P5900 无标号无根树计数 题解
不懂为啥都要对原式神秘转化之后再牛顿迭代，直接对原式牛顿迭代即可！完全不用转化！
设无标号有根树的组合类是 \(\mathcal T\)，则有 \(\mathcal T=\mathcal Z\times\mathrm{MSET}(\mathcal T)\)，即 \(T(x)=x\exp\sum\limits_{j\ge1}\dfrac{T(x^j)}j\)，
设 \(G(F(x))=F(x)-x\exp\sum\limits_{j\ge1}\dfrac{F(x^j)}j=0\)，要求 \(F(x)\bmod x^n\)，问题变为牛顿迭代的形式。
首先钦定 \(F(x)\bmod x^1=[x^0]F(x)=0\)（否则 \(\exp\sum\limits_{j\ge1}\dfrac{F(x^j)}j\) 的常数项不收敛），
然后假设已经求出模 \(x^{\frac n2}\) 意义下的解 \(F_0(x)\)，则模 \(x^n\) 意义下的解 \(F(x)\equiv F_0(x)-\dfrac{G(F_0(x))}{G'(F_0(x))}\pmod{x^n}\)，
考虑如何求 \(G'(F_0(x))\)。
观察到 \(G(F_0(x))=F_0(x)-x\exp F_0(x)\exp\sum\limits_{j\ge2}\dfrac{F_0(x^j)}j\)，
设最终答案是 \(H(x)\)（这里 \(H(x)\) 是与 \(F_0(x)\) 无关的常量），
则 \(\forall j\ge2\)，有 \(F_0(x^j)\equiv H(x^j)\pmod{x^n}\)，则 \(\exp\sum\limits_{j\ge2}\dfrac{F_0(x^j)}j\equiv\exp\sum\limits_{j\ge2}\dfrac{H(x^j)}j\pmod{x^n}\)，
于是 \(\exp\sum\limits_{j\ge2}\dfrac{F_0(x^j)}j\) 是与 \(F_0(x)\) 无关的常量，
则 \(G'(F_0(x))=1-x\exp F_0(x)\exp\sum\limits_{j\ge2}\dfrac{F_0(x^j)}j=1-x\exp\sum\limits_{j\ge1}\dfrac{F_0(x^j)}j\)，
于是 \(F(x)\equiv F_0(x)-\dfrac{G(F_0(x))}{G'(F_0(x))}\equiv F_0(x)-\dfrac{F_0(x)-x\exp\sum\limits_{j\ge1}\dfrac{F_0(x^j)}j}{1-x\exp\sum\limits_{j\ge1}\dfrac{F_0(x^j)}j}\pmod{x^n}\)。
现在我们会算无标号有根树了，设 \(f_i\) 表示 \(i\) 个点的无标号有根树个数，考虑怎么算无标号无根树。
\(n\) 为奇数时，每个无标号无根树都有唯一的重心，所以只需统计根为重心的无标号有根树个数，
考虑用 \(f_n\) 减去根不为重心的无标号有根树个数，这些树一定有一棵大小超过 \(\left\lfloor\dfrac n2\right\rfloor\) 的子树，
枚举这棵子树的大小，则答案为 \(f_n-\sum\limits_{i=\left\lfloor\frac n2\right\rfloor+1}^{n-1}f_i\times f_{n-i}\)。
\(n\) 为偶数时，考虑用同样的计数方法，
发现有两个重心的树，即有一棵大小为 \(\dfrac n2\) 的子树的树，可能会被重复统计，
设这棵大小为 \(\dfrac n2\) 的子树为 \(T\)，若原树去掉 \(T\) 后剩下的树与 \(T\) 完全一致，
则原树分别以两个重心为根形成的有根树完全一致，这样的树不会被重复统计，
反之若原树去掉 \(T\) 后剩下的树与 \(T\) 不同，这样的树就会被重复统计，
所以从大小为 \(\dfrac n2\) 的有根树中任选出两棵，就可以组合出一种被重复统计的树，
所以有 \(f_{\frac n2}\choose2\) 种被重复统计的树，答案减去 \(f_{\frac n2}\choose2\) 即可。
没有封装多项式类，所以代码比较混乱邪恶，建议谨慎阅读。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define int long long
#define G 3
#define _G 332748118
#define M 998244353
using namespace std;
int n, l, r[6000050], f[6000050], g[6000050], h[6000050], x[6000050], y[6000050], z[6000050], k[6000050], v[6000050];
int P(int x, int y)
{
int q = 1;
for (; y; y >>= 1, x = x * x % M)
if (y & 1)
q = q * x % M;
return q;
}
void F(int *f, int n, int v)
{
for (int i = 0; i < n; ++i)
if (i < r[i])
swap(f[i], f[r[i]]);
for (int L = 2, m; L <= n; L <<= 1)
{
m = L >> 1;
int W = P(v == 1 ? G : _G, (M - 1) / L);
for (int l = 0, r = L - 1; r <= n; l += L, r += L)
{
int o = 1;
for (int p = l; p < l + m; ++p)
{
int x = f[p], y = f[p + m];
f[p] = (x + o * y) % M, f[p + m] = (x + M - o * y % M) % M;
o = o * W % M;
}
}
}
}
void I(int *f, int *g, int n)
