#3456. 城市规划（生成函数，多项式求逆）

我们枚举111所在的点的连通块可得：
gn=∑i=1nCn−1i−1fign−i2(2n)=∑i=1n(i−1n−1)fi2(2n−i)2(2n)=∑i=1n(n−1)!(i−1)!(n−i)!fi2(2n−i)2(2n)(n−1)!=∑i=1nfi(i−1)!2(2n−i)(n−i)!设G(x)=∑n=1∞2(2n)(n−1)!xnF(x)=∑n=1∞fn(n−1)!H(x)=∑n=0∞2(2n)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[xn]项系数乘上(n−1)!即是答案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;
}

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#3456. 城市规划（生成函数，多项式求逆）

#3456. 城市规划

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