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^{lceilfrac{n}{2}
ceil})\
F(x)(G(x)-H(x))equiv 0 (mod x^{lceilfrac{n}{2}
ceil})\
G(x)-H(x)equiv 0 (mod x^{lceilfrac{n}{2}
ceil})\
(G(x)-H(x))^2equiv 0 (mod x^{lceilfrac{n}{2}
ceil})\
G^2(x)+H^2(x)-2*G(x)H(x)equiv 0 (mod x^{lceilfrac{n}{2}
ceil})\
]
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)
]