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142 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1 << 20, mod = 998244353;
int a[maxn], b[maxn], g[maxn], gg[maxn];
int qpow(int x, int y) { // 快速幂
int ans = 1;
while (y) {
if (y & 1) {
ans = (long long)1 * ans * x % mod;
}
x = (long long)1 * x * x % mod;
y >>= 1;
}
return ans;
}
int inv2 = qpow(2, mod - 2); // 逆元
inline void change(int *f, int len) {
for (int i = 1, j = len / 2; i < len - 1; i++) {
if (i < j) {
swap(f[i], f[j]);
}
int k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if (j < k) {
j += k;
}
}
}
inline void NTT(int *f, int len, int type) { // NTT
change(f, len);
for (int q = 2; q <= len; q <<= 1) {
int nxt = qpow(3, (mod - 1) / q);
for (int i = 0; i < len; i += q) {
int w = 1;
for (int k = i; k < i + (q >> 1); k++) {
int x = f[k];
int y = (long long)1 * w * f[k + (q / 2)] % mod;
f[k] = (x + y) % mod;
f[k + (q / 2)] = (x - y + mod) % mod;
w = (long long)1 * w * nxt % mod;
}
}
}
if (type == -1) {
reverse(f + 1, f + len);
int iv = qpow(len, mod - 2);
for (int i = 0; i < len; i++) {
f[i] = (long long)1 * f[i] * iv % mod;
}
}
}
inline void inv(int deg, int *f, int *h) { // 求逆元
if (deg == 1) {
h[0] = qpow(f[0], mod - 2);
return;
}
inv(deg + 1 >> 1, f, h);
int len = 1;
while (len < deg * 2) { // 倍增
len *= 2;
}
copy(f, f + deg, gg);
fill(gg + deg, gg + len, 0);
NTT(gg, len, 1);
NTT(h, len, 1);
for (int i = 0; i < len; i++) {
h[i] = (long long)1 * (2 - (long long)1 * gg[i] * h[i] % mod + mod) % mod *
h[i] % mod;
}
NTT(h, len, -1);
fill(h + deg, h + len, 0);
}
int n, t[maxn];
// deg:次数
// f:被开根数组
// h:答案数组
inline void sqrt(int deg, int *f, int *h) {
if (deg == 1) {
h[0] = 1;
return;
}
sqrt(deg + 1 >> 1, f, h);
int len = 1;
while (len < deg * 2) { // 倍增
len *= 2;
}
fill(g, g + len, 0);
inv(deg, h, g);
copy(f, f + deg, t);
fill(t + deg, t + len, 0);
NTT(t, len, 1);
NTT(g, len, 1);
NTT(h, len, 1);
for (int i = 0; i < len; i++) {
h[i] = (long long)1 * inv2 *
((long long)1 * h[i] % mod + (long long)1 * g[i] * t[i] % mod) % mod;
}
NTT(h, len, -1);
fill(h + deg, h + len, 0);
}
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
sqrt(n, a, b);
for (int i = 0; i < n; i++) {
printf("%d ", b[i]);
}
return 0;
}
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