BZOJ 3456: 城市规划 [多项式求逆元 组合数学 | 生成函数 多项式求ln]

3456: 城市规划

题意：n个点组成的无向连通图个数

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以前做过，今天复习一下

令(f[n])为n个点的无向连通图个数

n个点的完全图个数为(2^{inom{n}{2}})

和Bell数的推导很类似，枚举第一个cc的点的个数

[2^{inom{n}{2}} = sum_{i=1}^n inom{n-1}{i-1} f(i) 2^{inom{n-i}{2}} ]

整理后

[frac{2^{inom{n}{2}}}{(n-1)!} = sum_{i=1}^n frac{f(i)}{(i-1)!}frac{2^{inom{n-i}{2}}}{(n-i)!} ]

这是卷积的形式

[C(x) = A(x)B(x) ightarrow A(x) = C(x)B^{-1}(x) ]

多项式求逆就可做了

注意(b_0=1)

其实就是EGP求ln......

简单的有标号集合计数

[G(x) = sum_{ige 0} 2^{inom{i}{2}}frac{x^i}{i!} = sum_{i ge 0} frac{F(x)^i}{i!} = e^{F(x)}\ F(x) = ln G(x) = int frac{G'(x)}{G(x)}dx ]

第一份是多项式求ln的代码，注意EGP要除以阶乘逆元

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<18) + 5;
inline int read(){
	char c=getchar();int x=0,f=1;
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

int P = 1004535809, inv2 = (P+1)/2;
inline int Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b>>=1, a=a*a%P)
		if(b&1) ans=ans*a%P;
	return ans;
}

ll inv[N];
namespace ntt {
	int g = 3, rev[N];
	void dft(int *a, int n, int flag) {
		int k = 0; while((1<<k) < n) k++;
		for(int i=0; i<n; i++) {
			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
			if(i < rev[i]) swap(a[i], a[rev[i]]);
		}
		for(int l=2; l<=n; l<<=1) {
			int m = l>>1, wn = Pow(g, flag == 1 ? (P-1)/l : P-1-(P-1)/l);
			for(int *p = a; p != a+n; p += l)
				for(int k=0, w=1; k<m; k++, w = (ll)w*wn %P) {
					int t = (ll) w * p[k+m] %P, r = p[k];
					p[k+m] = (r - t + P) %P;
					p[k] = (r + t) %P;
				}
		}
		if(flag == -1) {
			ll inv = Pow(n, P-2);
			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
		}
	}
	
	void inverse(int *a, int *b, int l) {
		static int t[N];
		if(l == 1) {b[0] = Pow(a[0], P-2); return;}
		inverse(a, b, l>>1);
		int n = l<<1;
		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0;
		dft(t, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;
		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
	}

	void ln(int *a, int *b, int l) {
		static int da[N], ia[N];
		int n = l<<1;
		for(int i=0; i<n; i++) da[i] = ia[i] = 0;
		for(int i=0; i<l-1; i++) da[i] = (ll) (i+1) * a[i+1] %P;
		inverse(a, ia, l);
		dft(da, n, 1); dft(ia, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) da[i] * ia[i] %P;
		dft(b, n, -1);
		for(int i=l-1; i>0; i--) b[i] = (ll) inv[i] * b[i-1] %P; b[0] = 0;
		for(int i=l; i<n; i++) b[i] = 0;
	}
	
}

ll fac[N], facInv[N];
int n, a[N], b[N];
int main() {
	freopen("in", "r", stdin);
	n = read();
	int len = 1;
	while(len <= n) len <<= 1;
	inv[1] = 1; fac[0] = facInv[0] = 1;
	for(int i=1; i<=n; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}
	a[0] = 1;
	for(int i=1; i<=n; i++) a[i] = Pow(2, (ll) i * (i - 1) /2 %(P-1)) * facInv[i] %P;
	ntt::ln(a, b, len);
	//for(int i=0; i<=n; i++) printf("%d ", b[i]); puts("");
	int ans = b[n] * fac[n] %P;
	printf("%d", ans);
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<18) + 5;
inline int read(){
	char c=getchar();int x=0,f=1;
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

ll P = 1004535809;
inline ll Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b>>=1, a=a*a%P)
		if(b&1) ans=ans*a%P;
	return ans;
}
inline void mod(int &x) {if(x>=P) x-=P; else if(x<0) x+=P;}
namespace fnt {
	int n, g=3, rev[N];
	void dft(int *a, int n, int flag=1) {
		for(int i=0; i<n; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);

		for(int l=2; l<=n; l<<=1) {
			int m = l>>1;
			ll wn = Pow(g, flag==1 ? (P-1)/l : P-1-(P-1)/l);
			for(int *p=a; p!=a+n; p+=l) {
				ll w = 1;
				for(int k=0; k<m; k++) {
					ll t = p[k+m] * w %P;
					mod(p[k+m] = p[k] - t);
					mod(p[k] = p[k] + t);
					w = w * wn %P;
				}
			}
		}
		if(flag == -1) {
			ll inv = Pow(n, P-2);
			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
		}
	}
	int t[N];
	void inverse(int *a, int *b, int l) { // mod x^l
		if(l == 1) {b[0] = Pow(a[0], P-2); return;}
		inverse(a, b, (l+1)>>1);
		int n = 1, k = 0; while(n < l<<1) n<<=1, k++;
		for(int i=0; i<n; i++) rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
		for(int i=0; i<l; i++) t[i] = a[i]; for(int i=l; i<n; i++) t[i] = 0;
		dft(t, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;
		dft(b, n, -1);
		for(int i=l; i<n; i++) b[i] = 0;
	}

	void mul(int *a, int *b, int l) {
		int n = 1; while(n < l<<1) n<<=1;
		dft(a, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) a[i] = (ll) a[i] * b[i] %P;
		dft(a, n, -1);
	}
}

int n, a[N], b[N], bi[N];
ll inv[N], fac[N], facInv[N];
int main() {
	freopen("in", "r", stdin);
	n=read();
	inv[1] = 1; fac[0] = facInv[0] = 1;
	for(int i=1; i<=n; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}
	for(int i=1; i<=n; i++) {
		ll mi = Pow(2, (ll) i * (i-1) / 2 %(P-1));
		a[i] = mi * facInv[i-1] %P;
		b[i] = mi * facInv[i] %P;
	}
	b[0] = 1;

	fnt::inverse(b, bi, n+1); 
	fnt::mul(a, bi, n+1);
	ll ans = a[n] * fac[n-1] %P;
	printf("%lld", ans);
}