AIOJ

This page covers common elementary operations on polynomials. Specifically, it includes the following:

Polynomial inversion
Polynomial square roots
Polynomial division
Polynomial remainder
Polynomial exponentials
Polynomial logarithms
Polynomial trigonometric functions
Polynomial inverse trigonometric functions

Elementary and non-elementary functions

The definition of an elementary function is as follows¹:

If there exists a mapping $u\to \partial u$ in a field $F$ satisfying:

$\partial(u+v)=\partial u+\partial v$
$\partial(uv)=u\partial v+v\partial u$

then this field is called a differential field.

If a function $u$ on the differential field $F$ satisfies any one of the following conditions, then $u$ is called an elementary function:

$u$ is an algebraic function over $F$ .
$u$ is an exponential function over $F$ , meaning that there exists $a\in F$ such that $\partial u=u\partial a$ .
$u$ is a logarithmic function over $F$ , meaning that there exists $a\in F$ such that $\partial u=\frac{\partial a}{a}$ .

The following are common elementary functions:

Algebraic functions: functions $f(x)$ for which there exists a finite-degree polynomial $P$ such that $P(f(x))=0$ , such as $2x+1$ , $\sqrt{x}$ , $(1+x^2)^{-1}$ , $|x|$ .
Exponential functions
Logarithmic functions
Trigonometric functions
Inverse trigonometric functions
Hyperbolic functions
Inverse hyperbolic functions
Compositions of the above functions, such as:
$\frac{\mathrm{e}^{\tan x}}{1+x^2}\sin\left(\sqrt{1+\ln^2 x}\right)$ $-\mathrm{i} \ln\left(x+\mathrm{i}\sqrt{1-x^2}\right)$

The following is a common non-elementary function:

The error function:
$\operatorname{erf}(x):=\frac{2}{\sqrt{\pi}}\int_{0}^{x}\exp\left(-t^2\right)\mathrm{d}t$

Polynomial Inversion

Given a polynomial $f\left(x\right)$ , find $f^{-1}\left(x\right)$ .

Solution

Doubling Method

First, it is easy to see that

\left[x^{0}\right]f^{-1}\left(x\right)=\left(\left[x^{0}\right]f\left(x\right)\right)^{-1}

Assume that the inverse $f^{-1}_{0}\left(x\right)$ of $f\left(x\right)$ modulo $x^{\left\lceil\frac{n}{2}\right\rceil}$ has already been found. Then:

\begin{aligned} f\left(x\right)f^{-1}_{0}\left(x\right)&\equiv 1 &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ f\left(x\right)f^{-1}\left(x\right)&\equiv 1 &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ f^{-1}\left(x\right)-f^{-1}_{0}\left(x\right)&\equiv 0 &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}} \end{aligned}

Squaring both sides gives:

f^{-2}\left(x\right)-2f^{-1}\left(x\right)f^{-1}_{0}\left(x\right)+f^{-2}_{0}\left(x\right)\equiv 0 \pmod{x^{n}}

Multiplying both sides by $f\left(x\right)$ and rearranging gives:

f^{-1}\left(x\right)\equiv f^{-1}_{0}\left(x\right)\left(2-f\left(x\right)f^{-1}_{0}\left(x\right)\right) \pmod{x^{n}}

The result follows by recursion.

Time complexity

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

Newton's Method

See Newton's Method.

Graeffe's Method

To find $f^{-1}(x)\bmod x^{2n}$ , consider

\begin{aligned} f^{-1}(x)\bmod x^{2n}&= f(-x)(f(x)f(-x))^{-1}\bmod x^{2n}\\ &=f(-x)g^{-1}(x^2)\bmod x^{2n} \end{aligned}

It is enough to find $g^{-1}(x)\bmod x^n$ to recover $g^{-1}(x^2)\bmod x^{2n}$ , because $f(x)f(-x)$ is an even function. The time complexity is the same as above.

