有两个非负整数序列，我们称其为(a_1cdots a_n)和(b_1cdots b_m)。每次游戏中玩家会从(a)序列和(b)序列中分别随机地抽取一个数，假设抽出的数为(a_i,b_j)，则定义这次游戏的(k)次价值为：((a_i+b_j)^k)。要求对于每个(kin [1,t])的(k)次价值的期望，对(998244353)取模。(n,m,tleq 10^5,a_i,b_jleq 998244352)。
显然有答案=(frac{sum_{i=1}^nsum_{j=1}^m(a_i+b_j)^k}{nm})。我们只关注分子的式子，把它用二项式定理展开一下就有：
[sum_{i=1}^{n}sum_{j=1}^msum_{p=0}^k{kchoose p}a_i^pb_j^{k-p}
\=k!sum_{p=0}^ksum_{i=1}^nfrac{a_i^p}{p!}sum_{j=1}^mfrac{b_j^{k-p}}{(k-p)!}
]
观察到后面的两个求和式中分子的指数和分母的阶乘是一样的，那么可以写成指数型生成函数的形式，下面以第一个求和式为例：
[A(x)=sum_{p=0}(sum_{i=1}^na_i^p)frac{x^p}{p!}
]
那么答案式子就可以写成(k![x^k]A(x)B(x))。现在问题就是要计算(A(x))和(B(x))的系数。
仍然写出系数的生成函数，以(A(x))为例，就有：
[F_A(x)=sum_{p=0}(sum_{i=1}^na_i^p)x^p = sum_{i=1}^nsum_{p=0}(a_ix)^p = sum_{i=1}^nfrac{1}{1-a_ix}
]
怎么计算最后这个多项式呢？普通的做法就是通分，分母的多项式相乘，分子与分母交叉相乘再相加。但如果从左往右加过去开销就很大。我们可以用分治的思想，每次把问题折半，先算左边和右边的(frac{n}{2})规模的问题，然后左右两边通分加起来。复杂度就是(O(nlog^2n))的。
或者有一个常数更小的做法。注意到((ln{(1-a_ix)})'=-frac{a_i}{1-a_ix}=frac{1}{1-a_ix}-1)，那么设(G_A(x)=sum_{i=1}^n-frac{a_i}{1-a_ix})，就有(F_A(x)=-xG_A(x)+n)，于是：
[G_A(x)=sum_{i=1}^n(ln{(1-a_ix)})'=(sum_{i=1}^nln{(1-a_ix)})'=(ln{prod_{i=1}^n(1-a_ix)})'
]
分治NTT即可。
#include<bits/stdc++.h>
#define rg register
#define il inline
#define cn const
#define gc getchar()
#define fp(i,a,b) for(rg int i=(a),ed=(b);i<=ed;++i)
#define fb(i,a,b) for(rg int i=(a),ed=(b);i>=ed;--i)
#define go(u) for(rg int i=head[u];~i;i=e[i].nxt)
using namespace std;
typedef cn int cint;
typedef long long LL;
il int rd(){
rg int x(0),f(1); rg char c(gc);
while(c<'0'||'9'<c){if(c=='-')f=-1;c=gc;}
while('0'<=c&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=gc;
return x*f;
}
cint maxn = 1e5+10, mod = 998244353, G = 3, invG = (mod+1)/3, inf = 0x3f3f3f3f;
int n, m, t, a[maxn], b[maxn], fac[maxn], ifac[maxn];
il void sum(int &a, int b){if((a += b) >= mod)a -= mod;}
il int fpow(int a, int b, int ans = 1){
for(; b; b >>= 1, a = 1ll*a*a%mod)if(b&1)ans = 1ll*ans*a%mod;
return ans;
}
int rev[maxn<<4];
struct poly{
vector<int> ary;
int len;
};
void print(cn poly &a){fp(i, 0, a.len-1)printf("%d ", a.ary[i]);puts("");}
il void crt(poly &po, cint &len){po.ary.resize(len), po.len = len;}
il void ntt(poly &a, cint &f){
fp(i, 0, a.len-1)if(i < rev[i])swap(a.ary[i], a.ary[rev[i]]);
for(rg int md = 1; md < a.len; md <<= 1){
rg int len = md<<1, Gn=fpow(f ? G : invG, (mod-1)/len);
