有兩個非負整數序列，我們稱其爲\(a_1\cdots a_n\)和\(b_1\cdots b_m\)。每次遊戲中玩家會從\(a\)序列和\(b\)序列中分別隨機地抽取一個數，假設抽出的數爲\(a_i,b_j\)，則定義這次遊戲的\(k\)次價值爲：\((a_i+b_j)^k\)。要求對於每個\(k\in [1,t]\)的\(k\)次價值的期望，對\(998244353\)取模。\(n,m,t\leq 10^5,a_i,b_j\leq 998244352\)。
顯然有答案=\(\frac{\sum_{i=1}^n\sum_{j=1}^m(a_i+b_j)^k}{nm}\)。我們只關注分子的式子，把它用二項式定理展開一下就有：
\[\sum_{i=1}^{n}\sum_{j=1}^m\sum_{p=0}^k{k\choose p}a_i^pb_j^{k-p}
\\=k!\sum_{p=0}^k\sum_{i=1}^n\frac{a_i^p}{p!}\sum_{j=1}^m\frac{b_j^{k-p}}{(k-p)!}
\]
觀察到後面的兩個求和式中分子的指數和分母的階乘是一樣的，那麼可以寫成指數型生成函數的形式，下面以第一個求和式爲例：
\[A(x)=\sum_{p=0}(\sum_{i=1}^na_i^p)\frac{x^p}{p!}
\]
那麼答案式子就可以寫成\(k![x^k]A(x)B(x)\)。現在問題就是要計算\(A(x)\)和\(B(x)\)的係數。
仍然寫出係數的生成函數，以\(A(x)\)爲例，就有：
\[F_A(x)=\sum_{p=0}(\sum_{i=1}^na_i^p)x^p = \sum_{i=1}^n\sum_{p=0}(a_ix)^p = \sum_{i=1}^n\frac{1}{1-a_ix}
\]
怎麼計算最後這個多項式呢？普通的做法就是通分，分母的多項式相乘，分子與分母交叉相乘再相加。但如果從左往右加過去開銷就很大。我們可以用分治的思想，每次把問題折半，先算左邊和右邊的\(\frac{n}{2}\)規模的問題，然後左右兩邊通分加起來。複雜度就是\(O(nlog^2n)\)的。
或者有一個常數更小的做法。注意到\((\ln{(1-a_ix)})'=-\frac{a_i}{1-a_ix}=\frac{1}{1-a_ix}-1\)，那麼設\(G_A(x)=\sum_{i=1}^n-\frac{a_i}{1-a_ix}\)，就有\(F_A(x)=-xG_A(x)+n\)，於是：
\[G_A(x)=\sum_{i=1}^n(\ln{(1-a_ix)})'=(\sum_{i=1}^n\ln{(1-a_ix)})'=(\ln{\prod_{i=1}^n(1-a_ix)})'
\]
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\n", 1ll*fac[i]*ans.ary[i]%mod*tot%mod);
return 0;
}