题目链接
题解
之前我们用分治(ntt)在(O(nlog^2n))的复杂度下做了这题，今天我们使用多项式求逆
设(f_n)表示(n)个点带标号无向连通图数
设(g_n)表示(n)个点图的数量，显然(g_n = 2^{{n choose 2}})
枚举(1)号点所在联通块大小，我们有
[g_n = sumlimits_{i = 1}^{n} {n - 1 choose i - 1}f_{i}g_{n - i}
]
代入(g_n)
[2^{{n choose 2}} = sumlimits_{i = 1}^{n} frac{(n - 1)!}{(n - i)!(i - 1)!}f_{i}2^{{n - i choose 2}}
]
整理一下：
[frac{2^{{n choose 2}}}{(n - 1)!} = sumlimits_{i = 1}^{n} frac{f_i}{(i - 1)!} * frac{2^{{n - i choose 2}}}{(n - i)!}
]
发现是一个卷积的形式
令
[F(x) = sumlimits_{n = 1}^{+infty} frac{f_n}{(n - 1)!} x^n
]
[G(x) = sumlimits_{n = 0}^{+infty} frac{2^{{n choose 2}}}{n!} x^n
]
[H(x) = sumlimits_{n = 1}^{+infty} frac{2^{{n choose 2}}}{(n - 1)!} x^n
]
则有
[H(x) = F(x)G(x)
]
那么
[F(x) = H(x)G^{-1}(x)
]
多项式求逆后再求一次卷积就可求得答案
复杂度(O(nlogn))，快了不少
#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 = 300005,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 f[maxn],g[maxn],h[maxn],gv[maxn],N;
int fac[maxn],inv[maxn],fv[maxn];
int c[maxn],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;
}
void work(int deg,int* a,int* b){
if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
work((deg + 1) >> 1,a,b);
int L = 0,n = 1;
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) % P * b[i] % P;
NTT(b,n,-1);
for (int i = deg; i < n; i++) b[i] = 0;
}
void init(){
fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
for (int i = 2; i <= N; 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;
}
g[0] = 1;
for (int i = 1; i <= N; i++){
h[i] = 1ll * qpow(2,1ll * i * (i - 1) / 2) * fv[i - 1] % P;
g[i] = 1ll * qpow(2,1ll * i * (i - 1) / 2) * fv[i] % P;
}
}
int main(){
N = read();
init();
work(N + 1,g,gv);
int L = 0,n = 1;
while (n <= (N << 1)) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
NTT(h,n,1); NTT(gv,n,1);
for (int i = 0; i < n; i++)
f[i] = 1ll * h[i] * gv[i] % P;
NTT(f,n,-1);
int ans = 1ll * f[N] * fac[N - 1] % P;
printf("%d
",ans);
return 0;
}