集合划分计数 多项式EXP模板

题链

定义贝尔数$B_n=sum _{k=1}{n}S(n,k)$，其中$S(n,k)$为第二类斯特林数，我们有$B_{n+1}=sum_{i=0}^{n}B_i* binom{n}{i}$,

    ​ $frac{B_{n+1}}{n!}=sum_{i=1}^nfrac{1}{i!}*frac{B_{n-i}}{(n-i)!}$

同乘 $x^n$

​     $frac{B_{n+1}}{n!}x^n=sum_{i=1}^nfrac{x^i}{i!}*frac{B_{n-i}x^{n-i}}{(n-i)!}$​

    $frac{dfrac{B{n+1}}{(n+1)!}x^{n+1}}{dx}=sum_{i=1}^nfrac{xi}{i!}*frac{B_{n-i}x^{n-i}}{(n-i)!}$

考虑指数型生成函数$F(x)=sum_{x=0}^{infty}frac{B_i}{i!}$

有$frac {dF}{dx}=e^x$,$ln|F|=e^x+C$,将$x=1$带入得$F=e^{e^x-1}$，这样这道题就做完了。

接下来是多项式EXP复习时间：

   1. 多项式逆：考虑倍增，当前有$b(x)(mod x^{n})$，求 $B(x)(mod x^{2n})$，

             注意到$0=(B(x)-b(x))^2*A(x) (mod x^{2n})$

             整理有$B(x)=b(x)(2-A(x)) (mod x^{2n})$

   2.多项式ln：注意到$frac{d(ln(A(x))}{dx}=frac{A'(x)}{A(x)}$

               $ln(A(x))=int frac{A'(x)}{A(x)}dx$

  3.多项式exp: 考虑牛顿迭代(下图by miskcoo)

    

我们关心$F(x)=e^{A(x)}$，带入易得$F(x)=F_0(x)(1-lnF_0(x)+A(x))(mod x^n)$

代码：

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define getchar nc
#define sight(c) ('0'<=c&&c<='9')
#define swap(a,b) a^=b,b^=a,a^=b
#define LL long long
#define debug(a) cout<<#a<<" is "<<a<<"
"
#define dput(a) puts("a")
#define eho(x) for(int i=head[x];i;i=net[i])
#define fi first
#define se second
#define mo 998244353
inline char nc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template <class T>
inline void read(T &x){// unsigned
    static char c;
    for (c=getchar();!sight(c);c=getchar());
    for (x=0;sight(c);c=getchar())x=x*10+c-48;
}
template <class T> void write(T x){if (x<10) {putchar('0'+x); return;} write(x/10); putchar('0'+x%10);}
template <class T> inline void writeln(T x){ if (x<0) putchar('-'),x*=-1; write(x); putchar('
'); }
template <class T> inline void writel(T x){ if (x<0) putchar('-'),x*=-1; write(x); putchar(' '); }
using namespace std;
LL qsm(LL x,LL y=mo-2){
    static LL anw;
    for (anw=1,x=x%mo;y;y>>=1,x=x*x%mo) if (y&1) anw=anw*x%mo;
    return anw;
}
#define N 1000005
#define G 3
int rev[N];LL a[N],b[N]; 
void NTT(LL *a,int ln,int op){
    int lim=1<<ln;
    for (int i=0;i<lim;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<ln-1);
    for (int i=0;i<lim;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=1;i<lim;i<<=1) {
        LL wn=qsm(G,(mo-1)/(i<<1));
        if (op==-1) wn=qsm(wn);
        for (int j=0;j<lim;j+=i<<1) {
            LL w=1,X,Y;
            for (int k=j;k<i+j;k++,w=w*wn%mo) {
                X=a[k]; Y=w*a[k+i]%mo;
                a[k]=(X+Y)%mo; a[k+i]=(X-Y+mo)%mo;
            }
        }
    }
    if (op==-1) {
        LL sn=qsm(lim);
        for (int i=0;i<lim;i++) a[i]=a[i]*sn%mo;
    }
}
void mul(LL* a,LL *b,LL *c,int ln){//input array of len 1<<ln, return 1<<ln 
    static LL A[N],B[N];
    int lim=(1<<ln);
    for (int i=0;i<lim;i++) A[i]=a[i],B[i]=b[i];
    for (int i=lim;i<2*lim;i++) A[i]=B[i]=0;
    NTT(A,ln+1,1); NTT(B,ln+1,1);
    for (int i=0;i<2*lim;i++)
    c[i]=A[i]*B[i]%mo;
    NTT(c,ln+1,-1); 
}
void inv(LL *a,LL *b,int LG){
    static LL C[N];
    b[0]=qsm(a[0]);
    for (int k=1;k<=LG;k++) {
        mul(b,a,C,k);
        for (int i=0;i<(1<<k);i++) C[i]=C[i]?mo-C[i]:0;
        C[0]=(C[0]+2)%mo;
        mul(C,b,b,k);
    }
}
void Dif(LL *a,LL *b,int lim){ //传lim而不是LG 
    for (int i=0;i<lim;i++) b[i]=a[i+1]*(i+1)%mo;
    b[lim-1]=0; 
}
void Int(LL *a,LL *b,int lim){ //复杂度O(nlogn) 
    for (int i=lim-1;i;i--) b[i]=a[i-1]*qsm(i)%mo;
    b[0]=0; 
}
void Ln(LL *a,LL *b,int LG){
    static LL C[N];
    inv(a,b,LG);
    Dif(a,C,1<<LG);
    mul(b,C,b,LG);
    Int(b,b,1<<LG);
}
void exp(LL *a,LL *b,int LG){
    static LL D[N];
    b[0]=1;
    for (int k=1;k<=LG;k++) {
        Ln(b,D,k);
        for (int i=0;i<(1<<(k));i++) D[i]=(a[i]-D[i]+mo)%mo;
        D[0]=(D[0]+1)%mo;
        mul(b,D,b,k); 
    }
}
int n,T,X; LL fac[N];
signed main() {
    #ifdef LOCAL
    //    freopen("test.in", "r", stdin);
    //    freopen("test.out", "w", stdout);
    #endif
    a[1]=1; 
    exp(a,b,17);
    b[0]=0;
    exp(b,a,17);
    fac[0]=1; for (int i=1;i<N;i++) fac[i]=fac[i-1]*i%mo;
    read(T);
    while(T--) {
        read(X); writeln(a[X]*fac[X]%mo);
    }
    return 0;
}
/*
3
1 3 3 1
1 3 3 1

*/