多项式除法，多项式取模

多项式除法

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

R ( x ) R(x) R(x)是一个 n − m n - m n−m阶多项式， Q ( x ) Q(x) Q(x)是一个小于 m m m阶的多项式。
有 F ( x ) ≡ R ( x ) G ( x ) + Q ( x ) ( m o d x n + 1 ) F ( 1 x ) ≡ R ( 1 x ) G ( 1 x ) + Q ( 1 x ) ( m o d x n + 1 ) 同 时 乘 上 一 个 x n , F r e v ( x ) ≡ ( x m R r e v ( x ) ) ( x n − m G r e v ( x ) ) + x n − d e g Q Q r e v ( x ) ( m o d x n + 1 ) F r e v ( x ) ≡ R r e v ( x ) G r e v ( x ) + Q r e v ( x ) x n − d e g Q ( m o d x n + 1 ) 有 d e g Q < m , n − d e g Q > = n − m + 1 , 所 以 有 F r e v ( x ) ≡ R r e v ( x ) G r e v ( x ) ( m o d x n − m + 1 ) 有F(x) \equiv R(x) G(x) + Q(x) \pmod{x ^ {n + 1}}\\ F(\frac{1}{x}) \equiv R(\frac{1}{x})G(\frac{1}{x}) + Q(\frac{1}{x}) \pmod {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 - deg_Q} Q ^{rev} (x) \pmod{x ^{n + 1}} \\ F^{rev}(x) \equiv R^{rev}(x) G^{rev}(x) + Q^{rev}(x) x ^{n - deg_Q}\pmod{x ^{n + 1}}\\ 有deg_Q < m, n - deg_Q >= n - m + 1,所以有F^{rev}(x) \equiv R^{rev}(x) G^{rev}(x) \pmod{x ^{n - m + 1}}\\ 有F(x)≡R(x)G(x)+Q(x)(modxn+1)F(x1​)≡R(x1​)G(x1​)+Q(x1​)(modxn+1)同时乘上一个xn,Frev(x)≡(xmRrev(x))(xn−mGrev(x))+xn−degQ​Qrev(x)(modxn+1)Frev(x)≡Rrev(x)Grev(x)+Qrev(x)xn−degQ​(modxn+1)有degQ​<m,n−degQ​>=n−m+1,所以有Frev(x)≡Rrev(x)Grev(x)(modxn−m+1)
只要多项式求逆，即可得到 R ( x ) R(x) R(x)，然后代入原式求得 Q ( x ) Q(x) 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;
}