Code

Polynomial Inversion

constexpr int MAXN = 262144;
constexpr int mod = 998244353;

using i64 = long long;
using poly_t = int[MAXN];
using poly = int *const;

void polyinv(const poly &h, const int n, poly &f) {
  /* f = 1 / h = f_0 (2 - f_0 h) */
  static poly_t inv_t;
  std::fill(f, f + n + n, 0);
  f[0] = fpow(h[0], mod - 2);
  for (int t = 2; t <= n; t <<= 1) {
    const int t2 = t << 1;
    std::copy(h, h + t, inv_t);
    std::fill(inv_t + t, inv_t + t2, 0);

    DFT(f, t2);
    DFT(inv_t, t2);
    for (int i = 0; i != t2; ++i)
      f[i] = (i64)f[i] * sub(2, (i64)f[i] * inv_t[i] % mod) % mod;
    IDFT(f, t2);

    std::fill(f + t, f + t2, 0);
  }
}

Example

Counting labeled simple undirected connected graphs: 「POJ 1737」Connected Graph

Polynomial Square Roots

Given a polynomial $g\left(x\right)$ , find $f\left(x\right)$ satisfying:

f^{2}\left(x\right)\equiv g\left(x\right) \pmod{x^{n}}

Solution

Doubling Method

First consider the case where $\left[x^0\right]g(x)$ is not $0$ .

It is easy to see that:

\left[x^0\right]f(x) = \sqrt{\left[x^0\right]g(x)}

If $\left[x^0\right]g(x)$ has no square root, then the polynomial $g(x)$ has no square root.

$\left[x^0\right]g(x)$ may have multiple square roots; choosing different roots yields different $f(x)$ .

Assume that the square root $f_{0}\left(x\right)$ of $g\left(x\right)$ modulo $x^{\left\lceil\frac{n}{2}\right\rceil}$ has already been found. Then:

\begin{aligned} f_{0}^{2}\left(x\right)&\equiv g\left(x\right) &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ f_{0}^{2}\left(x\right)-g\left(x\right)&\equiv 0 &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ \left(f_{0}^{2}\left(x\right)-g\left(x\right)\right)^{2}&\equiv 0 &\pmod{x^{n}}\\ \left(f_{0}^{2}\left(x\right)+g\left(x\right)\right)^{2}&\equiv 4f_{0}^{2}\left(x\right)g\left(x\right) &\pmod{x^{n}}\\ \left(\frac{f_{0}^{2}\left(x\right)+g\left(x\right)}{2f_{0}\left(x\right)}\right)^{2}&\equiv g\left(x\right) &\pmod{x^{n}}\\ \frac{f_{0}^{2}\left(x\right)+g\left(x\right)}{2f_{0}\left(x\right)}&\equiv f\left(x\right) &\pmod{x^{n}}\\ 2^{-1}f_{0}\left(x\right)+2^{-1}f_{0}^{-1}\left(x\right)g\left(x\right)&\equiv f\left(x\right) &\pmod{x^{n}} \end{aligned}

Compute by doubling.

Time complexity

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

Another version with a smaller constant is to maintain $f^{-1}\left(x\right)$ alongside $f\left(x\right)$ during doubling, instead of computing the inverse each time.

When $\left[x^{0}\right]g\left(x\right)\neq 1$ , a quadratic-residue algorithm may be needed to compute $\left[x^{0}\right]f\left(x\right)$ .

The method above requires knowing the inverse of $f_{0}(x)$ , so its constant term cannot be $0$ .

If $\left[x^0\right]g(x) = 0$ , factor $g(x)$ as $x^{k}h(x)$ , where $\left[x^0\right]h(x) \not = 0$ .

If $k$ is odd, then $g(x)$ has no square root.
If $k$ is even, find the square root $\sqrt{h(x)}$ of $h(x)$ , and then obtain $f(x) \equiv x^{k/2} \sqrt{h(x)} \pmod{x^{n}}$ .

