LibreOJ 6363.「地底蔷薇」

拉格朗日反演基本操作的一点改造?(

\(H(x)\) 为有标号有根无向连通图个数的 EGF。不会求这个的建议出门左转城市规划。

然后设 \(b_i\) 表示 \(i+1\) 个点的有标号点双个数,设 \(B(x)\) 是它的 EGF。
考虑一个有根连通图,删去其根应当形成若干连通块,而每个连通块中都有一部分点与根本在同一个点双。
考虑删去根再删去所有包含根的点双的边,那么这些点双被拆散以后会剩下一些以它们为根的连通图。
于是 \[ H(x) = x{\rm e}^{B(H(x))} \]

考虑求出 \(G(x) = \sum\limits_{i+1\in S} \frac{b_i x^i}{i!}\)
可以试着对这个形式构造复合逆 \[ H^{-1}(x) = \frac{x}{ {\rm e}^{B(x)} } \]

反过来用复合逆表示 \(B\) \[ B(x) = \ln \frac x{H^{-1}(x)} \]

构造 \[ A(x) = \ln \frac{H(x)}x \]

根据扩展拉格朗日反演有 \[ [x^n] B(x) = [x^n] A(H^{-1}(x)) = \frac1n [x^{n-1}] A'(x) \left(\frac x{H(x)}\right)^n \]

有了 \(G\) 后,设 \(F\) 表示答案的 EGF,那么同理可得 \[ F(x) = x{\rm e}^{G(F(x))} \]

从而 \[ F^{-1}(x) = \frac x{ {\rm e}^{G(x)} } \\ [x^n] F(x) = \frac 1n[x^{n-1}] \left(\frac x{F^{-1}(x)}\right)^n = \frac 1n[x^{n-1}] {\rm e}^{nG(x)} \]

复杂度 \(O(n \log n)\)(视 \(n,\sum\limits_{i\in S}i\) 同阶)。

代码:

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#include <cstdio>
#include <vector>
#include <cstring>
#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 = 1e5;
const int mod = 998244353;
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;
}
int n,m;
int a[N + 5];
namespace Poly
{
const int N = 1 << 18;
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 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 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 G,H,A;
int ans;
int main()
{
init();
scanf("%d%d",&n,&m);
H.resize(n + 2),G.resize(n + 1);
for(register int i = 0;i < H.size();++i)
H.a[i] = (long long)fpow(2,(long long)i * (i - 1) / 2 % (mod - 1)) * Poly::ifac[i] % mod;
H = H.log(n + 2);
for(register int i = 0;i < H.size();++i)
H.a[i] = (long long)H[i] * i % mod;
A.resize(n + 1);
for(register int i = 0;i < A.size();++i)
A.a[i] = H[i + 1];
A = (A.deriv() * (H = A.inver(n + 1))).modxn(n + 1);
for(register int i = 1;i <= m;++i)
{
scanf("%d",a + i);
if(a[i] > 1 && a[i] <= n)
--a[i],
G.a[a[i]] = (long long)Poly::inv[a[i]] * (A.modxn(a[i]) * H.pow(a[i],a[i]))[a[i] - 1] % mod;
}
ans = (long long)Poly::inv[n] * (n * G).exp(n)[n - 1] % mod * Poly::fac[n - 1] % mod;
printf("%d\n",ans);
}