Codeforces 1349F2.Slime and Sequences (Hard Version)

由于心情不好,做了一晚上的 dirty works……

首先是神仙映射。
打个暴力会发现 \(|s_n| = n!\),这意味着一个好序列很可能存在着某种精妙的与排列的映射。
结论:对于 \(n\) 阶排列 \(a\),存在好序列 \(p\),满足: \[ \newcommand{\euler}[2]{ \genfrac{\langle}{\rangle}{0pt}{}{ #1 }{ #2 } } \def\e{ {\rm e} } p_{a_i} = 1+\sum\limits_{j=1}^{i - 1} [a_j < a_{j+1}] \]

\(p\) 为好序列是显然的。
通过察看 \(p_n\dots p_1\) 可以确定 \(n\dots1\) 在原排列 \(a\) 中的的位置,显然这也是唯一对应的。

考虑欧拉数 \(\euler nk\) 表示长度为 \(n\) 的排列,含有 \(k\) 个「升高」的排列个数。
\[ S_k = \sum\limits_{p\in s_n} f_p(k+1) \]

则枚举 \(k+1\) 在原排列中产生的位置,显然 \[ S_k = \sum\limits_{i=\max(1,k)}^n \euler ik \binom ni (n-i)! \]

考虑求出长度为 \(n\) 的序列中钦定 \(k\) 个「升高」的方案数,二项式反演即可得到 \(\euler nk\)
那么考虑到钦定 \(k\) 个「升高」会出现 \(n-k\) 个钦定的「升高」段,升高段内的 EGF 是显然的,因此 \[ \euler nk = n! \sum\limits_{i=\max(1,k)}^n (-1)^{i-k} \binom ik [x^n] (\e^x-1)^{n-i} \]

继续一波代入 \[ \begin{aligned} S_k &= \sum\limits_{i=\max(1,k)}^n \euler ik \binom ni (n-i)! \\ &= \sum\limits_{i=\max(1,k)}^n \binom ni (n-i)! i! \sum\limits_{j=k}^i (-1)^{j-k} \binom jk [x^i] (\e^x-1)^{i-j} \\ &= \frac{n!}{k!} \sum\limits_{j=k}^n \frac{(-1)^{j-k}j!}{(j-k)!} \sum\limits_{i=\max(1,j)}^n [x^i](\e^x-1)^{i-j} \end{aligned} \]

看出了卷积形式,令 \[ W_k = \sum\limits_{i=k}^n [x^i] (\e^x-1)^{i-k} \]

处理一下需要提取系数的项 \[ \begin{aligned} W_k &= \sum\limits_{i=k}^n [x^i] (\e^x-1)^{i-k} \\ &= [x^k] \sum\limits_{i=k}^n \left(\frac{\e^x-1}x\right)^{i-k} \\ &= [x^k] \sum\limits_{i=0}^{n-k} \left(\frac{\e^x-1}x\right)^i \end{aligned} \]

\(F(x) = \frac{\e^x-1}x\),则 \[ W_k = [x^k] \frac{1-F^{n-k+1}(x)}{1-F(x)} = [x^k] \frac1{1-F(x)} - [x^k] \frac{F^{n-k+1}(x)}{1-F(x)} \]

前一项考虑如此处理求逆 \[ [x^k] \frac1{1-F(x)} = [x^{k+1}] \frac x{1-F(x)} \]

后一项先推导亿下 \[ [x^k] \frac{F^{n-k+1}(x)}{1-F(x)} = [x^{n+1}] \frac{ (xF(x))^{n-k+1} }{1-F(x)} \]

鏼爷的论文启发我们使用二元生成函数 \[ [x^{n+1}] \frac{ (xF(x))^{n-k+1} }{1-F(x)} = [x^{n+1}w^{n-k+1}] \frac1{(1-F(x))(1-wxF(x))} \]

那么到此我们将求 \(W_k\) 转为求以上生成函数的 \(x^{n+1}\) 项系数。
尝试探求扩展拉格朗日反演的科技,考虑构造某两个函数的复合 \(H(G(x))\) 恰为以上生成函数,且 \(G(x)\) 的复合逆容易求得。

\(G(x)=xF(x)=\e^x-1\),为了构造 \(H\) 同时需要构造 \(P(G(x))=F(x)\)
显然 \[ P(G(x)) = F(x) \iff P(xF(x)) = F(x) \iff P(x) = \frac{x}{G^{-1}(x)} = \frac x{\ln(1+x)} \]

注意指数上的 \(-1\) 表示函数幂。

从而有 \[ H(x) = \frac1{(1-P(x))(1-wx)} \]

根据扩展拉格朗日反演有 \[ [x^{n+1}] H(G(x)) = \frac1{n+1} [x^n] H'(x) \left(\frac x{G^{-1}(x)}\right)^{n+1} = \frac1{n+1} [x^n] H'(x) P^{n+1}(x) \]