Luogu Template Problem P5205 【Template】Polynomial Square Root Reference Code

#include <algorithm>
#include <iostream>
using namespace std;

constexpr int MAXN = 1 << 20, mod = 998244353;

int a[MAXN], b[MAXN], g[MAXN], gg[MAXN];

int qpow(int x, int y) {  // fast exponentiation
  int ans = 1;

  while (y) {
    if (y & 1) {
      ans = (long long)1 * ans * x % mod;
    }
    x = (long long)1 * x * x % mod;
    y >>= 1;
  }
  return ans;
}

int inv2 = qpow(2, mod - 2);  // modular inverse

void change(int *f, int len) {
  for (int i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) {
      swap(f[i], f[j]);
    }

    int k = len / 2;
    while (j >= k) {
      j -= k;
      k /= 2;
    }
    if (j < k) {
      j += k;
    }
  }
}

void NTT(int *f, int len, int type) {  // NTT
  change(f, len);

  for (int q = 2; q <= len; q <<= 1) {
    int nxt = qpow(3, (mod - 1) / q);
    for (int i = 0; i < len; i += q) {
      int w = 1;

      for (int k = i; k < i + (q >> 1); k++) {
        int x = f[k];
        int y = (long long)1 * w * f[k + (q / 2)] % mod;

        f[k] = (x + y) % mod;
        f[k + (q / 2)] = (x - y + mod) % mod;
        w = (long long)1 * w * nxt % mod;
      }
    }
  }

  if (type == -1) {
    reverse(f + 1, f + len);
    int iv = qpow(len, mod - 2);

    for (int i = 0; i < len; i++) {
      f[i] = (long long)1 * f[i] * iv % mod;
    }
  }
}

void inv(int deg, int *f, int *h) {  // polynomial inversion
  if (deg == 1) {
    h[0] = qpow(f[0], mod - 2);
    return;
  }

  inv((deg + 1) >> 1, f, h);

  int len = 1;
  while (len < deg * 2) {  // doubling
    len *= 2;
  }

  copy(f, f + deg, gg);
  fill(gg + deg, gg + len, 0);

  NTT(gg, len, 1);
  NTT(h, len, 1);
  for (int i = 0; i < len; i++) {
    h[i] = (long long)1 * (2 - (long long)1 * gg[i] * h[i] % mod + mod) % mod *
           h[i] % mod;
  }

  NTT(h, len, -1);
  fill(h + deg, h + len, 0);
}

int n, t[MAXN];

// deg: degree
// f: input array whose square root is taken
// h: output array
void sqrt(int deg, int *f, int *h) {
  if (deg == 1) {
    h[0] = 1;
    return;
  }

  sqrt((deg + 1) >> 1, f, h);

  int len = 1;
  while (len < deg * 2) {  // doubling
    len *= 2;
  }
  fill(g, g + len, 0);
  inv(deg, h, g);
  copy(f, f + deg, t);
  fill(t + deg, t + len, 0);
  NTT(t, len, 1);
  NTT(g, len, 1);
  NTT(h, len, 1);

  for (int i = 0; i < len; i++) {
    h[i] = (long long)1 * inv2 *
           ((long long)1 * h[i] % mod + (long long)1 * g[i] * t[i] % mod) % mod;
  }
  NTT(h, len, -1);
  fill(h + deg, h + len, 0);
}

int main() {
  cin >> n;

  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sqrt(n, a, b);

  for (int i = 0; i < n; i++) {
    cout << b[i] << ' ';
  }

  return 0;
}

Newton's Method

See Newton's Method.

Example

「Codeforces Round #250」E. The Child and Binary Tree

Polynomial Division & Modulo

Given polynomials $f\left(x\right),g\left(x\right)$ , find the quotient $Q\left(x\right)$ and remainder $R\left(x\right)$ when $f\left(x\right)$ is divided by $g\left(x\right)$ .

