bzoj3456 城市规划 多项式求In

$n$个点的无向联通图的个数

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打着好累啊

一定要封装一个板子

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记$C(x)$为无向图个数的指数型生成函数，(C(0) = 1)

记$G(x)$为无向联通图个数的指数型生成函数，(G(0) = 0)

那么$G(x) = e^{C(x)}$

从而，(C(x) = In(G(x)))

复杂度$O(n log n)$

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 270000;
const int mod = 1004535809;

inline int Inc(int a, int b) { return (a + b >= mod) ? a + b - mod : a + b; }
inline int Dec(int a, int b) { return (a - b < 0) ? a - b + mod : a - b; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int fp(int a, int k) {
	int ret = 1;
	for( ; k; k >>= 1, a = mul(a, a))
		if(k & 1) ret = mul(ret, a);
	return ret;
}

int N_, N, n, lg;
int ans[sid], ivfac[sid], fac[sid], inv[sid], rev[sid];

inline void init(int Maxn, int &rn, int &rlg, int opt = 0) {
	rn = 1; rlg = 0;
	while(rn < Maxn) rn <<= 1, rlg ++;
	if(opt) rep(i, 0, rn) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (rlg - 1));
}

inline int NTT(int *a, int n, int opt) {
	for(ri i = 0; i < n; i ++) 
		if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(ri i = 1; i < n; i <<= 1)
	for(ri j = 0, g = fp(3, (mod - 1) / (i << 1)); j < n; j += (i << 1))
	for(ri k = j, G = 1; k < i + j; k ++, G = 1ll * G * g % mod) {
		int x = a[k], y = mul(G, a[i + k]);
		a[k] = (x + y >= mod) ? x + y - mod : x + y;
		a[i + k] = (x - y < 0) ? x - y + mod : x - y; 
	}
	if(opt == -1) {
		reverse(a + 1, a + n);
		for(ri i = 0; i < n; i ++) a[i] = mul(a[i], inv[n]);
	}
}

int iva[sid], ivb[sid];
inline void get_inv(int *a, int n, int *ret) {
	if(n == 1) { ret[0] = inv[a[0]]; return; }
	get_inv(a, n >> 1, ret);
	init(n + n, N, lg, 1);
	for(ri i = 0; i < N; i ++) iva[i] = ivb[i] = 0;
	for(ri i = 0; i < n; i ++) iva[i] = a[i], ivb[i] = ret[i];
	NTT(iva, N, 1); NTT(ivb, N, 1);
	for(ri i = 0; i < N; i ++) iva[i] = Dec(Inc(ivb[i], ivb[i]), mul(iva[i], mul(ivb[i], ivb[i])));
	NTT(iva, N, -1);
	for(ri i = 0; i < n; i ++) ret[i] = iva[i];
}

inline void wf(int *a, int n, int *ret) { for(ri i = 1; i < n; i ++) ret[i - 1] = mul(a[i], i); }
inline void jf(int *a, int n, int *ret) { for(ri i = 1; i < n; i ++) ret[i] = mul(a[i - 1], inv[i]); }

int ivf[sid], df[sid];
inline void get_ln(int *a, int n, int *ret) {
	int N = 1, lg = 0;
	init(n + n, N, lg);
	wf(a, n, df); get_inv(a, n, ivf);
	init(n + n, N, lg, 1);
	NTT(df, N, 1); NTT(ivf, N, 1);
	for(ri i = 0; i < N; i ++) df[i] = mul(df[i], ivf[i]);
	NTT(df, N, -1); jf(df, n, ret);
}

int C2[sid], f[sid];
int main() {
	freopen("3456.in", "r", stdin);
	freopen("3456.out", "w", stdout);
	cin >> n;
	init(n + 5, N_, lg);
	
	N_ <<= 1;
	inv[0] = inv[1] = 1;
	fac[0] = fac[1] = 1;
	ivfac[0] = ivfac[1] = 1;
	rep(i, 2, N_) {
		fac[i] = mul(fac[i - 1], i);
		inv[i] = mul(inv[mod % i], mod - mod / i);
	}
	rep(i, 2, N_) ivfac[i] = mul(inv[i], ivfac[i - 1]);
	rep(i, 0, N_) C2[i] = 1ll * i * (i - 1) / 2 % (mod - 1);
	rep(i, 0, N_) f[i] = mul(fp(2, C2[i]), ivfac[i]); 
	N_ >>= 1;
	
	get_ln(f, N_, ans);
	printf("%d
", mul(ans[n], fac[n]));
	return 0;
}