题目链接
题解
真是一道经典好题，至此已经写了分治(NTT)，多项式求逆，多项式求(ln)三种写法
我们发现我们要求的是大小为(n)无向联通图的数量
而(n)个点的无向图是由若干个无向联通图构成的
那么我们设(F(x))为无向联通图数量的指数型生成函数
设(G(x))为无向图数量的指数型生成函数
(G(x))很好求
而
[G(x) = frac{F(x)}{1!} + frac{F^2(x)}{2!} + frac{F^3(x)}{3!} + dots = e^{F(x)}
]
则
[F(x) = lnG(x)
]
多项式求(ln)即可
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
const int G = 3,P = 1004535809;
int R[maxn];
inline int qpow(int a,LL b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
void NTT(int* a,int n,int f){
for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k],y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
}
}
}
if (f == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int n,c[maxn],f[maxn],g[maxn],Fv[maxn];
int fac[maxn],fv[maxn],inv[maxn];
void init(){
fac[0] = fac[1] = fv[0] = fv[1] = inv[0] = inv[1] = 1;
for (int i = 2; i <= (n << 1); i++){
fac[i] = 1ll * fac[i - 1] * i % P;
inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
fv[i] = 1ll * fv[i - 1] * inv[i] % P;
}
}
void Inv(int* a,int* b,int deg){
if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
Inv(a,b,(deg + 1) >> 1);
int n = 1,L = 0;
while (n < (deg << 1)) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (int i = 0; i < deg; i++) c[i] = a[i];
for (int i = deg; i < n; i++) c[i] = 0;
NTT(c,n,1); NTT(b,n,1);
for (int i = 0; i < n; i++)
b[i] = 1ll * ((2ll - 1ll * c[i] * b[i] % P) + P) % P * b[i] % P;
NTT(b,n,-1);
for (int i = deg; i < n; i++) b[i] = 0;
}
void Der(int* a,int& n){
n--;
for (int i = 0; i <= n; i++) a[i] = 1ll * a[i + 1] * (i + 1) % P;
}
void Int(int* a,int& n){
n++;
for (int i = n; i; i--) a[i] = 1ll * a[i - 1] * inv[i] % P;
}
void Getln(int* a,int* b){
int deg = n;
Inv(a,Fv,n + 1);
Der(a,deg);
int m = deg + n,n = 1,L = 0;
while (n <= m) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
NTT(a,n,1); NTT(Fv,n,1);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * Fv[i] % P;
NTT(a,n,-1);
deg = m;
Int(a,deg);
for (int i = 0; i <= deg; i++) b[i] = a[i];
}
int main(){
n = read();
init();
f[0] = f[1] = 1;
for (int i = 2; i <= n; i++)
f[i] = 1ll * qpow(2,1ll * i * (i - 1) / 2) * fv[i] % P;
Getln(f,g);
printf("%lld
",1ll * g[n] * fac[n] % P);
return 0;
}