Solution

If the effect of $R\left(x\right)$ can be eliminated, the problem can be solved directly using polynomial inversion.

Consider constructing the transformation

f^{R}\left(x\right)=x^{\operatorname{deg}{f}}f\left(\frac{1}{x}\right)

It can be seen that this simply reverses the coefficients of $f\left(x\right)$ .

Let $n=\operatorname{deg}{f},m=\operatorname{deg}{g}$ .

Substitute $\frac{1}{x}$ for $x$ in $f\left(x\right)=Q\left(x\right)g\left(x\right)+R\left(x\right)$ and multiply both sides by $x^{n}$ to obtain:

\begin{aligned} x^{n}f\left(\frac{1}{x}\right)&=x^{n-m}Q\left(\frac{1}{x}\right)x^{m}g\left(\frac{1}{x}\right)+x^{n-m+1}x^{m-1}R\left(\frac{1}{x}\right)\\ f^{R}\left(x\right)&=Q^{R}\left(x\right)g^{R}\left(x\right)+x^{n-m+1}R^{R}\left(x\right) \end{aligned}

Notice that the coefficients of $R^{R}\left(x\right)$ in the equation above are multiplied by $x^{n-m+1}$ , so taking the equation modulo $x^{n-m+1}$ eliminates the effect of $R^{R}\left(x\right)$ .

Also, since the degree of $Q^{R}\left(x\right)$ is $\left(n-m\right)<\left(n-m+1\right)$ , $Q^{R}\left(x\right)$ is unaffected.

Therefore:

f^{R}\left(x\right)\equiv Q^{R}\left(x\right)g^{R}\left(x\right)\pmod{x^{n-m+1}}

The quotient $Q\left(x\right)$ can be obtained using polynomial inversion, and substituting it back gives $R\left(x\right)$ .

Time complexity $O\left(n\log{n}\right)$ .

Polynomial Logarithms & Exponentials

Given a polynomial $f(x)$ , find $\ln{f(x)}$ and $\exp{f(x)}$ modulo $x^{n}$ .

Solution

Ordinary Method

Polynomial logarithm

First, for a polynomial $f(x)$ , if $\ln{f(x)}$ exists, then by its definition, it must satisfy:

[x^{0}]f(x)=1

Differentiate $\ln{f(x)}$ and then integrate to obtain:

\begin{aligned} \frac{\mathrm{d} \ln{f(x)}}{\mathrm{d} x} & \equiv \frac{f'(x)}{f(x)} & \pmod{x^{n}} \\ \ln{f(x)} & \equiv \int \mathrm{d} \ln{f(x)} \equiv \int\frac{f'(x)}{f(x)} \mathrm{d} x & \pmod{x^{n}} \end{aligned}

Polynomial differentiation and integration each take $O(n)$ time, while inversion takes $O(n\log{n})$ time. Thus, computing the polynomial $\ln$ takes $O(n\log{n})$ time.

Polynomial exponential

First, for a polynomial $f(x)$ , if $\exp{f(x)}$ exists, then it must satisfy:

[x^{0}]f(x)=0

Otherwise, the constant term of $\exp{f(x)}$ does not converge.

Differentiate $\exp{f(x)}$ to obtain:

\frac{\mathrm{d} \exp{f(x)}}{\mathrm{d} x} \equiv \exp{f(x)}f'(x)\pmod{x^{n}}

Comparing coefficients on both sides gives:

[x^{n-1}]\frac{\mathrm{d} \exp{f(x)}}{\mathrm{d} x} = \sum_{i = 0}^{n - 1} \left([x^{i}]\exp{f(x)}\right) \left([x^{n-i-1}]f'(x)\right)

n[x^{n}]\exp{f(x)} = \sum_{i = 0}^{n - 1} \left([x^{i}]\exp{f(x)}\right) \left((n - i)[x^{n - i}]f(x)\right)

This can be solved using divide-and-conquer FFT.

