3456: 城市规划
题意：n个点组成的无向连通图个数
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\(2^{\binom{n}{2}}\)
和Bell数的推导很类似，枚举第一个cc的点的个数
\[2^{\binom{n}{2}} = \sum_{i=1}^n \binom{n-1}{i-1} f(i) 2^{\binom{n-i}{2}}
\]
\]
整理后
\[\frac{2^{\binom{n}{2}}}{(n-1)!} = \sum_{i=1}^n \frac{f(i)}{(i-1)!}\frac{2^{\binom{n-i}{2}}}{(n-i)!}
\]
\]
这是卷积的形式
\[C(x) = A(x)B(x) \rightarrow A(x) = C(x)B^{-1}(x)
\]
\]
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
\[G(x) = \sum_{i\ge 0} 2^{\binom{i}{2}}\frac{x^i}{i!} = \sum_{i \ge 0} \frac{F(x)^i}{i!} = e^{F(x)}\\
F(x) = \ln G(x) = \int \frac{G'(x)}{G(x)}dx
\]
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);
}