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设fnf_nfn为nnn个点的的点的简单无向连通图数目#xff0c;gng_ngn为nnn个点的简单无向图个数#xff08;不要求联通#xff09;。
对于gng_ngn显然有gn2n(n−1)2g_n 2 ^{\frac{n(n - 1)}{2}}gn22n(n−1)#xff0c;共有n(n1)2\frac{n(n 1)}…#3456. 城市规划
设fnf_nfn为nnn个点的的点的简单无向连通图数目gng_ngn为nnn个点的简单无向图个数不要求联通。
对于gng_ngn显然有gn2n(n−1)2g_n 2 ^{\frac{n(n - 1)}{2}}gn22n(n−1)共有n(n1)2\frac{n(n 1)}{2}2n(n1)条边然后每条边可选可不选。
我们枚举111所在的点的连通块可得 gn∑i1nCn−1i−1fign−i2(2n)∑i1n(i−1n−1)fi2(2n−i)2(2n)∑i1n(n−1)!(i−1)!(n−i)!fi2(2n−i)2(2n)(n−1)!∑i1nfi(i−1)!2(2n−i)(n−i)!设G(x)∑n1∞2(2n)(n−1)!xnF(x)∑n1∞fn(n−1)!H(x)∑n0∞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\\ gni1∑nCn−1i−1fign−i2(2n)i1∑n(i−1n−1)fi2(2n−i)2(2n)i1∑n(i−1)!(n−i)!(n−1)!fi2(2n−i)(n−1)!2(2n)i1∑n(i−1)!fi(n−i)!2(2n−i)设G(x)n1∑∞(n−1)!2(2n)xnF(x)n1∑∞(n−1)!fnH(x)n0∑∞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.husing 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_distributionint 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lne存放多项式指数expt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;
}
