https://www.luogu.com.cn/problem/P4705
多项式毒瘤题...
由于是随机选的，所以每次选到(a_i)和(b_j)的概率就是(frac{1}{nm})，那么期望就是
[Ans_k=frac{1}{nm}sumlimits_{i=1}^nsumlimits_{j=1}^m (a_i+b_j)^k
]
先不管(frac{1}{nm})，我们要求(sum_{i=1}^nsum_{j=1}^m(a_i+b_j)^k)，大力二项式定理
[egin{aligned}Ans_k imes nm &=sumlimits_{i=1}^nsumlimits_{j=1}^m sumlimits_{t=0}^k inom{k}{t}a_i^{t}b_j^{k-t}\&=sumlimits_{t=0}^kfrac{k!}{t!(k-t)!}sumlimits_{i=1}^na_i^{t}sumlimits_{j=1}^ma_j^{k-t}\&=k!sumlimits_{t=0}^k left(sumlimits_{i=1}^nfrac{a_i^t}{t!}
ight)left(sumlimits_{j=1}^mfrac{b_j^{k-t}}{(k-t)!}
ight)end{aligned}
]
令(A_k=frac{sum_{i=1}^na_i^k}{k!})，(B_k=frac{sum_{i=1}^m b_i^k}{k!})，那么
[Ans_k imes nm=k!sumlimits_{i=0}^k A_iB_{k-i}
]
如果能快速求出(A,B)，卷积就好了，考虑求这个数列
(f_k=sumlimits_{i=1}^n a_i^k)
如果能求出(f)的话，用(A_i=frac{f_i}{i!})算(A,B)就好了。
构造其生成函数(F(x)=sumlimits_{i=0}^{infty}f_ix^i=sumlimits_{i=0}^{infty}x^isumlimits_{j=1}^n a_j^i=sumlimits_{j=1}^nsumlimits_{i=0}^{infty}a_j^ix^i=sumlimits_{j=1}^n frac{1}{1-a_jx})
然后就推不下去了2333
我们要求的是(sum_{i=1}^n frac{1}{1-a_ix})
注意到((ln u)'=frac{u'}{u})，那么(ln (1-a_ix)'=frac{-a_i}{1-a_ix})，惊讶的发现
[frac{1}{1-a_ix}=1-frac{-a_ix}{1-a_ix}=1-xln(1-a_ix)'
]
太神奇辣~~
所以
[F(x)=sumlimits_{i=1}^n frac{1}{1-a_ix}=sumlimits_{i=1}^n 1-xln(1-a_ix)'=n-xsumlimits_{i=1}^n ln(1-a_ix)'
]
然后把求导拎到外面，在把求和移到(ln)里面变成乘积，所以
[F(x)=n-xleft(lnprodlimits_{i=1}^n(1-a_ix)
ight)'
]
里面乘法分治算一下就好了。
把(F(x))求出来了过后就可以把(A)求出来，类似的(B)也这样做就好了。
最后别忘除以nm...
但是说得简单这代码写的是真的爽......