Time complexity $O(n\log^{2}{n})$ .

Newton's Method

Using Newton's Method, the polynomial $\exp$ can be computed in $O(n\log{n})$ time.

Code

Polynomial ln/exp

constexpr int MAXN = 262144;
constexpr int mod = 998244353;

using i64 = long long;
using poly_t = int[MAXN];
using poly = int *const;

void derivative(const poly &h, const int n, poly &f) {
  for (int i = 1; i != n; ++i) f[i - 1] = (i64)h[i] * i % mod;
  f[n - 1] = 0;
}

void integrate(const poly &h, const int n, poly &f) {
  for (int i = n - 1; i; --i) f[i] = (i64)h[i - 1] * inv[i] % mod;
  f[0] = 0; /* C */
}

void polyln(const poly &h, const int n, poly &f) {
  /* f = ln h = ∫ h' / h dx */
  assert(h[0] == 1);
  static poly_t ln_t;
  const int t = n << 1;

  derivative(h, n, ln_t);
  std::fill(ln_t + n, ln_t + t, 0);
  polyinv(h, n, f);

  DFT(ln_t, t);
  DFT(f, t);
  for (int i = 0; i != t; ++i) ln_t[i] = (i64)ln_t[i] * f[i] % mod;
  IDFT(ln_t, t);

  integrate(ln_t, n, f);
}

void polyexp(const poly &h, const int n, poly &f) {
  /* f = exp(h) = f_0 (1 - ln f_0 + h) */
  assert(h[0] == 0);
  static poly_t exp_t;
  std::fill(f, f + n + n, 0);
  f[0] = 1;
  for (int t = 2; t <= n; t <<= 1) {
    const int t2 = t << 1;

    polyln(f, t, exp_t);
    exp_t[0] = sub(pls(h[0], 1), exp_t[0]);
    for (int i = 1; i != t; ++i) exp_t[i] = sub(h[i], exp_t[i]);
    std::fill(exp_t + t, exp_t + t2, 0);

    DFT(f, t2);
    DFT(exp_t, t2);
    for (int i = 0; i != t2; ++i) f[i] = (i64)f[i] * exp_t[i] % mod;
    IDFT(f, t2);

    std::fill(f + t, f + t2, 0);
  }
}

Example

Computing $f^{k}(x)$

The ordinary approach is polynomial fast exponentiation, with time complexity $O(n\log{n}\log{k})$ .

When $[x^{0}]f(x)=1$ , we have:
$f^{k}(x)=\exp{\left(k\ln{f(x)}\right)}$
When $[x^{0}]f(x)\neq 1$ , let the lowest-degree term of $f(x)$ be $f_{i}x^{i}$ . Then:
$f^{k}(x)=f_{i}^{k}x^{ik}\exp{\left(k\ln{\frac{f(x)}{f_{i}x^{i}}}\right)}$
Time complexity $O(n\log{n})$ .

Polynomial Trigonometric Functions

Given a polynomial $f\left(x\right)$ , find $\sin{f\left(x\right)}, \cos{f\left(x\right)}$ , and $\tan{f\left(x\right)}$ modulo $x^{n}$ .

Solution

By Euler's formula, $\left(\mathrm{e}^{\mathrm{i}x} = \cos{x} + \mathrm{i}\sin{x}\right)$ , we obtain another expression for trigonometric functions:

\begin{aligned} \sin{x} &= \frac{\mathrm{e}^{\mathrm{i}x} - \mathrm{e}^{-\mathrm{i}x}}{2\mathrm{i}} \\ \cos{x} &= \frac{\mathrm{e}^{\mathrm{i}x} + \mathrm{e}^{-\mathrm{i}x}}{2} \end{aligned}

