多項式求逆

$f(x)g(x)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$，稱$f(x)$為$g(x)$或者$g(x)$為$f(x)$膜$x^n$意義下的逆元。

下面我們討論給定$f(x)$，求其逆$f^{?1}(x)$。

倍增求解

假設我們已經求得$f(x)$膜$x^{?\frac{n}{2}}?$下的逆元$f_0^{?1}(x)$，要求$f^{?1}(x)$，即膜$x^n$下的逆元，則

$f(x)f_0^{?1}(x)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{?\frac{n}{2}?})$

顯然$f(x)f^{?1}(x)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{?\frac{n}{2}?})$也是成立的

對兩邊同時乘以$f_0^{?1}(x)$并移項有

$f^{?1}(x)?f_0^{?1}(x)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{?\frac{n}{2}?})$

對兩邊同時開方得到

$f^{?2}(x)?2f^{?1}{f}_0^{?1}(x)+f_0^{?2}(x)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$

我們再對兩邊乘上一個$f(x)$，則有

$f^{?1}(x)?2f_0^{?1}+f(x)f_0^{?2}(x)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$

再對其進行移項可得

$f^{?1}(x)\equiv f_0^{?1}(x)\left(2?f(x)f_0^{?1}(x)\right)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$

由此我們遞歸求解即可。

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 5e6 + 10, mod = 998244353;int a[N], b[N], c[N], r[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % 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;}} }void polyinv(int *a, int *b, int n) {if (n == 1) {b[0] = quick_pow(a[0], mod - 2);return ;}polyinv(a, b, n + 1 >> 1);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i < n; i++) {c[i] = a[i];}for (int i = n; i < lim; i++) {c[i] = 0;}NTT(b, lim, 1);NTT(c, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * c[i] * b[i] % mod + mod) % mod;b[i] = 1ll * b[i] * cur % mod;}NTT(b, lim, -1);for (int i = n; i < lim; i++) {b[i] = 0;} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &a[i]);}polyinv(a, b, n);for (int i = 0; i < n; i++) {printf("%d%c", b[i], i + 1 == n ? '\n' : ' ');}return 0; }

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