BZOJ4555 HEOI2016/TJOI2016求和（NTT+斯特林数）

　　S(i,j)=Σ(-1)(1/j!)·C(j,k)·k=Σ(-1)·k/k!/(j-k)!。原式=ΣΣ(-1)·k·2·j!/k!/(j-k)! (i,j=0~n)。可以发现i只在式中出现了一次且与j不相关，如果对每个k求出其剩余部分的答案，各自乘一下即可。而剩余部分显然是一个卷积。

#include<bits/stdc++.h>

using namespace std;

int getbit(){char c=getchar();while (c<''||c>'') c=getchar();return c^;}

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

#define N 100010

#define P 998244353

#define inv3 332748118

int n,r[N<<],fac[N],inv[N],f[N<<],g[N<<];

int ksm(int a,int k)

{

    int s=;

    for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;

    return s;

}

int C(int n,int m){if (m>n) return ;return 1ll*fac[n]*inv[m]%P*inv[n-m]%P;}

void DFT(int *a,int n,int g)

{

    for (int i=;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);

    for (int i=;i<=n;i<<=)

    {

        int wn=ksm(g,(P-)/i);

        for (int j=;j<n;j+=i)

        {

            int w=;

            for (int k=j;k<j+(i>>);k++,w=1ll*w*wn%P)

            {

                int x=a[k],y=1ll*w*a[k+(i>>)]%P;

                a[k]=(x+y)%P,a[k+(i>>)]=(x-y+P)%P;

            }

        }

    }

}

void mul(int *f,int *g)

{

    int t=;while (t<=(n<<)) t<<=;

    for (int i=;i<t;i++) r[i]=(r[i>>]>>)|(i&)*(t>>);

    DFT(f,t,),DFT(g,t,);

    for (int i=;i<t;i++) f[i]=1ll*f[i]*g[i]%P;

    DFT(f,t,inv3);

    int u=ksm(t,P-);

    for (int i=;i<t;i++) f[i]=1ll*f[i]*u%P;

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("b.in","r",stdin);

    freopen("b.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    n=read();

    fac[]=;for (int i=;i<=n;i++) fac[i]=1ll*fac[i-]*i%P;

    inv[]=inv[]=;for (int i=;i<=n;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;

    for (int i=;i<=n;i++) inv[i]=1ll*inv[i]*inv[i-]%P;

    for (int i=;i<=n;i++) g[i]=1ll*ksm(,i)*fac[i]%P;reverse(g,g+n+);

    for (int i=;i<=n;i++) if (i&) f[i]=P-inv[i];else f[i]=inv[i];

    mul(f,g);

    reverse(f,f+n+);

    for (int i=;i<=n;i++) f[i]=1ll*f[i]*inv[i]%P;

    f[]=1ll*f[]*(n+)%P;

    for (int k=;k<=n;k++)

    f[k]=1ll*f[k]*(ksm(k,n+)-)%P*ksm(k-,P-)%P;

    int ans=;for (int k=;k<=n;k++) ans=(ans+f[k])%P;

    cout<<ans;

    return ;

}