拓展欧几里得算法模板

版权声明：个人笔记，仅供复习 https://blog.csdn.net/weixin_41162823/article/details/83720523

代码示例：求出ax + by = c的所有解

#include<cstdio>
int exgcd(int a,int b,int& x,int& y){
	if(b == 0){
		x = 1,y = 0;
		return a;
	}
	int d = exgcd(b,a%b,x,y);
	int t = x;
	x = y,y = t - a/b*y;
	return d;
}
int main(){
	int a,b,c,x,y;
	scanf("%d%d%d",&a,&b,&c);
	int d = exgcd(a,b,x,y);
	int x0 = c/d*x;		//特解 
	int y0 = c/d*y;		//特解 
	printf("%d %d\n",x0,y0);
	for(int k = 1;k < 100;k++){
		printf("%d %d\n",x0+b/d*k,y0-a/d*k);
	} 
	return 0;
}