多項式除法

給定多項式$F(x)$和$G(x)$，求$F(x)$除以$G(x)$的商$Q(x)$和餘數$R(x)$。即
$$F(x)=Q(x)G(x)+R(x)Deg(R)<Deg(G)<Deg(F)$$
首先構造
$$F^R(x)=x^{Deg(F)}F(\frac{1}{x})$$
這個東西就是將係數反轉。比如
$$\begin{array}{rl}F(x)& =x^4+2x^3+3x^2+4x+5\\ F^R(x)& =5x^4+4x^3+3x^2+2x1\end{array}$$
令$n=Deg(F),m=Deg(G)$，則$n-m=Deg(Q)$
那麼就有
$$\begin{array}{rl}F(x)& =Q(x)G(x)+R(x)\\ x^{n}F(x)& =x^{n}Q(x)G(x)+x^{n}R(x)\\ x^{n}F(\frac{1}{x})& =x^{n-m}Q(\frac{1}{x})x^{m}G(\frac{1}{x})+x^{n-m+1}{x}^{m-1}R(\frac{1}{x})\\ F^R(x)& =Q^R(x)G^R(x)+x^{n-m+1}{x}^{m-1}R(\frac{1}{x})\\ F^R(x)& \equiv Q^R(x)G^R(x)mod{x}^{n-m+1}\\ F^R(x)[G^R(x){]}^{-1}& \equiv Q^R(x)mod{x}^{n-m+1}\end{array}$$
現在我們求出了在模$x^{n-m+1}$意義下$Q^R(x)$的解，則
$$Q^R(x)=F^R(x)[G^R(x){]}^{-1}mod{x}^{n-m+1}+a_{n-m+1}{x}^{n-m+1}+a_{n-m+2}{x}^{n-m+2}+...$$
但又因為$Deg(G^R)=Deg(G)=n-m<n-m+1$，所以
$$Q^R(x)=F^R(x)[G^R(x){]}^{-1}mod{x}^{n-m+1}$$