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i=(a);i<(b);++i)
#define per(i,a,b) for (int i=(a)-1;i>=(b);--i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef double db;
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const int N=3e5+10,P=998244353,gn=3,ign=332748118;
inline int add(int x,int y) {return (x+=y)>=P?x-P:x;}
inline int sub(int x,int y) {return (x-=y)<0?x+P:x;}
inline int fpow(int x,int y) {
int ret=1; for (;y;y>>=1,x=1ll*x*x%P)
if (y&1) ret=1ll*ret*x%P;
return ret;
}
namespace Poly {
int rev[N];
void init(int n) {rep(i,0,n) rev[i]=rev[i>>1]>>1|((i&1)?n>>1:0);}
void ntt(int *f,int n,int flg) {
rep(i,0,n) if (rev[i]<i) swap(f[i],f[rev[i]]);
for (int len=2,k=1;len<=n;len<<=1,k<<=1) {
int wn=fpow(flg==1?gn:ign,(P-1)/len);
for (int i=0;i<n;i+=len)
for (int j=i,w=1;j<i+k;j++,w=1ll*w*wn%P) {
int tmp=1ll*f[j+k]*w%P;
f[j+k]=sub(f[j],tmp),f[j]=add(f[j],tmp);
}
}
if (flg==-1) {
int inv=fpow(n,P-2);
rep(i,0,n) f[i]=1ll*f[i]*inv%P;
}
}
void dao(int *f,int n,int *g) {
static int F[N]; rep(i,0,n) F[i]=f[i];
rep(i,1,n) g[i-1]=1ll*F[i]*i%P;g[n]=0;
}
void jifen(int *f,int n,int *g) {
static int F[N]; rep(i,0,n) F[i]=f[i];
rep(i,0,n) g[i+1]=1ll*F[i]*fpow(i+1,P-2)%P;g[0]=0;
}
void getinv(int *f,int n,int *G) {
if (n==1) {G[0]=fpow(f[0],P-2);return;}
getinv(f,(n+1)>>1,G);
static int F[N]; int limit=1;
while(limit<=2*(n-1)) limit<<=1; init(limit);
rep(i,0,limit) F[i]=i>=n?0:f[i],G[i]=i>=n?0:G[i];
ntt(F,limit,1),ntt(G,limit,1);
rep(i,0,limit) G[i]=1ll*G[i]*sub(2,1ll*F[i]*G[i]%P)%P;
ntt(G,limit,-1);
rep(i,n,limit) G[i]=0;
}
void getln(int *f,int n,int *G) {
static int derv[N],inv[N];
getinv(f,n,inv),dao(f,n,derv);
int limit=1; while(limit<=2*(n-1)) limit<<=1; init(limit);
rep(i,n,limit) derv[i]=inv[i]=0;
ntt(derv,limit,1),ntt(inv,limit,1);
rep(i,0,limit) G[i]=1ll*derv[i]*inv[i]%P;
ntt(G,limit,-1);
jifen(G,n,G);
rep(i,n,limit) G[i]=0;
}
}
using Poly::ntt;
using Poly::getinv;
using Poly::getln;
using Poly::dao;
int tn,ifac[N],fac[N],inv[N];
void init(int n) {
ifac[0]=ifac[1]=fac[0]=fac[1]=inv[1]=1;
rep(i,2,n+1) {
inv[i]=1ll*inv[P%i]*(P-P/i)%P;
ifac[i]=1ll*ifac[i-1]*inv[i]%P;
fac[i]=1ll*fac[i-1]*i%P;
}
}
void fff(int *a,int l,int r,int *F) {
if (l==r) {F[0]=1,F[1]=(P-a[l])%P;return;}
int mid=(l+r)>>1;
int Fl[N],Fr[N]; // 这里分治乘就直接把数组开在函数里了...不知道有没有更妙的做法qwq
fff(a,l,mid,Fl),fff(a,mid+1,r,Fr);
int limit=1; while(limit<=r-l+1) limit<<=1; Poly::init(limit);
rep(i,mid-l+2,limit) Fl[i]=0;
rep(i,r-mid+1,limit) Fr[i]=0;
ntt(Fl,limit,1),ntt(Fr,limit,1);
rep(i,0,limit) F[i]=1ll*Fl[i]*Fr[i]%P;
ntt(F,limit,-1);
rep(i,r-l+2,limit) F[i]=0;
}
void getF(int *a,int n,int *F) {
fff(a,1,n,F); static int G[N];
getln(F,tn+1,G);
dao(G,tn+1,F);
per(i,tn+1,1) F[i]=P-F[i-1];
F[0]=n;
rep(i,0,tn+1) F[i]=1ll*F[i]*ifac[i]%P;
}
int n,m,K,a[N],b[N],A[N],B[N];
int main() {
#ifdef LOCAL
freopen("a.in","r",stdin);
#endif
scanf("%d%d",&n,&m);
rep(i,1,n+1) scanf("%d",&a[i]);
rep(i,1,m+1) scanf("%d",&b[i]);
scanf("%d",&K); init(tn=max(max(n,m),K));
getF(a,n,A),getF(b,m,B);
int limit=1; while(limit<=2*tn) limit<<=1; Poly::init(limit);
ntt(A,limit,1),ntt(B,limit,1);
rep(i,0,limit) A[i]=1ll*A[i]*B[i]%P;
ntt(A,limit,-1); int iv=fpow(1ll*n*m%P,P-2);
rep(i,1,K+1) printf("%d
",1ll*iv*fac[i]%P*A[i]%P);
return 0;
}