多项式除法

给定一个$n$次多项式$F(x)$和$m$次多项式$G(x)$，要求$R(x),Q(x)$，满足$F(x)=R(x)G(x)+Q(x)$。

$R(x)$是一个$n-m$阶多项式，$Q(x)$是一个小于$m$阶的多项式。
$$\begin{gathered} 有F(x)\equiv R(x)G(x)+Q(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{n+1}) \\ F(\frac{1}{x})\equiv R(\frac{1}{x})G(\frac{1}{x})+Q(\frac{1}{x})(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{n+1}) \\ 同时乘上一个x^n,F^{rev}(x)\equiv \left(x^m{R}^{rev}(x)\right)\left(x^{n-m}{G}^{rev}(x)\right)+x^{n-de{g}_Q}{Q}^{rev}(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{n+1}) \\ {F}^{rev}(x)\equiv R^{rev}(x)G^{rev}(x)+Q^{rev}(x)x^{n-de{g}_Q}(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{n+1}) \\ 有de{g}_Q<m,n-de{g}_Q>=n-m+1,所以有F^{rev}(x)\equiv R^{rev}(x)G^{rev}(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^{n-m+1}) \end{gathered}$$
只要多项式求逆，即可得到$R(x)$，然后代入原式求得$Q(x)$。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = mod + 1 >> 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex &a, Complex &b) {return Complex((1ll * a.r * b.r % mod  + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans = Complex(1, 0);while (n) {if (n & 1) {ans = ans * a;}a = a * a;n >>= 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n == 0) {return 0;}if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {return -1;}uniform_int_distribution<int> r(0, mod - 1);int a = r(e);while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {a = r(e);}I2 = (1ll * a * a % mod - n + mod) % mod;int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;if(x > y) swap(x, y);return x;}
}const int N = 6e5 + 10;int r[N], inv[N], b[N], c[N], d[N], e[N], t[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 get_inv(int n) {inv[1] = 1;for (int i = 2; i <= n; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;}
}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 *f, int *g, int n) {if (n == 1) {g[0] = quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n + 1 >> 1);for (int i = 0; i < n; i++) {t[i] = f[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;g[i] = 1ll * g[i] * cur % mod;t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}void polysqrt(int *f, int *g, int n) {if (n == 1) {g[0] = Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n + 1 >> 1);polyinv(g, b, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i < n; i++) {t[i] = f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] = t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}void derivative(int *a, int *b, int n) {for (int i = 0; i < n; i++) {b[i] = 1ll * a[i + 1] * (i + 1) % mod;}
}void integrate(int *a, int n) {for (int i = n - 1; i >= 1; i--) {a[i] = 1ll * a[i - 1] * inv[i] % mod;}a[0] = 0;
}void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * b[i] % mod;b[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}integrate(g, n);
}void polyexp(int *f, int *g, int n) {if (n == 1) {g[0] = 1;return ;}polyexp(f, g, n + 1 >> 1);int lim = 1;while (lim < 2 * n) {lim <<= 1;}polyln(g, d, n);for (int i = 0; i < n; i++) {t[i] = (f[i] - d[i] + mod) % mod;}t[0] = (t[0] + 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * t[i] % mod;t[i] = d[i] =  0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}/*b存放多项式逆，c存放多项式开根，d存放多项式对数ln，e存放多项式指数exp，t作为中间转移数组,如果要用到polyinv，得提前调用get_inv(n)先预先得到我们想要得到的逆元范围。
*/int f[N], fr[N], g[N], gr[N], rr[N], n, m;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &m);for (int i = 0; i <= n; i++) {scanf("%d", &f[i]);fr[n - i] = f[i];}for (int i = 0; i <= m; i++) {scanf("%d", &g[i]);gr[m - i] = g[i];}for (int i = n - m + 1; i <= n; i++) {fr[i] = gr[i] = 0;}polyinv(gr, b, n - m + 1);for (int i = 0; i < n - m + 1; i++) {gr[i] = b[i];b[i] = 0;}int lim = 1;while (lim < 2 * (n - m + 1)) {lim <<= 1;}get_r(lim);NTT(fr, lim, 1);NTT(gr, lim, 1);for (int i = 0; i < lim; i++) {fr[i] = 1ll * fr[i] * gr[i] % mod;}NTT(fr, lim, -1);for (int i = 0; i <= n - m; i++) {rr[i] = fr[n - m - i];}for (int i = 0; rr[i]; i++) {printf("%d ", rr[i]);}puts("");lim = 1;while (lim <= 2 * n) {lim <<= 1;}get_r(lim);NTT(rr, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * rr[i] % mod;}NTT(g, lim, -1);for (int i = 0; i < m; i++) {f[i] = (f[i] - g[i] + mod) % mod;}for (int i = 0; i < m; i++) {printf("%d ", f[i]);}puts("");return 0;
}