exgcd学习笔记

目录

• 扩欧用途
• 扩欧推导过程

本文为oiwiki内容整理，主要是过一遍思路，也权当LaTex数学公式练习，侵权紫衫
LaTex公式表
oiwiki上的扩欧

扩欧用途

扩欧(exgcd)，常用来解决形如$$ax+by=gcd(a,b),其中\ a,b,x,y\in\Z$$的方程可行解。

扩欧推导过程

\[Let\: \begin{cases} ax_1+by_1=\gcd(a,b)\\ bx_2+(a\bmod b)y_2=\gcd(b,a\bmod b) \end{cases}\\ \]

\[\begin{align*} \because \gcd (a,b)&=\gcd (b,a\bmod b)\\ \therefore ax_1+by_1&=bx_2+(a\bmod b)y_2\\ \text{and that}\,\,\,a\bmod b&=a-(\lfloor{\dfrac{a}{b}}\rfloor\times b)\\ \therefore ax_1+by_1&=ay_2+b(x_2-\lfloor{\dfrac{a}{b}}\rfloor y_2)\\ \because a=&a,b=b\\ \therefore\begin{cases} x_1=y_2\\y_1=x_2-\lfloor{\dfrac{a}{b}}\rfloor y_2\end{cases} &\text{ is a legit solution for recursing} \end{align*} \]

Therefore, we just have to keep recursing until \(b=0\) ,and we have \(\begin{cases}x_0=1\\y_0=0\end{cases}\) as the boundary of recursion.
To solve for the general solution, let \(\begin{cases}x=x_0\\y=y_0\end{cases}\) be the special solution we have found.
Then, the general solution shall be:

\[\begin{cases} x=x_0+\frac{b}{\gcd(a,b)}\\ y=y_0-\frac{a}{\gcd(a,b)} \end{cases} \]

Proof:
It is obvious that \(\forall x,y \text{ mentioned above},\text{LHS(left-hand-side)}=ax+by=ax_0+by_0+\frac{ab}{\gcd(a,b)}-\frac{ab}{\gcd(a,b)}=\text{RHS}\)

\[Q.E.D. \]