好的那么先来一波喜闻乐见的 dirty works:关于 \(x\)\(H(x)\) 求导: \[ \begin{aligned} H'(x) &= \left(\frac1{(1-P(x))(1-wx)}\right)' \\ &= \left(\frac1{1-P(x)}\right)'\frac1{1-wx}+\frac1{1-P(x)}\left(\frac1{1-wx}\right)' \\ &= \frac{P'(x)}{(1-P(x))^2(1-wx)}+\frac w{(1-P(x))(1-wx)^2} \end{aligned} \]

不用再推下去了,因为接下来只需要这样的形式,刚刚好。

还在本蒟蒻可接受范围之内(


好的那么接下来又是一波 dirty works:xjb 代入 \[ \begin{aligned}[] [x^{n+1}] H(G(x)) &= \frac1{n+1} [x^n] H'(x)P^{n+1}(x) \\ &= \frac1{n+1} [x^n] \left(\frac{P'(x)}{(1-P(x))^2(1-wx)}+\frac w{(1-P(x))(1-wx)^2}\right)P^{n+1}(x) \\ &= \frac1{n+1} [x^n] \left(\frac{P'(x)}{(1-P(x))^2}\sum\limits_{i\ge 0} w^ix^i + \frac1{1-P(x)}\sum\limits_{i\ge 0} (i+1)w^{i+1}x^i\right)P^{n+1}(x) \end{aligned} \]

提取你的系数,令 \(i=n-k+1\)\[ \begin{aligned}[] [w^i x^n] H'(x) P^{n+1}(x) &= [w^i x^n] \left(\frac{P'(x)}{(1-P(x))^2}\sum\limits_{i\ge 0} w^ix^i + \frac1{1-P(x)}\sum\limits_{i\ge 0} (i+1)w^{i+1}x^i\right)P^{n+1}(x) \\ &= [x^n] \left(\frac{P'(x)x^i}{(1-P(x))^2} + \frac{ ix^{i-1} }{1-P(x)}\right)P^{n+1}(x) \\ &= [x^{n-i+2}] \frac{x^2P'(x)P^{n+1}(x)}{(1-P(x))^2}+[x^{n-i+2}] \frac{ixP^{n+1}(x)}{1-P(x)} \\ &= [x^{k+1}] \frac{x^2P'(x)P^{n+1}(x)}{(1-P(x))^2}+[x^{k+1}] \frac{(n-k+1)xP^{n+1}(x)}{1-P(x)} \end{aligned} \]

这东西推出式子之后实现也是很容易出锅的……建议好好整理一下流程再写。

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 = 1e5;
const int mod = 998244353;
int n;
int w[N + 5],s[N + 5];
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 << 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 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;
inline int C(int n,int m)
{
return n < m ? 0 : (long long)Poly::fac[n] * Poly::ifac[m] % mod * Poly::ifac[n - m] % mod;
}
poly f,p,pd,p2,pw,t;
poly T1,T2;
int main()
{
init();
scanf("%d",&n),f.resize(n + 2);
for(register int i = 0;i < f.size();++i)
f.a[i] = dec(0,Poly::ifac[i + 2]);
f = f.inver(n + 2);
for(register int i = 0;i <= n;++i)
w[i] = f[i + 1];
p.resize(n + 3);
for(register int i = 0;i < p.size();++i)
p.a[i] = (long long)(i & 1 ? mod - 1 : 1) * Poly::inv[i + 1] % mod;
p = p.inver(n + 3),pd = p.deriv(),pw = p.pow(n + 2,n + 1);
for(register int i = 0;i <= n + 1;++i)
p.a[i] = dec(0,p[i + 1]);
p = p.inver(n + 2),p2 = (p * p).modxn(n + 2);
t = ((p2 * pd).modxn(n + 2) * pw).modxn(n + 2);
for(register int i = 0;i <= n;++i)
w[i] = (w[i] - (long long)t[i + 1] * Poly::inv[n + 1] % mod + mod) % mod;
t = (pw * p).modxn(n + 2);
for(register int i = 0;i <= n;++i)
w[i] = (w[i] - (long long)t[i + 1] * (n - i + 1) % mod * Poly::inv[n + 1] % mod + mod) % mod;
w[0] = dec(w[0],1),T1.resize(n + 1),T2.resize(n + 1);
for(register int i = 0;i <= n;++i)
T1.a[i] = (long long)(i & 1 ? mod - 1 : 1) * Poly::ifac[i] % mod,
T2.a[n - i] = (long long)w[i] * Poly::fac[i] % mod;
T1 = (T1 * T2).modxn(n + 1);
for(register int i = 0;i <= n;++i)
s[i] = (long long)Poly::fac[n] * Poly::ifac[i] % mod * T1[n - i] % mod;
for(register int i = 0;i < n;++i)
printf("%d%c",s[i]," \n"[i == n - 1]);
}