【LG4841】城市规划
题面
题解
记(t_i)表示(i)个点的无向图个数，显然(t_i=2^{C_i^2})。
设(f_i)表示(i)个点的无向连通图个数，容斥一下，枚举(1)号点所在连通块的大小，再让剩下的点随便构成联通图，
则有：
[f_i=t_i-sum_{j=1}^{i-1}f_j*C_{i-1}^{j-1}*t_{i-j}
]
展开组合数：
[f_i=t_i-sum_{j=1}^{i-1}f_j*t_{i-j}*frac {(i-1)!}{(i-j)!(j-1)!}\
Leftrightarrow frac {f_i}{(i-1)!}=frac {t_i}{(i-1)!}-sum_{j=1}^{i-1}frac {f_j}{(j-1)!}*frac {t_{i-j}}{(i-j)!}
]
设(g_i=frac {f_i}{(i-1)!},h_i=frac {t_i}{i!})，那么
[g_i=i*h_i-sum_{j=1}^{i-1}g_i*h_i
]
分治+(NTT)即可。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int Mod = 1004535809;
int fpow(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % Mod;
x = 1ll * x * x % Mod;
y >>= 1;
}
return res;
}
const int G = 3, iG = fpow(G, Mod - 2);
const int MAX_N = 4e5 + 5;
int Limit, rev[MAX_N];
void NTT(int *p, int op) {
for (int i = 0; i < Limit; i++) if (i < rev[i]) swap(p[i], p[rev[i]]);
for (int i = 1; i < Limit; i <<= 1) {
int rot = fpow(op == 1 ? G : iG, (Mod - 1) / (i << 1));
for (int j = 0; j < Limit; j += (i << 1)) {
int w = 1;
for (int k = 0; k < i; k++, w = 1ll * w * rot % Mod) {
int x = p[j + k], y = 1ll * w * p[i + j + k] % Mod;
p[j + k] = (x + y) % Mod, p[i + j + k] = (x - y + Mod) % Mod;
}
}
}
if (op == -1) {
int inv = fpow(Limit, Mod - 2);
for (int i = 0; i < Limit; i++) p[i] = 1ll * p[i] * inv % Mod;
}
}
int g[MAX_N], h[MAX_N];
void Div(int l, int r) {
if (l == r) return ;
static int A[MAX_N], B[MAX_N], C[MAX_N];
int mid = (l + r) >> 1;
Div(l, mid);
int p = 0;
for (Limit = 1; Limit <= (r - l) * 2; Limit <<= 1, ++p) ;
for (int i = 0; i < Limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (p - 1));
for (int i = 0; i <= mid - l; i++) A[i] = g[i + l];
for (int i = 0; i < r - l; i++) B[i] = h[i + 1];
for (int i = mid - l + 1; i < Limit; i++) A[i] = 0;
for (int i = r - l; i < Limit; i++) B[i] = 0;
NTT(A, 1); NTT(B, 1);
for (int i = 0; i < Limit; i++) C[i] = 1ll * A[i] * B[i] % Mod;
NTT(C, -1);
for (int i = mid + 1; i <= r; i++) g[i] = (g[i] - C[i - 1 - l] + Mod) % Mod;
Div(mid + 1, r);
}
int N, fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
if (m > n) return 0;
else return 1ll * fac[n] * ifc[m] % Mod * ifc[n - m] % Mod;
}
int main () {
#ifndef ONLINE_JUDGE
freopen("cpp.in", "r", stdin);
#endif
cin >> N;
fac[0] = 1; for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
ifc[N] = fpow(fac[N], Mod - 2); for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
for (int i = 1; i <= N; i++) h[i] = 1ll * fpow(2, 1ll * i * (i - 1) / 2 % (Mod - 1)) * ifc[i] % Mod;
for (int i = 1; i <= N; i++) g[i] = 1ll * ifc[i - 1] * fpow(2, 1ll * i * (i - 1) / 2 % (Mod - 1)) % Mod;
Div(1, N);
printf("%d
", (int)(1ll * g[N] * fac[N - 1] % Mod));
return 0;
}