多项式开根

多项式开根

给定多项式 g ( x ) g(x) g(x)，求 f ( x ) f(x) f(x)，满足 f 2 ( x ) = g ( x ) f ^ 2(x) = g(x) f2(x)=g(x)。

假设我们已经得到了 g ( x ) g(x) g(x)，膜 x ⌈ n 2 ⌉ x ^{\lceil \frac{n}{2} \rceil} x⌈2n​⌉下的根 f 0 ( x ) f_0 (x) f0​(x)，要求膜 x n x ^ n xn下的根 f ( x ) f(x) f(x)

有 f 0 2 ( x ) ≡ g ( x ) ( m o d x ⌈ n 2 ⌉ ) f_0 ^2(x) \equiv g(x) \pmod {x ^{\lceil \frac{n}{2} \rceil}} f02​(x)≡g(x)(modx⌈2n​⌉)

移项再开方有 ( f 0 2 ( x ) − g ( x ) ) 2 ≡ 0 ( m o d x n ) \left(f_0 ^2(x) - g(x) \right) ^ 2 \equiv 0 \pmod {x ^ n} (f02​(x)−g(x))2≡0(modxn)

则， ( f 0 2 ( x ) + g ( x ) ) 2 ≡ 4 f 0 2 ( x ) g ( x ) ( m o d x n ) \left( f_0 ^ 2(x) + g(x) \right) ^ 2 \equiv 4 f_0 ^ 2(x) g(x) \pmod {x ^ n} (f02​(x)+g(x))2≡4f02​(x)g(x)(modxn)

g ( x ) ≡ ( f 0 2 ( x ) + g ( x ) 2 f 0 ( x ) ) 2 ( m o d x n ) g(x) \equiv \left(\frac{f_0 ^ 2(x) + g(x)}{2f_0 (x)} \right) ^ 2 \pmod {x ^ n} g(x)≡(2f0​(x)f02​(x)+g(x)​)2(modxn)

所以 f ( x ) ≡ f 0 2 ( x ) + g ( x ) 2 f 0 ( x ) ( m o d x n ) f(x) \equiv \frac{f_0 ^ 2(x) + g(x)}{2f_0(x)} \pmod {x ^ n} f(x)≡2f0​(x)f02​(x)+g(x)​(modxn)。

所以有 f ( x ) ≡ 2 − 1 f 0 ( x ) + 2 − 1 f 0 − 1 ( x ) g ( x ) ( m o d x n ) f(x) \equiv 2 ^{-1} f_0 (x) + 2 ^{-1} f_0 ^{-1}(x) g(x) \pmod {x ^ n} f(x)≡2−1f0​(x)+2−1f0−1​(x)g(x)(modxn)，

对于 g ( 0 ) = 1 g(0) = 1 g(0)=1的特殊情况

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e6 + 10, mod = 998244353, inv2 = mod + 1 >> 1;

int a[N], b[N], c[N], d[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 polyinv1(int *a, int *b, int n) {
  if (n == 1) {
    b[0] = quick_pow(a[0], mod - 2);
    return ;
  }
  polyinv1(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;
  }
}

void polysqrt(int *a, int *b, int n) {
  if (n == 1) {
    b[0] = 1;
    return ;
  }
  polysqrt(a, b, n + 1 >> 1);
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  for (int i = 0; i < lim; i++) {
    d[i] = 0;
  }
  polyinv1(b, d, n);
  for (int i = 0; i < n; i++) {
    c[i] = a[i];
  }
  for (int i = n; i < lim; i++) {
    c[i] = 0;
  }
  get_r(lim);
  NTT(b, lim, 1);
  NTT(c, lim, 1);
  NTT(d, lim, 1);
  for (int i = 0; i < lim; i++) {
    b[i] = (1ll * inv2 * b[i] % mod + 1ll * inv2 * d[i] % mod * c[i] % mod) % 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]);
  }
  polysqrt(a, b, n);
  for (int i = 0; i < n; i++) {
    printf("%d%c", b[i], i + 1 == n ? '\n' : ' ');
  }
  return 0;
}

二次剩余解一般情况

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e6 + 10, mod = 998244353, inv2 = mod + 1 >> 1;

int a[N], b[N], c[N], d[N], r[N];

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;
  }
}

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 polyinv1(int *a, int *b, int n) {
  if (n == 1) {
    b[0] = quick_pow(a[0], mod - 2);
    return ;
  }
  polyinv1(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;
  }
}

void polysqrt(int *a, int *b, int n) {
  if (n == 1) {
    b[0] = Quadratic_residue::get_residue(a[0]);
    return ;
  }
  polysqrt(a, b, n + 1 >> 1);
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  for (int i = 0; i < lim; i++) {
    d[i] = 0;
  }
  polyinv1(b, d, n);
  for (int i = 0; i < n; i++) {
    c[i] = a[i];
  }
  for (int i = n; i < lim; i++) {
    c[i] = 0;
  }
  get_r(lim);
  NTT(b, lim, 1);
  NTT(c, lim, 1);
  NTT(d, lim, 1);
  for (int i = 0; i < lim; i++) {
    b[i] = (1ll * inv2 * b[i] % mod + 1ll * inv2 * d[i] % mod * c[i] % mod) % 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]);
  }
  polysqrt(a, b, n);
  for (int i = 0; i < n; i++) {
    printf("%d%c", b[i], i + 1 == n ? '\n' : ' ');
  }
  return 0;
}