{
memset(g, 0, n << 4);
memset(x, 0, n << 4);
g[0] = P(f[0], M - 2);
int L;
for (L = 4;; L <<= 1)
{
memcpy(x, f, L << 2);
memcpy(y, g, L << 3);
l = __lg(L);
for (int i = 0; i < L; ++i)
r[i] = r[i >> 1] >> 1 | (i & 1) << l - 1;
F(x, L, 1);
F(y, L, 1);
for (int i = 0; i < L; ++i)
x[i] = x[i] * y[i] % M;
F(x, L, -1);
int _ = P(L, M - 2);
for (int i = 0; i < L; ++i)
x[i] = (M - x[i] * _ % M) % M;
x[0] = (x[0] + 2) % M;
memset(x + (L >> 1), 0, L << 2);
F(g, L, 1);
F(x, L, 1);
for (int i = 0; i < L; ++i)
g[i] = g[i] * x[i] % M;
F(g, L, -1);
for (int i = 0; i < L; ++i)
g[i] = g[i] * _ % M;
if (L >> 1 >= n)
break;
}
memset(g + n, 0, L - n << 3);
}
void LN(int *f, int *g, int n)
{
memset(h, 0, n << 4);
for (int i = 0; i < n - 1; ++i)
h[i] = (i + 1) * f[i + 1] % M;
I(f, g, n);
int L = 1;
while (L >> 1 < n)
L <<= 1;
l = __lg(L);
for (int i = 0; i < L; ++i)
r[i] = r[i >> 1] >> 1 | (i & 1) << l - 1;
F(g, L, 1);
F(h, L, 1);
for (int i = 0; i < L; ++i)
h[i] = g[i] * h[i] % M;
F(h, L, -1);
int _ = P(L, M - 2);
for (int i = 0; i < L; ++i)
h[i] = h[i] * _ % M;
g[0] = 0;
for (int i = 1; i < n; ++i)
g[i] = h[i - 1] * P(i, M - 2) % M;
memset(g + n, 0, L - n << 3);
}
void EXP(int *f, int *g, int n)
{
memset(g, 0, n << 4);
g[0] = 1;
int L;
for (L = 4;; L <<= 1)
{
LN(g, z, L >> 1);
for (int i = 0; i < L >> 1; ++i)
z[i] = (f[i] + M - z[i]) % M;
z[0] = (z[0] + 1) % M;
l = __lg(L);
for (int i = 0; i < L; ++i)
r[i] = r[i >> 1] >> 1 | (i & 1) << l - 1;
F(g, L, 1);
F(z, L, 1);
for (int i = 0; i < L; ++i)
g[i] = g[i] * z[i] % M;
F(g, L, -1);
int _ = P(L, M - 2);
for (int i = 0; i < L; ++i)
g[i] = g[i] * _ % M;
memset(g + (L >> 1), 0, L << 2);
if (L >> 1 >= n)
break;
}
memset(g + n, 0, L - n << 3);
}
signed main()
{
v[1] = 1;
for (int i = 2; i <= 6e6; ++i)
v[i] = (M - M / i) * v[M % i] % M;
scanf("%lld", &n), ++n;
for (int L = 4;; L <<= 1)
{
memset(g, 0, L << 2);
for (int i = 1; i < L >> 2; ++i)
for (int j = i; j < L >> 1; j += i)
g[j] = (g[j] + f[i] * i % M * v[j]) % M;
EXP(g, k, L >> 1);
for (int i = (L >> 1) - 1; i >= 1; --i)
k[i] = (M - k[i - 1]) % M;
k[0] = 1;
I(k, g, L >> 1);
k[0] = 0;
for (int i = 1; i < L >> 2; ++i)
k[i] = (k[i] + f[i]) % M;
F(g, L, 1);
F(k, L, 1);
for (int i = 0; i < L; ++i)
g[i] = g[i] * k[i] % M;
F(g, L, -1);
int _ = P(L, M - 2);
for (int i = 0; i < L; ++i)
g[i] = g[i] * _ % M;
for (int i = 0; i < L >> 1; ++i)
f[i] = (f[i] + M - g[i]) % M;
if (L >> 1 >= n)
break;
}
--n;
int q = f[n];
for (int i = (n >> 1) + 1; i < n; ++i)
q = (q + M - f[i] * f[n - i] % M) % M;
if (!(n & 1))
{
int u = f[n >> 1];
q = (q + M - u * (u - 1) % M * (M + 1 >> 1) % M) % M;
}
printf("%lld", q);
return 0;
}
