#3456. 城市规划（生成函数，多项式求逆）
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#3456. 城市规划
设 f n f_n fn为 n n n个点的的点的简单无向连通图数目， g n g_n gn为 n n n个点的简单无向图个数（不要求联通）。
对于 g n g_n gn显然有 g n = 2 n ( n − 1 ) 2 g_n = 2 ^{\frac{n(n - 1)}{2}} gn=22n(n−1)，共有 n ( n + 1 ) 2 \frac{n(n + 1)}{2} 2n(n+1)条边，然后每条边可选可不选。
我们枚举
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答
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g_n = \sum\limits_{i = 1} ^{n}C_{n - 1} ^{i - 1}f_ig_{n - i}\\ 2 ^{(_2 ^ n)} = \sum_{i = 1} ^{n}(_{i - 1} ^{n - 1})f_i 2 ^{(_2 ^{n - i})}\\ 2 ^{(_2 ^ n)} = \sum_{i = 1} ^{n} \frac{(n - 1)!}{(i - 1)!(n - i)!} f_i 2 ^{(_2 ^{n - i})}\\ \frac{2 ^{(_2 ^ n)}}{(n - 1)!} = \sum_{i = 1} ^{n} \frac{f_i}{(i - 1)!} \frac{2^{(_2 ^{n - i})}}{(n - i)!}\\ 设G(x) = \sum_{n = 1} ^{\infty} \frac{2 ^{(_2 ^n)}}{(n - 1)!} x ^ n\\ F(x) = \sum_{n = 1} ^{\infty} \frac{f_n}{(n - 1)!}\\ H(x) = \sum_{n = 0} ^{\infty} \frac{2 ^{(_2 ^n)}}{n!}\\ G(x) = F(x) H(x)\\ F(x) = G(x)H^{-1}(x)\\ 构造多项式，多项式求逆，把[x ^ n]项系数乘上(n - 1)!即是答案\\
gn=i=1∑nCn−1i−1fign−i2(2n)=i=1∑n(i−1n−1)fi2(2n−i)2(2n)=i=1∑n(i−1)!(n−i)!(n−1)!fi2(2n−i)(n−1)!2(2n)=i=1∑n(i−1)!fi(n−i)!2(2n−i)设G(x)=n=1∑∞(n−1)!2(2n)xnF(x)=n=1∑∞(n−1)!fnH(x)=n=0∑∞n!2(2n)G(x)=F(x)H(x)F(x)=G(x)H−1(x)构造多项式，多项式求逆，把[xn]项系数乘上(n−1)!即是答案
#include <bits/stdc++.h>
using namespace std;
const int mod = 1004535809, inv2 = mod + 1 >> 1;
namespace Quadratic_residue {
struct Complex {
int r, i;
Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}
};
int I2;
Complex operator * (const Complex &a, Complex &b) {
return Complex((1ll * a.r * b.r % mod + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);
}
Complex quick_pow(Complex a, int n) {
Complex ans = Complex(1, 0);
while (n) {
if (n & 1) {
ans = ans * a;
}
a = a * a;
n >>= 1;
}
return ans;
}
int get_residue(int n) {
mt19937 e(233);
if (n == 0) {
return 0;
}
if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {
return -1;
}
uniform_int_distribution<int> r(0, mod - 1);
int a = r(e);
while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {
a = r(e);
}
I2 = (1ll * a * a % mod - n + mod) % mod;
int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;
if(x > y) swap(x, y);
return x;
}
}
const int N = 1e6 + 10;
int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];
int quick_pow(int a, int n) {
int ans = 1;
while (n) {
if (n & 1) {
ans = 1ll * a * ans % mod;
}
a = 1ll * a * a % mod;
n >>= 1;
}
return ans;
}
void get_r(int lim) {
for (int i = 0; i < lim; i++) {
r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
}
}
void get_inv(int n) {
inv[1] = 1;
for (int i = 2; i <= n; i++) {
inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
}
}
void NTT(int *f, int lim, int rev) {
for (int i = 0; i < lim; i++) {
if (i < r[i]) {
swap(f[i], f[r[i]]);
}
}
for (int mid = 1; mid < lim; mid <<= 1) {
int wn = quick_pow(3, (mod - 1) / (mid << 1));
for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
int w = 1;
for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
}
}
}
if (rev == -1) {
int inv = quick_pow(lim, mod - 2);
reverse(f + 1, f + lim);
