多项式对数函数|指数函数

Description

给定多项式 f\left(x\right) ,求模 x^{n} 意义下的 \ln{f\left(x\right)} \exp{f\left(x\right)}

Methods

普通方法


首先,对于多项式 f\left(x\right) ,若 \ln{f\left(x\right)} 存在,则由其 定义 ,其必须满足:

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

\ln{f\left(x\right)} 求导再积分,可得:

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

多项式的求导,积分时间复杂度为 O\left(n\right) ,求逆时间复杂度为 O\left(n\log{n}\right) ,故多项式求 \ln 时间复杂度 O\left(n\log{n}\right)


首先,对于多项式 f\left(x\right) ,若 \exp{f\left(x\right)} 存在,则其必须满足:

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

否则 \exp{f\left(x\right)} 的常数项不收敛。

\exp{f\left(x\right)} 求导,可得:

\exp'{f\left(x\right)}\equiv\exp{f\left(x\right)}f'\left(x\right)\pmod{x^{n}}

比较两边系数可得:

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

\left[x^{0}\right]f\left(x\right)=0 ,则:

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

使用分治 FFT 即可解决。

时间复杂度 O\left(n\log^{2}{n}\right)

Newton's Method

使用 Newton's Method 即可在 O\left(n\log{n}\right) 的时间复杂度内解决多项式 \exp

Code

多项式 ln/exp
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constexpr int maxn = 262144;
constexpr int mod = 998244353;

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

inline 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;
}

inline 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);
  }
}

Examples

  1. 计算 f^{k}\left(x\right)

普通做法为多项式快速幂,时间复杂度 O\left(n\log{n}\log{k}\right)

\left[x^{0}\right]f\left(x\right)=1 时,有:

f^{k}\left(x\right)=\exp{\left(k\ln{f\left(x\right)}\right)}

\left[x^{0}\right]f\left(x\right)\neq 1 时,设 f\left(x\right) 的最低次项为 f_{i}x^{i} ,则:

f^{k}\left(x\right)=f_{i}^{k}x^{ik}\exp{\left(k\ln{\frac{f\left(x\right)}{f_{i}x^{i}}}\right)}

时间复杂度 O\left(n\log{n}\right)


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