[集训队作业2013]城市规划

P4841 [集训队作业2013]城市规划

套路题，设\(f_i\)表示无标号联通图方案数，设\(g_i\)表示无标号图方案数，易得\(g_i = 2^{\binom{n}{2}}\)。
考虑固定 \(1\) 得

\[g_n = \sum_{i = 1}^n\dbinom{n - 1}{i - 1}f_{i}g_{n - i} \]

将组合数拆开化成\(EGF\)的形式

\[\frac{2^{\binom{n}{2}}}{(n - 1)!} = \sum_i \frac{f_i}{(i - 1)!} * \frac{2^{\binom{n - i}{2}}}{(n - i)!} \]

显然的卷积形式 \(H = F * G\)，所以\(F = H * G^{-1}\)。
最后答案为 \(n! * [x^n]F\)，多项式求逆即可。

Code

#include<cstdio>
#include<iostream>
#include<cstring>
#define IN inline
#define LL long long
using namespace std;
const int N = 2e5 + 5, P = 1004535809, G = 3;
int n, m, a[N << 2], b[N << 2], inv[N], pw[N];

LL fpow(LL x, LL y) {
	LL res = 1;
	for (; x; x >>= 1, y = y * y % P)
		if (x & 1) res = res * y % P;
	return res;
}
namespace Poly{
	int rev[N << 2], insF[N << 2], sginv[N << 2];
	void NTT(int *f, int len, int fl) {
		if (len == 1) return;
		for (int i = 0; i < len; i++)
			if (i < rev[i]) swap(f[i], f[rev[i]]);
		for (int l = 1; l < len; l <<= 1) {
			int I = fpow((P - 1) / (l << 1), G);
			if (fl == -1) I = fpow(P - 2, I);
			for (int i = 0; i < len; i += (l << 1)) {
				int W = 1;
				for (int j = 0; j < l; j++, W = (LL)W * I % P) {
					int x = f[i + j], y = (LL)f[i + j + l] * W % P;
					f[i + j] = (x + y) % P, f[i + j + l] = (x - y + P) % P;
				}
			}
		}
	}
	void Mulpoly(int *f, int *g, int lenF, int lenG) {  
		int len = 1, bit = 0;
		while (len <= lenF + lenG) len <<= 1, bit++;
		for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
		NTT(f, len, 1), NTT(g, len, 1);
		for (int i = 0; i < len; i++) f[i] = (LL)f[i] * g[i] % P;
		NTT(f, len, -1); int IV = fpow(P - 2, len);
		for (int i = 0; i <= lenF + lenG; i++) f[i] = (LL)f[i] * IV % P;
		for (int i = lenF + lenG + 1; i < len; i++) f[i] = 0;
	}
	void Invpoly(int *f, int lenF, int *g) { // [0, lenF)
		static int Flen[100]; int cnt = 0;
		for (int i = lenF; i > 1; i = i + 1 >> 1) Flen[++cnt] = i;
		for (int i = 1; i <= (cnt >> 1); i++) swap(Flen[i], Flen[cnt - i + 1]);	
		g[0] = fpow(P - 2, f[0]), Flen[0] = 1;
		for (int j = 1; j <= cnt; j++) {
			int len = 1, bit = 0;
			while (len < (Flen[j] << 1)) len <<= 1, bit++;
			for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
			for (int i = Flen[j - 1]; i < len; i++) g[i] = 0;
			
			for (int i = 0; i < Flen[j]; i++) insF[i] = f[i];
			for (int i = Flen[j]; i < len; i++) insF[i] = 0;
			
			NTT(g, len, 1), NTT(insF, len, 1);
			for (int i = 0; i < len; i++) g[i] = (LL)g[i] * (2LL - (LL)insF[i] * g[i] % P + P) % P;
			NTT(g, len, -1); int IV = fpow(P - 2, len);
			for (int i = 0; i < Flen[j]; i++) g[i] = (LL)g[i] * IV % P;
		}
	}
	void Getsginv(int len) {
		sginv[0] = sginv[1] = 1;
		for (int i = 2; i <= len; i++) sginv[i] = (LL)sginv[P % i] * (P - P / i) % P;
	}
	void Dpoly(int *f, int lenF) {
		for (int i = 0; i < lenF; i++) f[i] = (LL)f[i + 1] * (i + 1) % P;
	}
	void Jfpoly(int *f, int lenF) {
		for (int i = lenF; i >= 0; i--) f[i + 1] = (LL)f[i] * sginv[i + 1] % P;
		f[0] = 0;
	}
	void Lnpoly(int *f, int lenF) { // [0, lenF)
		static int Inv_f[N << 2];
		Invpoly(f, lenF, Inv_f), Dpoly(f, lenF - 1), Mulpoly(f, Inv_f, lenF - 1, lenF - 1), Jfpoly(f, lenF - 1);
		for (int i = lenF; i <= (lenF << 1); i++) f[i] = 0;
	}
	void Exppoly(int *f, int lenF, int *g) {
		static int s[N << 2];
		int limit = 1;
		while (limit < lenF) limit <<= 1;
		g[0] = 1;
		for (int len = 2; len <= limit; len <<= 1) {
			for (int i = 0; i < (len >> 1); i++) s[i] = g[i];
			for (int i = (len >> 1); i < len; i++) s[i] = 0;
			Lnpoly(s, len);
			for (int i = 0; i < len; i++) s[i] = (f[i] - s[i] + P) % P;
			s[0] = (s[0] + 1) % P;
			Mulpoly(g, s, len - 1, len - 1);
			for (int i = len; i < (len << 1); i++) g[i] = 0;
		}
		for (int i = lenF; i < limit; i++) g[i] = 0;
		for (int i = 0; i < limit; i++) s[i] = 0;
	}
}
IN int read() {
	int t = 0,res = 0; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) t |= (ch == '-');
	for (; isdigit(ch); ch = getchar()) res = (res << 3) + (res << 1) + (ch ^ 48);
	return t ? -res : res;
}
int main() {
	n = read(), Poly::Getsginv(n), inv[0] = inv[1] = 1;
	for (int i = 2; i <= n; i++) inv[i] = (LL)inv[i - 1] * Poly::sginv[i] % P;
	for (int i = 0; i <= n; i++) pw[i] = fpow(((LL)i * (i - 1LL) / 2LL) % (P - 1), 2LL);
	for (int i = 0; i <= n; i++) a[i] = (LL)pw[i] * inv[i] % P;
	Poly::Invpoly(a, n + 1, b), memset(a, 0, sizeof a);
	for (int i = 1; i <= n; i++) a[i] = (LL)pw[i] * inv[i - 1] % P;
	Poly::Mulpoly(a, b, n, n); int ans = a[n];
	for (int i = 1; i < n; i++) ans = (LL)ans * i % P;
	printf("%d\n", ans);
}