洛谷 5900.无标号无根树

鸽子写题了!!!

有个东西叫 Euler 变换。
似乎也叫做 Polya Exp 或者 Multiset 构造?

具体地,对于 \(\{f_i\}\) 的 OGF \(F(x)\),其 Euler 变换是这样的: \[ \mathcal E(F(x)) = \prod\limits_{i=1}^{\infty} \frac1{ (1-x^i)^{-f_i} } \]

做过付公主的背包或者遗忘的集合的同学应该很熟悉这个东西?
因为实际上就是背包,所以相当于 OGF 的 exp。

然后通过一些代数推导 \[ \begin{aligned} \mathcal E(F(x)) &= \prod\limits_{i=1}^{\infty} \frac1{ (1-x^i)^{-f_i} } \\ &= \exp\left(\sum\limits_{i=1}^{\infty} -f_i \ln(1-x^i) \right) \\ &= \exp\left(\sum\limits_{i=1}^{\infty} f_i \sum\limits_{t=1}^{\infty} \frac{ x^{it} }{t} \right) \\ &= \exp\left(\sum\limits_{t=1}^{\infty} \frac1t \sum\limits_{i=1}^{\infty} f_i x^{it} \right) \\ &= \exp\left(\sum\limits_{t=1}^{\infty} \frac{F(x^t)}{t}\right) \end{aligned} \]

Polya 的推导还不会……什么时候把群论学好了再回来填坑。

设无标号根树的 OGF 为 \(F(x)\),可以列出方程 \[ F(x) = x\mathcal E(F(x)) \]

可以两边求导得到 \[ \begin{aligned} F'(x) &= \left(x\mathcal E(F(x))\right)' \\ &= \mathcal E(F(x)) + x \left(\mathcal E(F(x))\right)' \\ &= \mathcal E(F(x)) + x \exp\left(\sum\limits_{t=1}^{\infty} \frac{F(x^t)}{t}\right) \left(\sum\limits_{t=1}^{\infty} \frac{F(x^t)}{t}\right)' \\ &= \frac{F(x)}x + F(x) \sum\limits_{t=1}^{\infty} F'(x^t) x^{t-1} \end{aligned} \]

然后可以直接分治 NTT 解决。

但这样不好玩。
考虑牛顿迭代。

为了规避常数爆炸的 exp,可以将方程化为以下形式 \[ \ln\frac{F(x)}x - \sum\limits_{t=1}^{\infty} \frac{F(x^t)}t = 0 \]

考虑设 \(G(x) = \frac{F(x)}x\)\(H(G(x)) = \ln G(x) - \sum\limits_{t=1}^{\infty} \frac{x^t G(x^t)}t\),那么便是要解 \[ H(G(x)) = 0 \]

但是你可能会说后面那一大坨怎么办呢?
考虑平时牛顿迭代的过程:求得 \(G_0(x)\) 满足 \[ H(G_0(x)) \equiv 0 \pmod{x^n} \]

然后求 \[ G(x) \equiv G_0(x) - \frac{H(G(x))}{H'(G(x))} \pmod{ x^{2n} } \]

于是注意到,在我们求 \(G(x)\) 时,\(\sum\limits_{t=2}^{\infty} \frac{x^t G(x^t)}{t}\) 是可以得知的。
所以用 \(\sum\limits_{t=2}^{\infty} \frac{x^t G_0(x^t)}{t}\) 替换之,并视为一个常数。
从而 \[ H(G(x)) = \ln G(x) - xG(x) - \sum\limits_{t=2}^{\infty} \frac{x^t G_0(x^t)}{t} \]

\[ H'(G(x)) = \frac1{G(x)} - x \]

牛顿迭代即可。

但是题目要求的貌似是无根树呢……
考虑钦定一个特殊的点为根,比如重心。
然后用 \(f_n\) 减去根不是重心的情况。

根不是重心当且仅当其中一个子树的大小超过了 \(\left\lfloor\frac n2\right\rfloor\),于是答案为 \[ f_n - \sum\limits_{i=\left\lfloor\frac n2\right\rfloor+1}^{n-1} f_i f_{n-i} \]

不过,当 \(n\) 为偶数时,可能存在两个重心。
即存在一个子树大小恰为 \(\frac n2\)
如果这棵子树和其他部分完全同构,那么只会被计算一次。
否则会重复算,需要减去。
此时答案为 \[ f_n - \sum\limits_{i=\frac n2+1}^{n-1} f_i f_{n-i} - \binom{f_{\frac n2}}{2} \]

代码:

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#include <cstdio>
#include <vector>
#include <cstring>
#include <utility>
#include <algorithm>
#define add(a,b) (a + b >= mod ? a + b - mod : a + b)
#define dec(a,b) (a < b ? a - b + mod : a - b)
using namespace std;
const int N = 2e5;
const int mod = 998244353;
int n;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
namespace Poly
{
const int N = 1 << 19;
const int G = 3;
int lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init()
{
for(register int i = 2;i <= N;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / N);
rt[N >> 1] = 1;
for(register int i = (N >> 1) + 1;i <= N;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (N >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= N;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[N] = fpow(fac[N],mod - 2);
for(register int i = N;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= N;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
struct poly
{
vector<int> a;
inline poly(int x = 0)
{
x && (a.push_back(x),1);
}
inline poly(const vector<int> &o)
{
a = o,shrink();
}
inline poly(const poly &o)
{
a = o.a,shrink();
}
inline void shrink()
{
for(;!a.empty() && !a.back();a.pop_back());
}
inline int size() const
{
return a.size();
}
inline void resize(int x)
{
a.resize(x);
}
inline int operator[](int x) const
{
if(x < 0 || x >= size())
return 0;
return a[x];
}
inline void clear()
{
vector<int>().swap(a);
}
inline poly rever() const
{
return poly(vector<int>(a.rbegin(),a.rend()));
}
inline void ntt(int type = 1)
{
int n = size();
type == -1 && (reverse(a.begin() + 1,a.end()),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
friend inline poly operator+(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = add(a[i],b[i]);
return poly(ret);
}
friend inline poly operator-(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = dec(a[i],b[i]);
return poly(ret);
}
friend inline poly operator*(poly a,poly b)
{
if(a.a.empty() || b.a.empty())
return poly();
int lim = 1,tot = a.size() + b.size() - 1;
for(;lim < tot;lim <<= 1);
a.resize(lim),b.resize(lim);
a.ntt(),b.ntt();
for(register int i = 0;i < lim;++i)
a.a[i] = (long long)a[i] * b[i] % mod;
a.ntt(-1),a.shrink();
return a;
}
poly &operator+=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = add(a[i],o[i]);
return *this;
}
poly &operator-=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = dec(a[i],o[i]);
return *this;
}
poly &operator*=(poly o)
{
return (*this) = (*this) * o;
}
poly deriv() const
{
if(a.empty())
return poly();
vector<int> ret(size() - 1);
for(register int i = 0;i < size() - 1;++i)
ret[i] = (long long)(i + 1) * a[i + 1] % mod;
return poly(ret);
}
poly integ() const
{
if(a.empty())
return poly();
vector<int> ret(size() + 1);
for(register int i = 0;i < size();++i)
ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
return poly(ret);
}
inline poly modxn(int n) const
{
if(a.empty())
return poly();
n = min(n,size());
return poly(vector<int>(a.begin(),a.begin() + n));
}
inline poly inver(int m) const
{
poly ret(fpow(a[0],mod - 2));
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
return ret.modxn(m);
}
inline pair<poly,poly> div(poly o) const
{
if(size() < o.size())
return make_pair(poly(),*this);
poly f,g;
f = (rever().modxn(size() - o.size() + 1) * o.rever().inver(size() - o.size() + 1)).modxn(size() - o.size() + 1).rever();
g = (modxn(o.size() - 1) - o.modxn(o.size() - 1) * f.modxn(o.size() - 1)).modxn(o.size() - 1);
return make_pair(f,g);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ().modxn(m);
}
inline poly exp(int m) const
{
poly ret(1);
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
return ret.modxn(m);
}
inline poly pow(int m,int k1,int k2 = -1) const
{
if(a.empty())
return poly();
if(k2 == -1)
k2 = k1;
int t = 0;
for(;t < size() && !a[t];++t);
if((long long)t * k1 >= m)
return poly();
poly ret;
ret.resize(m);
int u = fpow(a[t],mod - 2),v = fpow(a[t],k2);
for(register int i = 0;i < m - t * k1;++i)
ret.a[i] = (long long)operator[](i + t) * u % mod;
ret = ret.log(m - t * k1);
for(register int i = 0;i < ret.size();++i)
ret.a[i] = (long long)ret[i] * k1 % mod;
ret = ret.exp(m - t * k1),t *= k1,ret.resize(m);
for(register int i = m - 1;i >= t;--i)
ret.a[i] = (long long)ret[i - t] * v % mod;
for(register int i = 0;i < t;++i)
ret.a[i] = 0;
return ret;
}
};
}
using Poly::init;
using Poly::poly;
poly f,g,x;
int ans;
int main()
{
init();
scanf("%d",&n),f.resize(1),f.a[0] = 1,x.resize(2),x.a[1] = 1;
for(register int k = 1;k < n;)
{
k <<= 1,g.resize(k);
for(register int i = 0;i < k;++i)
g.a[i] = 0;
for(register int t = 2;t < k;++t)
for(register int i = 0;i * t + t < k;++i)
g.a[i * t + t] = (g.a[i * t + t] + (long long)f[i] * Poly::inv[t]) % mod;
f = f - ((f.log(k) - (x * f).modxn(k) - g) * (f.inver(k) - x).inver(k)).modxn(k);
}
f.resize(n + 1);
for(register int i = n;~i;--i)
f.a[i] = f[i - 1];
ans = f[n];
for(register int i = (n >> 1) + 1;i < n;++i)
ans = (ans - (long long)f[i] * f[n - i] % mod + mod) % mod;
if(!(n & 1))
ans = (ans - (long long)f[n >> 1] * (f[n >> 1] - 1 + mod) % mod * Poly::inv[2] % mod + mod) % mod;
printf("%d\n",ans);
}