多项式乘法逆

多项式乘法逆

给定\(F(x)\)

求\(G(x)\)满足

\[G(x)F(x)\equiv 1\ (mod\ x^n)\\ \]

假设已知

\[H(x)F(x)\equiv 1\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\ F(x)(G(x)-H(x))\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\ G(x)-H(x)\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\ (G(x)-H(x))^2\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\ G^2(x)+H^2(x)-2*G(x)H(x)\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\ \]

两边同乘\(F(x)\)，且\(F(x)G(x)\equiv 1\)

\[G(x)=2H(x)-H(x)^2F(x)\ (mod\ x^n) \]

或硬上牛顿迭代

\[F(G(x))=\frac{1}{G_0(x)}-F(x)\equiv 0\ (mod\ x^n)\\ G(x)=G_0(x)-\frac{\frac{1}{G_0(x)}-F(x)}{-\frac{1}{G_0^2(x)}}\\ G(x)=2G_0(x)-F(x)G_0^2(x) \]