Comet OJ 1091 Permutation

析合树真是好东西（

首先特判一下 n≤2 的情况。

设 H(x) 表示排列的 OGF（此处令 0!=0），G(x) 表示根结点为合点且儿子序列为升序的排列个数。
那么有 G(x)=∞∑i=2(H(x)−G(x))i

因此有 G(x)=H2(x)H(x)+1

然后设 F(x) 表示答案，那么易知根结点为析点的排列个数即为 ∞∑i=4fiHi(x)=F(H(x))

那么有 F(H(x))+2G(x)+x=H(x)

设 P(x)=H(x)−2G(x)−x，那么有 F(H(x))=P(x)

代入 H−1(x) 可得 F(x)=P(H−1(x))=x−2x21+x−H−1(x)

根据扩展拉格朗日反演，有 [xn]H−1(x)=1n[xn−1](xH(x))n

直接多项式快速幂即可。
复杂度 O(nlogn)。

代码：

#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;
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;
poly H;
int ans;
int main()
{
    init();
    scanf("%d",&n);
    if(n <= 2)
    {
        printf("%d\n",n);
        return 0;
    }
    H.resize(n);
    for(register int i = 0;i < H.size();++i)
        H.a[i] = Poly::fac[i + 1];
    H = H.pow(n,mod - n,mod - 1 - n);
    ans = (long long)H[n - 1] * (mod - Poly::inv[n]) % mod,
    ans = n & 1 ? add(ans,2) : dec(ans,2);
    printf("%d\n",ans);
}