多项式求逆
问题
给定一个多项式\(F(x)\) ,请求出一个多项式\(G(x)\),满足\(F(x)\times G(x) \equiv 1(\mathrm{mod\:}x^n)\)
推导
假设我们已经求得了\(G_0(x)\)满足\(F(x)\times G_0(x) \equiv 1(\mathrm{mod\:}x^\frac{n}{2})\),现在求\(F(x)\times G(x) \equiv 1(\mathrm{mod\:}x^n)\)。
\[\begin{aligned}
F(x)\times G_0(x) &\equiv 1(\mathrm{mod\:}x^\frac{n}{2})\\
F(x)\times G_0(x)-1 &\equiv 0(\mathrm{mod\:}x^\frac{n}{2})\\
(F(x)\times G_0(x)-1)^2 &\equiv 0(\mathrm{mod\:}x^n)\\
(F(x)\times G_0(x))^2+1-2 \times F(x)\times G_0(x) &\equiv 0(\mathrm{mod\:}x^n)\\
F(x)^2\times G_0(x)^2+1-2 \times F(x)\times G_0(x) &\equiv 0(\mathrm{mod\:}x^n)\\
F(x)^2\times G_0(x)^2-2 \times F(x)\times G_0(x) &\equiv -1(\mathrm{mod\:}x^n)\\
2 \times F(x)\times G_0(x)-F(x)^2\times G_0(x)^2 &\equiv 1(\mathrm{mod\:}x^n)\\
F(x)\times(2 \times G_0(x)-F(x)\times G_0(x)^2) &\equiv 1(\mathrm{mod\:}x^n)
\end{aligned}\]
所以我们在已知\(G_0(x)\)的前提下,只要求一下\(2 \times G_0(x)-F(x)\times G_0(x)^2)\)就可以了。
时间复杂度
\(O(n\log n)\)