Substituting $f\left(x\right)$ gives:

\begin{aligned} \sin{f\left(x\right)} &= \frac{\exp{\left(\mathrm{i}f\left(x\right)\right)} - \exp{\left(-\mathrm{i}f\left(x\right)\right)}}{2\mathrm{i}} \\ \cos{f\left(x\right)} &= \frac{\exp{\left(\mathrm{i}f\left(x\right)\right)} + \exp{\left(-\mathrm{i}f\left(x\right)\right)}}{2} \end{aligned}

Writing a program directly from the expressions above gives $\sin{f\left(x\right)}$ and $\cos{f\left(x\right)}$ modulo $x^{n}$ . Then $\tan{f\left(x\right)}$ can be obtained from $\tan{f\left(x\right)} = \frac{\sin{f\left(x\right)}}{\cos{f\left(x\right)}}$ .

Code

Polynomial Trigonometric Functions

Notice that we are performing NTT over $\mathbb{Z}_{998244353}$ . Accordingly, the imaginary unit $\mathrm{i}$ should be replaced by $86583718$ or $911660635$ :

\begin{aligned} & \mathrm{i} = \sqrt{-1} \equiv \sqrt{998244352} \pmod{998244353} \\ \implies & \phantom{\text{or}} \quad \mathrm{i} \equiv 86583718 \pmod{998244353} \\ & \text{or} \quad \mathrm{i} \equiv 911660635 \pmod{998244353} \end{aligned}

constexpr int MAXN = 262144;
constexpr int mod = 998244353;
constexpr int imgunit = 86583718; /* sqrt(-1) = sqrt(998233452) */

using i64 = long long;
using poly_t = int[MAXN];
using poly = int *const;

void polytri(const poly &h, const int n, poly &sin_t, poly &cos_t) {
  /* sin(f) = (exp(i * f) - exp(- i * f)) / 2i */
  /* cos(f) = (exp(i * f) + exp(- i * f)) / 2 */
  /* tan(f) = sin(f) / cos(f) */
  assert(h[0] == 0);
  static poly_t tri1_t, tri2_t;

  for (int i = 0; i != n; ++i) tri2_t[i] = (i64)h[i] * imgunit % mod;
  polyexp(tri2_t, n, tri1_t);
  polyinv(tri1_t, n, tri2_t);

  if (sin_t != nullptr) {
    const int invi = fpow(pls(imgunit, imgunit), mod - 2);
    for (int i = 0; i != n; ++i)
      sin_t[i] = (i64)(tri1_t[i] - tri2_t[i] + mod) * invi % mod;
  }
  if (cos_t != nullptr) {
    for (int i = 0; i != n; ++i) cos_t[i] = div2(pls(tri1_t[i], tri2_t[i]));
  }
}

Polynomial Inverse Trigonometric Functions

Given a polynomial $f\left(x\right)$ , find $\arcsin{f\left(x\right)}, \arccos{f\left(x\right)}$ , and $\arctan{f\left(x\right)}$ modulo $x^{n}$ .

Solution

Following the method for computing the polynomial $\ln$ , differentiate and then integrate the inverse trigonometric functions to obtain:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x} \arcsin{x} &= \frac{1}{\sqrt{1 - x^{2}}} \\ \arcsin{x} &= \int \frac{1}{\sqrt{1 - x^{2}}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arccos{x} &= - \frac{1}{\sqrt{1 - x^{2}}} \\ \arccos{x} &= - \int \frac{1}{\sqrt{1 - x^{2}}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arctan{x} &= \frac{1}{1 + x^{2}} \\ \arctan{x} &= \int \frac{1}{1 + x^{2}} \mathrm{d} x \end{aligned}

Substituting $f\left(x\right)$ gives:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x} \arcsin{f\left(x\right)} &= \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \\ \arcsin{f\left(x\right)} &= \int \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arccos{f\left(x\right)} &= - \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \\ \arccos{f\left(x\right)} &= - \int \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arctan{f\left(x\right)} &= \frac{f'\left(x\right)}{1 + f^{2}\left(x\right)} \\ \arctan{f\left(x\right)} &= \int \frac{f'\left(x\right)}{1 + f^{2}\left(x\right)} \mathrm{d} x \end{aligned}

The result follows by evaluating these expressions directly.

