Euclid Problem-ACM

算法：假设a>b, ax1+by1=gcd(a,b), 那么bx2+(a-a/b*b)y2=gcd(b,a%b). 根据欧几里德定律可知，gcd(a,b)=gcd(b,a%b), 所以ax1+by1=bx2+(a-a/b*b)y2 =>ay2+bx2-a/b*by2. 又因为相应的项的系数是相等的，所以x1=y2, y1=x2-a/by2, 以些类推，运用递归方法，直到b的系数为0为止。

// Euclid Problem.cpp : Defines the entry point for the console application.//#include "stdafx.h"using namespace std; int x,y;int _tmain(int argc, _TCHAR* argv[]){void Euclid(unsigned int a,unsigned int b);unsigned int a=0;unsigned int b=0;while(cin>>a>>b){if(a==0&&b==0)break;else{if(b>a)swap(a,b);Euclid(a,b);cout<<x<<" "<<y<<" "<<a*x+b*y<<"/n";}}return 0;}void Euclid(unsigned int a,unsigned int b){if(b==0){x=1;y=0;}else{Euclid(b,a%b);int tmp=x;x=y;y=tmp-a/b*y;}}