P4091 [HEOI2016/TJOI2016]求和（第二类斯特林数，ntt）

题面:https://www.luogu.org/problem/P4091

题解:[egin{array}{l}
f(n) = sumlimits_{i = 0}^n {sumlimits_{j = 0}^i {{ m{S}}(i,j) cdot {2^{ m{j}}} cdot j!} } \
 = sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {{ m{S}}(i,j) cdot {2^{ m{j}}} cdot j!} }
end{array}]

把公式[S(n,m) = frac{{sumlimits_{i = 0}^m {{{( - 1)}^i}{ m{cdot}}C_m^i} { m{cdot}}{{(m - i)}^n}}}{{m!}}]带进去

则[{ m{ackslash begin{ array} { l} nackslash begin{ array} { *{ 20} { l} } }}f(n) = sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {{ m{S}}(i,j) cdot {2^{ m{j}}} cdot j!}  = sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {sumlimits_{k = 0}^j {frac{{{{( - 1)}^k}}}{{k!}} cdot frac{{{{(j - k)}^i}}}{{(j - k)!}}}  cdot {{ m{2}}^{ m{j}}}{ m{cdotj}}!} } } { m{n{ }} = sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdotj}}!sum { m{\_}}k = 0{ m{^}}j{ m{ { }}sum { m{\_}}i = 0{ m{^}}n{ m{ { }}frac{{{{( - 1)}^k}}}{{k!}} cdot frac{{{{(j - k)}^i}}}{{(j - k)!}} = sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdot}}j!sum { m{\_}}k = 0{ m{^}}j{ m{ { ackslash frac{ { { }}( - 1){ m{^{ ^}}k{ m{} } } } }}k! cdot { m{} }}frac{{sumlimits_{i = 0}^n {{{(j - k)}^i}} }}{{(j - k)!}}{ m{}  }  }  }  } nackslash end{ array} ackslash ackslash  = }}sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdot}}j!sum { m{\_}}k = 0{ m{^}}j{ m{ { ackslash frac{ { { }}( - 1){ m{^{ ^}}k{ m{} } } } }}k! cdot { m{} }}frac{{{{(j - k)}^{n + 1}} - 1}}{{(j - k)! cdot (j - k - 1)}}{ m{} nackslash end{ array} }}]

令[egin{array}{l}
h(x) = frac{{{{( - 1)}^x}}}{{x!}}\
g(x) = frac{{{x^{n + 1}} - 1}}{{x!{ m{cdot(x - 1)}}}}
end{array}]

后面的就是这两个的卷积，直接上ntt就行了

#include<bits/stdc++.h>
#define ms(x) memset(x,0,sizeof(x))
#define sws ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
const int maxn=5e5+5;
const double pi=acos(-1.0);
const ll mod=998244353;///通常情况下的模数,
const ll g=3;///模数的原根998244353,1004535809,469762049
ll qpow(ll a,ll n,ll p){
    ll ans=1;
    while(n){
        if(n&1) ans=ans*a%p;
        n>>=1;
        a=a*a%p;
    }
    return ans;
}
int rev[maxn];
void ntt(ll a[],int n,int len,int pd){
    rev[0]=0;
    for(int i=1;i<n;i++){
        rev[i]=(rev[i>>1]>>1 | ((i&1)<<(len-1)));
        if(i<rev[i]) swap(a[i],a[rev[i]]);
    }
    for(int mid=1;mid<n;mid<<=1){
        ll wn=qpow(g,(mod-1)/(mid*2),mod);///原根代替单位根
        if(pd==-1) wn=qpow(wn,mod-2,mod);///逆变换则改成逆元
        for(int j=0;j<n;j+=2*mid){
            ll w=1;
            for(int k=0;k<mid;k++){
                ll x=a[j+k],y=w*a[j+k+mid]%mod;
                a[j+k]=(x+y)%mod;
                a[j+k+mid]=(x-y+mod)%mod;
                w=w*wn%mod;
            }
        }
    }
    if(pd==-1){
        ll inv=qpow(n,mod-2,mod);
        for(int i=0;i<n;i++){
            a[i]=a[i]*inv%mod;

        }
    }
}
ll a[maxn],b[maxn],c[maxn];
void solve(int n,int m){
    int len=0,up=1;
    while(up<=n+m) up<<=1,len++;
    ntt(a,up,len,1);
    ntt(b,up,len,1);
    for(int i=0;i<up;i++) c[i]=1ll*a[i]*b[i]%mod;
    ntt(c,up,len,-1);
}
ll fa[maxn];
int main(){
    int n,m;
    sws;
    cin>>n;
    fa[0]=1;
    a[0]=1;
    b[0]=1;
    for(int i=1;i<=n;i++){
        fa[i]=1ll*fa[i-1]*i%mod;
        int t=(i&1)==1?-1:1;
        a[i]=(t*qpow(fa[i],mod-2,mod)+mod)%mod;
        if(i==1) b[1]=n+1;
        else {
            b[i]=(qpow(i,n+1,mod)-1+mod)%mod*qpow(1ll*(i-1)*fa[i]%mod,mod-2,mod)%mod;
        }
    }
    solve(n,n);
    ll ans=0;
    for(ll i=0;i<=n;i++){
        ans=(ans+qpow(2,i,mod)*fa[i]%mod*c[i]%mod)%mod;
    }
    cout<<ans<<endl;
}