首先是有一个结论，对多项式做任意多次 transformation ，其结果跟做一次 transformation $Tr(f,\sum _{i=1}^m{a}_i)$的结果是一样的，所以我们约定$a=-\sum _{i=1}^m{a}_i$。
$$\begin{gathered} g=f(x+a) \\ \sum _{i=0}^n{c}_i(x+a{)}^i \\ \sum _{i=0}^n\sum _{j=0}^i{c}_i\frac{i!}{j!(i-j)!}{x}^j{a}^{i-j} \\ 我们考虑只求这个多项式的x^j系数 \\ {g}_j=\sum _{i=j}^n{c}_i\frac{i!}{j!(i-j)!}{a}^{i-j} \\ {g}_j=\sum _{i=0}^{n-j}{c}_{i+j}\frac{(i+j)!}{j!i!}{a}^i \\ {g}_j=\frac{1}{j!}\sum _{i=0}^{n-j}(c_{i+j}(i+j)!)\times (\frac{a^i}{i!}) \\ 设f(n)=c_{n}n!,h(n)=\frac{a^n}{n!} \\ 不难发现后项像是多项式卷积的形式，考虑翻转\frac{a^i}{i}，即i=n-i \\ {g}_j=\frac{1}{j!}\sum _{i=0}^{n-j}f(i+j)h(n-i) \\ 这就是一个完美的卷积形式了，于是可以NTT求解 \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, mod = 998244353;int r[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * a * ans % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans;
}void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}void NTT(int *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}for (int mid = 1; mid < lim; mid <<= 1) {int wn = quick_pow(3, (mod - 1) / (mid << 1));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {int w = 1;for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;}}}if (rev == -1) {int inv = quick_pow(lim, mod - 2);reverse(f + 1, f + lim);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * inv % mod;}}
}int a[N], c[N], A[N], B[N], fac[N], ifac[N], n, m, sum;void init() {fac[0] = 1;for (int i = 1; i < N; i++) {fac[i] = 1ll * fac[i - 1] * i % mod;}ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);for (int i = N - 2; i >= 0; i--) {ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();while (scanf("%d", &n) != EOF) {for (int i = 0; i <= n; i++) {scanf("%d", &c[i]);}scanf("%d", &m);sum = 0;for (int i = 1; i <= m; i++) {scanf("%d", &a[i]);sum = (sum + a[i]) % mod;}sum = (mod - sum) % mod;int cur = 1;for (int i = 0; i <= n; i++) {A[i] = 1ll * c[i] * fac[i] % mod;B[n - i] = 1ll * ifac[i] * cur % mod;cur = 1ll * cur * sum % mod;}int lim = 1;while (lim <= 2 * n) {lim <<= 1;}get_r(lim);NTT(A, lim, 1);NTT(B, lim, 1);for (int i = 0; i < lim; i++) {A[i] = 1ll * A[i] * B[i] % mod;}NTT(A, lim, -1);for (int i = 0; i <= n; i++) {printf("%lld ", 1ll * ifac[i] * A[n + i] % mod);}puts("");for (int i = 0; i < lim; i++) {A[i] = B[i] = 0;}}return 0;
}