods;
}
int main()
{int n=read();Init(n+5);for (int i=0;i<=n;i++) a[i]=1ll*Poly::quick_pow(2,1ll*i*(i-1)/2%(mods-1))*inv[i]%mods;
//	for (int i=0;i<=n;i++) cout<<i<<":"<<a[i]<<endl;Poly::getln(a,n+1);
//	for (int i=0;i<=n;i++) cout<<i<<":"<<a[i]<<endl;printf("%d\n",1ll*a[n]*fac[n]%mods);return 0;
}