Codeforces 1349F2 Slime and Sequences (Hard Version)

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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。
令 \[ S_k = \sum\limits_{p\in s_n} f_p(k+1) \]

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 \[ \begin{aligned} W_k &= \sum\limits_{i=k}^n [x^i] (\e^x-1)^{i-k} \\ &= [x^k] \sum\limits_{i=k}^n \left(\frac{\e^x-1}x\right)^{i-k} \\ &= [x^k] \sum\limits_{i=0}^{n-k} \left(\frac{\e^x-1}x\right)^i \end{aligned} \]

令 \(F(x) = \frac{\e^x-1}x\)，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。

令 \(G(x)=xF(x)=\e^x-1\)，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 \(x\) 对 \(H(x)\) 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]);
}