for(rg int l = 0; l < a.len; l += len){
rg int Pow = 1;
for(rg int nw = 0; nw < md; ++nw, Pow = 1ll*Pow*Gn%mod){
rg int x = a.ary[l+nw], y = 1ll*Pow*a.ary[l+nw+md]%mod;
a.ary[l+nw] = (x+y)%mod, a.ary[l+nw+md] = (x-y+mod)%mod;
}
}
}
if(!f){
rg int inv = fpow(a.len, mod-2);
fp(i, 0, a.len - 1)a.ary[i] = 1ll*a.ary[i]*inv%mod;
}
}
il poly mul(cn poly &po1, cn poly &po2, int mx = inf){
poly a, b, c;
rg int len = po1.len+po2.len-1, lim = 1, hst = 0;
while(lim < len)lim <<= 1, ++hst;
fp(i, 1, lim-1)rev[i] = (rev[i>>1]>>1) | ((i&1)<<hst-1);
crt(a, lim), crt(b, lim);
fp(i, 0, po1.len-1)a.ary[i] = po1.ary[i];
fp(i, 0, po2.len-1)b.ary[i] = po2.ary[i];
ntt(a, 1), ntt(b, 1);
fp(i, 0, lim-1)a.ary[i] = 1ll*a.ary[i]*b.ary[i]%mod;
ntt(a, 0);
crt(c, min(mx, len));
fp(i, 0, c.len-1)c.ary[i] = a.ary[i];
return c;
}
poly divntt(int *a, int l, int r){
poly res;
if(l == r){
crt(res, 2), res.ary[0] = 1, res.ary[1] = mod-a[l];
return res;
}
int md = l+r>>1;
res = mul(divntt(a, l, md), divntt(a, md+1, r));
return res;
}
poly getinv(cn poly &a, int n){
poly res;
if(n == 1){
crt(res, 1), res.ary[0] = fpow(a.ary[0], mod-2);
return res;
}
poly f = getinv(a, n+1>>1), g;
crt(g, n);
fp(i, 0, n-1)if(i<a.len)g.ary[i] = a.ary[i];else g.ary[i] = 0;
g = mul(g, f, n);
fp(i, 0, g.len-1)g.ary[i] = mod-g.ary[i];
sum(g.ary[0], 2), res = mul(f, g, n);
return res;
}
il poly der(cn poly &a){
poly res;
crt(res, a.len-1);
fp(i, 1, a.len-1)res.ary[i-1] = 1ll*i*a.ary[i]%mod;
return res;
}
int main(){
n = rd(), m = rd();
fp(i, 1, n)a[i] = rd();
fp(i, 1, m)b[i] = rd();
t = rd();
poly GA = divntt(a, 1, n), GB = divntt(b, 1, m);
poly invGA = getinv(GA, max(n, t)+1), invGB = getinv(GB, max(m, t)+1);
GA = mul(der(GA), invGA), GB = mul(der(GB), invGB);
poly FA, FB;
crt(FA, GA.len+1), crt(FB, GB.len+1);
fp(i, 0, GA.len-1)FA.ary[i+1] = mod-GA.ary[i];
fp(i, 0, GB.len-1)FB.ary[i+1] = mod-GB.ary[i];
FA.ary[0] = n, FB.ary[0] = m;
fac[0] = 1; fp(i, 1, t)fac[i] = 1ll*fac[i-1]*i%mod;
ifac[t] = fpow(fac[t], mod-2); fb(i, t, 1)ifac[i-1] = 1ll*ifac[i]*i%mod;
fp(i, 0, FA.len-1)FA.ary[i] = 1ll*FA.ary[i]*ifac[i]%mod;
fp(i, 0, FB.len-1)FB.ary[i] = 1ll*FB.ary[i]*ifac[i]%mod;
poly ans = mul(FA, FB);
rg int tot = 1ll*fpow(n, mod-2)*fpow(m, mod-2)%mod;
fp(i, 1, t)printf("%lld
", 1ll*fac[i]*ans.ary[i]%mod*tot%mod);
return 0;
}