神tm題目，這幾天時間都耗在這道題目上了。

題意：一個森林的權值定義為其樹個數的 k 次方，求所有 n 個點的帶標號無根
森林的權值和。
對 998244353 取模。
n ≤ 20000 , k ≤ 10

寫完之後神清氣爽~（註意最後輸出的答案是B_k(x)的n次項系數乘上n!）

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ll long long
#define P 998244353
using namespace std;
const int maxn=100010;
const int g=3;
inline int read(){
    
 
int f=1,x=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘)f=-1;ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘){x=(x<<3)+(x<<1)+ch-‘0‘;ch=getchar();}
    return f*x;
}
 
inline int qpow(int x,int p){int ans=1;
    for(;p;p>>=1,x=1ll*x*x%P)if(p&1)ans=1ll*ans*x%P;
    
 
return ans;
}
int n,m,k;
int a[maxn],A[maxn],B[maxn],C[maxn],c[maxn],d[maxn],fac[maxn],tmp[maxn],s[11][maxn];
 
inline void NTT(int *a,int f,int k){  
    for (int i=0,j=0;i<k;i++){
        if (i>j)swap(a[i],a[j]);
        for (int l=k>>1;(j^=l)<l;l>>=1);
    }
    for (int i=1;i<k;i<<=1
 
){
        int w=qpow(g,(f*(P-1)/(i<<1)+P-1)%(P-1));
        for (int j=0;j<k;j+=(i<<1)){
            int e=1;
            for (int k=0;k<i;++k,e=1ll*e*w%P){
                int x,y;
                x=a[j+k];y=1ll*a[j+k+i]*e%P;
                a[j+k]=(x+y)%P;a[j+k+i]=(x-y+P)%P;        
            }
        }
    }
    if(f==-1){
        int _inv=qpow(k,P-2);
        for (int i=0;i<k;++i)a[i]=1ll*a[i]*_inv%P;
    }
}

inline void GetInv(int *a,int *b,int n){
      if (n==1) return void(b[0]=qpow(a[0],P-2));
      GetInv(a,b,n>>1);
      for (int i=0;i<n;++i)tmp[i]=a[i],tmp[n+i]=0;
      int k;for (k=1;k<=n;k<<=1);
      for (int i=n;i<k;++i)b[i]=0;
      NTT(tmp,1,k);NTT(b,1,k);
      for (int i=0;i<k;++i)
        tmp[i]=b[i]*(2+P-1ll*tmp[i]*b[i]%P)%P;
      NTT(tmp,-1,k);
      for (int i=0;i<n;++i)b[i]=tmp[i],b[n+i]=0;
}

inline void Getln(int *a,int *b,int *c,int n){
    for (int i=0;i<n;++i)b[i]=1ll*(i+1)*a[i+1]%P;
    GetInv(a,c,n);
    for (int i=0;i<n;++i)tmp[i]=c[i],tmp[n+i]=0;
    int k;for (k=1;k<=n;k<<=1);
    NTT(tmp,1,k);NTT(b,1,k);
    for (int i=0;i<k;++i)
        tmp[i]=1ll*b[i]*tmp[i]%P;
    NTT(tmp,-1,k);
    for (int i=1;i<=n;++i)b[i]=1ll*tmp[i-1]*qpow(i,P-2)%P,b[i+n]=0;b[0]=0;
}

inline void Getexp(int *a,int *b,int *c,int *d,int n){
    if (n==1)return void(b[0]=1);
    Getexp(a,b,c,d,n>>1);
    Getln(b,c,d,n);
    for (int i=0;i<n;++i)tmp[i]=(a[i]-c[i]+P)%P,tmp[n+i]=0;++tmp[0];
    int k;for (k=1;k<=n;k<<=1);
    NTT(tmp,1,k);NTT(b,1,k);
    for (int i=0;i<k;++i)
        tmp[i]=(ll)tmp[i]*b[i]%P;
    NTT(tmp,-1,k);
    for (int i=0;i<n;++i)b[i]=tmp[i],b[i+n]=0;
}

int main(){
    a[1]=s[0][0]=fac[0]=1;
    for (int i=1;i<=10;++i){
        for (int j=1;j<=i;++j)s[i][j]=(s[i-1][j-1]+1ll*s[i-1][j]*j%P)%P;
    }
    n=read();k=read();
    for (int i=2;i<=n;++i)a[i]=qpow(i,i-2);
    for (int i=1;i<=n;++i)fac[i]=1ll*fac[i-1]*i%P;
    for (int i=1;i<=n;++i)A[i]=1ll*a[i]*qpow(fac[i],P-2)%P;
    for (m=1;m<=n;m<<=1);
    Getexp(A,B,c,d,m);
    memset(tmp,0,sizeof(tmp));
    for (int i=0;i<=n;++i)tmp[i]=A[i];
    for (m=1;m<=2*n;m<<=1);
    NTT(tmp,1,m);
    for (int i=1;i<=k;++i){
        for (int j=0;j<=n;++j)C[j]=(C[j]+1ll*s[k][i]*A[j]%P)%P,A[j+n+1]=0;
        NTT(A,1,m);
        for (int j=0;j<=m;++j)A[j]=1ll*A[j]*tmp[j]%P;
        NTT(A,-1,m);
    }
    if (k){
        for (int i=n+1;i<=m;++i)B[i]=C[i]=0;
        NTT(B,1,m);NTT(C,1,m);
        for (int i=0;i<=m;++i)B[i]=1ll*B[i]*C[i]%P;
        NTT(B,-1,m);
    }B[n]=1ll*B[n]*fac[n]%P;
    printf("%d\n",B[n]);
}

玲瓏杯 1035 D-J