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Inv

描述

给定多项式 $f\left(x\right)$，求 $f^{-1}\left(x\right)$。

解法

倍增法

首先，易知

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

假设现在已经求出了 $f\left(x\right)$ 在模 $x^{\left\lceil\frac{n}{2}\right\rceil}$ 意义下的逆元 $f^{-1}_{0}\left(x\right)$。 有：

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

两边平方可得：

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

两边同乘 $f\left(x\right)$ 并移项可得：

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

递归计算即可。

时间复杂度

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

Newton's Method

参见 Newton's Method.

Graeffe 法

欲求 $f^{-1}(x)\bmod x^{2n}$ 考虑

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

只需求出 $g^{-1}(x)\bmod x^n$ 即可还原出 $g^{-1}(x^2)\bmod x^{2n}$ 因为 $f(x)f(-x)$ 是偶函数，时间复杂度同上。

代码

??? "多项式求逆" ```cpp 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);
  }
}
```

例题

  1. 有标号简单无向连通图计数：「POJ 1737」Connected Graph