qq_51509008的博客

gcd(A,0)=Agcd(A,0)=Agcd(A,0)=A 则有A∗x+0∗y=AA*x+0*y=AA∗x+0∗y=A 所以在递归终点 x为1，y可取任意数 $ gcd(a,b)=gcd(b,a;mod;b)$ a∗x1+b∗y1=b∗x0+(a−[ab]∗b)∗y0a*x_1+b*y_1=b*x_0+(a-[\frac{a}{b}]*b)*y_0a∗x1​+b∗y1​=b∗x0​+(a−[ba​]∗b)∗y0​ 整理得 a∗(x1−y0)+b∗(y1+[ab]∗y0−x0)=0a*(x_1-y_0)