Code

Polynomial Inverse Trigonometric Functions

constexpr int MAXN = 262144;
constexpr int mod = 998244353;

using i64 = long long;
using poly_t = int[MAXN];
using poly = int *const;

void derivative(const poly &h, const int n, poly &f) {
  for (int i = 1; i != n; ++i) f[i - 1] = (i64)h[i] * i % mod;
  f[n - 1] = 0;
}

void integrate(const poly &h, const int n, poly &f) {
  for (int i = n - 1; i; --i) f[i] = (i64)h[i - 1] * inv[i] % mod;
  f[0] = 0; /* C */
}

void polyarcsin(const poly &h, const int n, poly &f) {
  /* arcsin(f) = ∫ f' / sqrt(1 - f^2) dx  */
  static poly_t arcsin_t;
  const int t = n << 1;
  std::copy(h, h + n, arcsin_t);
  std::fill(arcsin_t + n, arcsin_t + t, 0);

  DFT(arcsin_t, t);
  for (int i = 0; i != t; ++i) arcsin_t[i] = sqr(arcsin_t[i]);
  IDFT(arcsin_t, t);

  arcsin_t[0] = sub(1, arcsin_t[0]);
  for (int i = 1; i != n; ++i)
    arcsin_t[i] = arcsin_t[i] ? mod - arcsin_t[i] : 0;

  polysqrt(arcsin_t, n, f);
  polyinv(f, n, arcsin_t);
  derivative(h, n, f);

  DFT(f, t);
  DFT(arcsin_t, t);
  for (int i = 0; i != t; ++i) arcsin_t[i] = (i64)f[i] * arcsin_t[i] % mod;
  IDFT(arcsin_t, t);

  integrate(arcsin_t, n, f);
}

void polyarccos(const poly &h, const int n, poly &f) {
  /* arccos(f) = - ∫ f' / sqrt(1 - f^2) dx  */
  polyarcsin(h, n, f);
  for (int i = 0; i != n; ++i) f[i] = f[i] ? mod - f[i] : 0;
}

void polyarctan(const poly &h, const int n, poly &f) {
  /* arctan(f) = ∫ f' / (1 + f^2) dx  */
  static poly_t arctan_t;
  const int t = n << 1;
  std::copy(h, h + n, arctan_t);
  std::fill(arctan_t + n, arctan_t + t, 0);

  DFT(arctan_t, t);
  for (int i = 0; i != t; ++i) arctan_t[i] = sqr(arctan_t[i]);
  IDFT(arctan_t, t);

  inc(arctan_t[0], 1);
  std::fill(arctan_t + n, arctan_t + t, 0);

  polyinv(arctan_t, n, f);
  derivative(h, n, arctan_t);

  DFT(f, t);
  DFT(arctan_t, t);
  for (int i = 0; i != t; ++i) arctan_t[i] = (i64)f[i] * arctan_t[i] % mod;
  IDFT(arctan_t, t);

  integrate(arctan_t, n, f);
}

References and Links

Footnotes

Elementary function——Wikipedia ↩

Elementary Polynomial Functions

Polynomial Inversion

Solution

Doubling Method

Newton's Method

Graeffe's Method

Code

Example

Polynomial Square Roots

Solution

Doubling Method

Newton's Method

Example

Polynomial Division & Modulo

Solution

Polynomial Logarithms & Exponentials

Solution

Ordinary Method

Newton's Method

Code

Example

Polynomial Trigonometric Functions

Solution

Code

Polynomial Inverse Trigonometric Functions

Solution

Code

References and Links

Footnotes