Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4841 [集训队作业2013]城市规划

P4841 [集训队作业2013]城市规划

求 $n$ 个点的无向连通图。如果不考虑连通,那么答案为 $2 ^ {\binom n 2}$ 。考虑 EGF ,那么答案的 EGF $f$ 有 $e ^ f = g = \sum _ {i = 0} 2 ^ {\binom n 2} \frac {x ^ i} {i!}$ 。多项式 $\ln$ 即可。

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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, mod = 1004535809, inv2 = mod + 1 >> 1;
int rev[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void ntt (int *x, int bit, int op)
{
int tot = 1 << bit;
for (int i = 1; i < tot; ++i)
if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
swap(x[rev[i]], x[i]);
for (int mid = 1; mid < tot; mid <<= 1)
{
int w1 = binpow(3, (mod - 1) / (mid << 1));
if (!~op) w1 = binpow(w1);
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
{
int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
}
}
if (~op) return;
int itot = binpow(tot);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
int bit = 0;
while (1 << bit < n + m - 1) ++bit;
int tot = 1 << bit;
for (int i = n; i < tot; ++i) f[i] = 0;
for (int i = m; i < tot; ++i) g[i] = 0;
ntt(f, bit, 1), ntt(g, bit, 1);
for (int i = 0; i < tot; ++i)
res[i] = (LL)f[i] * g[i] % mod;
ntt(res, bit, -1);
for (int i = nm; i < tot; ++i) res[i] = 0;
}
void PolyInv(int n, int *x, int *g)
{
if (n == 1) return void(g[0] = binpow(x[0]));
int m = n + 1 >> 1;
int bit = 0;
while (1 << bit < n + m + m - 2) ++bit;
int tot = 1 << bit;
PolyInv(m, x, g);
for (int i = m; i < tot; ++i) g[i] = 0;
static int A[N];
for (int i = 0; i < n; ++i) A[i] = x[i];
for (int i = n; i < tot; ++i) A[i] = 0;
ntt(g, bit, 1), ntt(A, bit, 1);
for (int i = 0; i < tot; ++i)
g[i] = (2 - (LL)g[i] * A[i]) % mod * g[i] % mod;
ntt(g, bit, -1);
for (int i = n; i < tot; ++i) g[i] = 0;
}
void PolyDerivate(int n, int *x, int *g)
{
for (int i = 1; i < n; ++i)
g[i - 1] = (LL)x[i] * i % mod;
g[n - 1] = 0;
}
void PolyIntegrate(int n, int *x, int *g)
{
for (int i = 1; i < n; ++i)
g[i] = (LL)x[i - 1] * binpow(i) % mod;
g[0] = 0;
}
void PolyLn(int n, int *x, int *g)
{
static int A[N], B[N];
PolyDerivate(n, x, A);
PolyInv(n, x, B);
PolyMul(n, A, n, B, n, A);
PolyIntegrate(n, A, g);
}
int main ()
{
static int n, A[N], B[N];
read(n);
for (int i = 0, t = 1; i <= n; t = (LL)t * ++i % mod)
A[i] = (LL)binpow(t) * binpow(2, i * (i - 1ll) / 2 % (mod - 1)) % mod;
PolyLn(n + 1, A, B);
int k = 1;
for (int i = 2; i <= n; ++i)
k = (LL)k * i % mod;
write(((LL)k * B[n] % mod + mod) % mod);
return 0;
}