传送门
游戏?
显然我们要求的是
考虑求
构建的生成函数
先不看
后面就是一个等比数列
就是
分治即可
然后每一项乘一个就把弄回来了
#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define poly vector<int>
#define bg begin
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(ll &a,ll b){a<b?a=b:0;}
inline void chemn(ll &a,ll b){a>b?a=b:0;}
cs int N=(1<<20|1),C=20;
poly w[C+1];
inline void init_w(){
for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
int wn=ksm(G,(mod-1)/(1<<C));
w[C][0]=1;
for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn);
for(int i=C-1;i;i--)
for(int j=0;j<(1<<(i-1));j++)w[i][j]=w[i+1][j<<1];
}
int rev[N<<2],inv[N<<2],fac[N<<2],ifac[N<<2];
inline void init_rev(int lim){
for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void init_inv(int len=N*4-5){
inv[0]=inv[1]=1;
for(int i=2;i<=len;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
ifac[0]=fac[0]=1;
for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
ifac[len]=ksm(fac[len],mod-2);
for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline void ntt(poly &f,int lim,int kd){
for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++)
for(int i=0;i<lim;i+=(mid<<1))
for(int j=0;j<mid;j++)
a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]),
f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
if(kd==-1){
reverse(f.bg()+1,f.bg()+lim);
for(int i=0;i<lim;i++)Mul(f[i],inv[lim]);
}
}
inline poly operator *(poly a,poly b){
int deg=a.size()+b.size()-1,lim=1;
if(deg<=128){
poly c(deg,0);
for(int i=0;i<a.size();i++)
for(int j=0;j<b.size();j++)
Add(c[i+j],mul(a[i],b[j]));
return c;
}
while(lim<deg)lim<<=1;init_rev(lim);
a.resize(lim),ntt(a,lim,1);
b.resize(lim),ntt(b,lim,1);
for(int i=0;i<lim;i++)Mul(a[i],b[i]);
ntt(a,lim,-1),a.resize(deg);
return a;
}
inline poly deriv(poly a){
for(int i=0;i<a.size()-1;i++)a[i]=mul(a[i+1],i+1);
a.pop_back();return a;
}
inline poly integ(poly a){
a.pb(0);
for(int i=a.size()-1;i;i--)a[i]=mul(a[i-1],inv[i]);
a[0]=0;
return a;
}
inline poly Inv(poly a,int deg){
poly c,b(1,ksm(a[0],mod-2));
for(int lim=4;lim<(deg<<2);lim<<=1){
init_rev(lim);
c=a,c.resize(lim>>1);
c.resize(lim),ntt(c,lim,1);
b.resize(lim),ntt(b,lim,1);
for(int i=0;i<lim;i++)Mul(b[i],dec(2,mul(b[i],c[i])));
ntt(b,lim,-1),b.resize(lim>>1);
}b.resize(deg);return b;
}
inline poly Ln(poly a,int lim){
a=integ(deriv(a)*Inv(a,lim)),a.resize(lim);
return a;
}
#define lc (u<<1)
#define rc ((u<<1)|1)
#define mid ((l+r)>>1)
poly f[N<<2];
inline void build(int u,int l,int r,int *v){
if(l==r){f[u].clear();f[u].pb(1),f[u].pb(mod-v[l]);return;}
build(lc,l,mid,v),build(rc,mid+1,r,v);
f[u]=f[lc]*f[rc];
}
#undef lc
#undef rc
#undef mid
int t;
inline poly Build(int n,int *v){
build(1,1,n,v);
poly res=deriv(Ln(f[1],t+1));
res.pb(0);
for(int i=res.size()-1;i;i--)res[i]=dec(0,res[i-1]);
res[0]=n;
for(int i=1;i<res.size();i++)Mul(res[i],ifac[i]);
return res;
}
int n,m;
int a[N],b[N];
int main(){
n=read(),m=read(),init_w(),init_inv();
for(int i=1;i<=n;i++)a[i]=read();
for(int i=1;i<=m;i++)b[i]=read();
t=read();
poly F=Build(n,a),G=Build(m,b);
F=F*G;int pt=mul(ksm(n,mod-2),ksm(m,mod-2));
for(int i=1;i<=t;i++)cout<<mul(F[i],mul(pt,fac[i]))<<'
';
}