扩展gcd求解二元不定方程及其证明

Posted 2020-09-18

#include <cstdio>
#include <iostream>

using namespace std;
/*扩展gcd证明
	由于当d = gcd(a,b)时；
	d = d1 = gcd(b,a%b);
	d1 = b1x1 + a%by1;
	d = ax+by = b1x1+a%by1。又由于a%b = a - a%b*b;
	上式变形能够有
	b1x1 + (a-b*a/b)*y1 = a*y1 + b*(x1-a/b*y1);
	也就是是说ax+by =  a*y1 + b*(x1-a/b*y1);
	所以当x=y1,y = x1-a/b*y1时。能够满足有d=ax+by; 
 */ 
int fun(int a,int b,int d,int &x,int &y){
	if(b == 0){
		x = 1;
		y = 0;
		return a;
	}
	else{
		d = fun(b,a%b,d,x,y);
		int t;
		t = x;
		x = y;
		y = t-a/b*y;
		return d;
	}
}
 
int main(){
	int a,b,d;
	cin >>a >> b >> d;
	int x,y;
	fun(a,b,d,x,y);
	printf("%d %d\n",x,y);
	return 0;
}