for (int i = 0; i < lim; i++) {
f[i] = 1ll * f[i] * inv % mod;
}
}
}
void polyinv(int *f, int *g, int n) {
if (n == 1) {
g[0] = quick_pow(f[0], mod - 2);
return ;
}
polyinv(f, g, n + 1 >> 1);
for (int i = 0; i < n; i++) {
t[i] = f[i];
}
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
get_r(lim);
NTT(t, lim, 1);
NTT(g, lim, 1);
for (int i = 0; i < lim; i++) {
int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;
g[i] = 1ll * g[i] * cur % mod;
t[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
}
void polysqrt(int *f, int *g, int n) {
if (n == 1) {
g[0] = Quadratic_residue::get_residue(f[0]);
return ;
}
polysqrt(f, g, n + 1 >> 1);
polyinv(g, b, n);
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
get_r(lim);
for (int i = 0; i < n; i++) {
t[i] = f[i];
}
NTT(g, lim, 1);
NTT(b, lim, 1);
NTT(t, lim, 1);
for (int i = 0; i < lim; i++) {
g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;
b[i] = t[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
}
void derivative(int *a, int *b, int n) {
for (int i = 0; i < n; i++) {
b[i] = 1ll * a[i + 1] * (i + 1) % mod;
}
}
void integrate(int *a, int n) {
for (int i = n - 1; i >= 1; i--) {
a[i] = 1ll * a[i - 1] * inv[i] % mod;
}
a[0] = 0;
}
void polyln(int *f, int *g, int n) {
polyinv(f, b, n);
derivative(f, g, n);
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
get_r(lim);
NTT(g, lim, 1);
NTT(b, lim, 1);
for (int i = 0; i < lim; i++) {
g[i] = 1ll * g[i] * b[i] % mod;
b[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
integrate(g, n);
}
void polyexp(int *f, int *g, int n) {
if (n == 1) {
g[0] = 1;
return ;
}
polyexp(f, g, n + 1 >> 1);
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
polyln(g, d, n);
for (int i = 0; i < n; i++) {
t[i] = (f[i] - d[i] + mod) % mod;
}
t[0] = (t[0] + 1) % mod;
get_r(lim);
NTT(g, lim, 1);
NTT(t, lim, 1);
for (int i = 0; i < lim; i++) {
g[i] = 1ll * g[i] * t[i] % mod;
t[i] = d[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
}
/*
b存放多项式逆，
c存放多项式开根，
d存放多项式对数ln，
e存放多项式指数exp，
t作为中间转移数组,
如果要用到polyln，得提前调用get_inv(n)先预先得到我们想要得到的逆元范围。
*/
int h[N], g[N], fac[N], ifac[N], n;
void init() {
fac[0] = 1;
for (int i = 1; i < N; i++) {
fac[i] = 1ll * i * fac[i - 1] % mod;
}
ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);
for (int i = N - 2; i >= 0; i--) {
ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
scanf("%d", &n);
init();
for (int i = 1; i <= n; i++) {
g[i] = 1ll * quick_pow(2, 1ll * i * (i - 1) / 2 % (mod - 1)) * ifac[i - 1] % mod;
}
for (int i = 0; i <= n; i++) {
h[i] = 1ll * quick_pow(2, 1ll * i * (i - 1) / 2 % (mod - 1)) * ifac[i] % mod;
}
polyinv(h, b, n + 1);
for (int i = 0; i <= n; i++) {
h[i] = b[i];
b[i] = 0;
}
int lim = 1;
while (lim <= 2 * n) {
lim <<= 1;
}
get_r(lim);
NTT(g, lim, 1);
NTT(h, lim, 1);
for (int i = 0; i < lim; i++) {
g[i] = 1ll * g[i] * h[i] % mod;
h[i] = 0;
}
NTT(g, lim, -1);
printf("%d\n", 1ll * g[n] * fac[n - 1] % mod);
return 0;
}
标签：求逆,3456,int,多项式,++,1ll,lim,quick,mod 来源： https://blog.csdn.net/weixin_45483201/article/details/113775039