拓展欧几里得(Exgcd)

ax+by=gcd(a,b)的一组整数解x_{1}y_{1}ab为常数)

首先:ax_{1}+by_{1}=gcd(a,b)

gcd(a,b)=gcd(b,a mod b)

(求解最大公约数的欧几里得算法中,我们就是利用这一原理迭代的,这里用到的是对欧几里得原理的理解)

则:bx_{2}+(a mod b)y_{2}=ax_{1}+by_{1}

即:bx_{2}+(a-b*(a/b))y_{2}=ax_{1}+by_{1}(注意“/”为整除)

则:ay_{2}+b(x_{2}-ay_{2}/b)=ax_{1}+by_{1}

\because gcd(a,b)=1

\therefore\left\{\begin{matrix} x_{1}=y_{2} & & \\ y_{1}=ay_{2}/b & & \end{matrix}\right.

所以只需不断迭代,求出最终的y_{2},即可求出x_{1},从而求出y_{1}.

详见一道例题:https://blog.csdn.net/weixin_39872717/article/details/78287843

猜你喜欢

转载自blog.csdn.net/weixin_39872717/article/details/81094722