P4721 【模板】分治 FFT
最具经济性的写法：\(\mathcal O(n^2)\) 暴力拿下 \(80\) 分，遂跑路。
一题多解了，分两部分：分治 和 多项式求逆。
分治
考虑 cdq 分治，每次把 \(f_{l \dots mid}\) 和 \(g_{1 \dots n - 1}\) 卷起来，贡献直接加到 \(f_{mid + 1 \dots r}\) 里，要注意一下顺序，先递归左区间，再算当前区间，最后递归右区间，否则不能确保正确性。
时间复杂度 \(T(n) = 2T(\dfrac n2) + \mathcal O(n \log n) = \mathcal O(n \log n)\)。常数较大。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int N = 1 << 17, MOD = 998244353;
int n, f[N], g[N], tf[N], tg[N];
int len, bits, rev[N], G[2][17];
ll qp(ll base, int e) {
ll res = 1;
while (e) {
if (e & 1) res = res * base % MOD;
base = base * base % MOD;
e >>= 1;
}
return res;
}
void NTT(int *A, bool I = 0) {
for (int i = 0; i < len; i++) if (i < rev[i]) swap(A[i], A[rev[i]]);
for (int i = 1; i < len; i <<= 1) {
ll wn = G[I][__builtin_ctz(i)];
for (int j = 0; j < len; j += (i << 1)) {
ll w = 1;
for (int k = j; k < j + i; k++) {
int t = w * A[k + i] % MOD;
A[k + i] = (A[k] - t + MOD) % MOD, A[k] = (A[k] + t) % MOD;
w = w * wn % MOD;
}
}
}
if (I) {
ll invlen = qp(len, MOD - 2);
for (int i = 0; i < len; i++) A[i] = A[i] * invlen % MOD;
}
}
void cdq(int l, int r) {
if (l == r) return;
int mid = (l + r) >> 1; cdq(l, mid);
len = 1, bits = -1; while (len <= r - l) len <<= 1, bits++;
for (int i = 0; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bits);
memset(tf, 0, len << 2), memcpy(tf, f + l, (mid - l + 1) << 2), memcpy(tg, g, (r - l + 1) << 2);
NTT(tf), NTT(tg); for (int i = 0; i < len; i++) tf[i] = (ll)tf[i] * tg[i] % MOD; NTT(tf, 1);
for (int i = mid + 1; i <= r; i++) f[i] = (f[i] + tf[i - l]) % MOD;
cdq(mid + 1, r);
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(nullptr), cout.tie(nullptr);
cin >> n;
for (int i = 1; i < n; i++) cin >> g[i];
for (int i = 0; i < 17; i++) G[0][i] = qp(3, (MOD - 1) / (1 << (i + 1))), G[1][i] = qp(G[0][i], MOD - 2);
f[0] = 1, cdq(0, n - 1);
for (int i = 0; i < n; i++) cout << f[i] << ' ';
return 0;
}
多项式求逆
设 \(F(x) = \sum\limits_{i = 0}^\infin f_ix^i, G(x) = \sum\limits_{i = 0}^\infin g_ix^i, g_0 = 0\)。
有
\[F(x) * G(x) = \sum\limits_{i = 0}^{\infin}\sum\limits_{j = 0}^if_{i - j}g_jx^i = F(x) - f_0 = F(x) - 1
\]
故
\[F(x) * [1 - G(x)] \equiv 1 \pmod{x^n}
\]
剩下的就是多项式乘法逆板子了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int N = 1 << 18, MOD = 998244353;
int n, f[N], g[N], fg[N];
int len, bits, rev[N], G[2][18];
ll qp(ll base, int e) {
ll res = 1;
while (e) {
if (e & 1) res = res * base % MOD;
base = base * base % MOD;
e >>= 1;
}
return res;
}
void NTT(int *A, bool I = 0) {
for (int i = 0; i < len; i++) if (i < rev[i]) swap(A[i], A[rev[i]]);
for (int i = 1; i < len; i <<= 1) {
ll wn = G[I][__builtin_ctz(i)];
for (int j = 0; j < len; j += (i << 1)) {
ll w = 1;
for (int k = j; k < j + i; k++) {
int t = w * A[k + i] % MOD;
A[k + i] = (A[k] - t + MOD) % MOD, A[k] = (A[k] + t) % MOD;
w = w * wn % MOD;
}
}
}
if (I) {
ll invlen = qp(len, MOD - 2);
for (int i = 0; i < len; i++) A[i] = A[i] * invlen % MOD;
}
}
void solve(int e) {
if (e == 1) return;
solve((e + 1) >> 1);
bits = -1, len = 1; while (len < (e << 1) - 1) len <<= 1, bits++;
for (int i = 0; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bits);
memcpy(fg, f, e << 2), fill(fg + e, fg + len, 0); NTT(fg), NTT(g);
for (int i = 0; i < len; i++) fg[i] = (ll)fg[i] * g[i] % MOD;
for (int i = 0; i < len; i++) g[i] = (ll)g[i] * (2 - fg[i] + MOD) % MOD;
NTT(g, 1); fill(g + e, g + len, 0);
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(nullptr), cout.tie(nullptr);
cin >> n; for (int i = 1; i < n; i++) cin >> f[i], f[i] = MOD - f[i];
for (int i = 0; i < 18; i++) G[0][i] = qp(3, (MOD - 1) / (1 << (i + 1))), G[1][i] = qp(G[0][i], MOD - 2);
g[0] = qp(f[0] = 1, MOD - 2); solve(n);
for (int i = 0; i < n; i++) cout << g[i] << ' ';
return 0;
}
