BZOJ 4555: [Tjoi2016&Heoi2016]求和 第二类斯特林数+NTT

code:

#include <bits/stdc++.h>   
#define G 3  
#define N 200001     
#define ll long long   
#define mod 998244353       
#define setIO(s) freopen(s".in","r",stdin)   
using namespace std;          
ll qpow(ll x,ll y) 
{
    ll tmp=1ll; 
    while(y)  
    { 
        if(y&1)  tmp=tmp*x%mod; 
        x=x*x%mod, y>>=1;  
    }  
    return tmp;   
}   
ll Inv(ll x)  { return qpow(x,mod-2);  }   
void NTT(ll *a,int len,int op) 
{
    int i,j,mid,k;   
    for(i=k=0;i<len;++i) 
    {
        if(i>k)   swap(a[i],a[k]);   
        for(j=len>>1;(k^=j)<j;j>>=1);   
    }    
    for(mid=1;mid<len;mid<<=1) 
    {
        ll wn=qpow(G,(mod-1)/(mid<<1)); 
        if(op==-1)   wn=Inv(wn);  
        for(i=0;i<len;i+=mid<<1) 
        { 
            ll w=1ll;    
            for(j=0;j<mid;++j) 
            {
                ll x=a[i+j],y=w*a[i+j+mid]%mod;     
                a[i+j]=(x+y)%mod, a[i+j+mid]=(x-y+mod)%mod;     
                w=w*wn%mod;     
            }
        }
    } 
    if(op==-1) 
    {
        ll rev=Inv(len); 
        for(i=0;i<len;++i)   a[i]=a[i]*rev%mod;         
    }
}       
ll fac[N],inv[N],f[N<<1],g[N<<1];        
int main() 
{ 
    // setIO("input");  
    int i,j,n,limit;   
    scanf("%d",&n);        
    fac[0]=inv[0]=1ll; 
    for(i=1;i<N;++i)    fac[i]=fac[i-1]*1ll*i%mod,   inv[i]=Inv(fac[i]);   
    f[0]=1ll, g[0]=1ll, g[1]=n+1;    
    for(i=1;i<=n;++i)   f[i]=(qpow(-1,i)*inv[i]+mod)%mod;   
    for(i=2;i<=n;++i)   g[i]=(qpow(i,n+1)-1+mod)%mod*inv[i]%mod*Inv(i-1)%mod;       
    for(limit=1;limit<=2*n;limit<<=1);          
    NTT(f,limit,1),  NTT(g,limit,1);                   
    for(i=0;i<limit;++i)    f[i]=f[i]*g[i]%mod;   
    NTT(f,limit,-1);    
    ll ans=0ll;   
    for(i=0;i<=n;++i)    (ans+=qpow(2,i)*fac[i]%mod*f[i]%mod)%=mod;       
    printf("%lld\n",ans);   
    return 0; 
}