扩展欧几里得算法

紫书：

求直线ax+by+c=0上的整数点（x，y）；

先求ax+by+c=gcd(a,b)的解

void gcd(int a,int b,int &d,int &x,int &y)//d为a，b的最大公因数
{
    if(!b)
    {
        d=a;
        x=1;
        y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}

若方程的一组解为（x0,y0)，则通解为（x0+kb',y0+ka')（a'=a/gcd(a,b),b'=b/gcd(a,b)）;k为任意整数

例题：

Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

 

Input

The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

 

Output

output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.

 

Sample Input

 77 51
10 44
34 79 

#include<cstdio>
#include<iostream>
using namespace std;
void gcd(int a,int b,int &d,int &x,int &y)
{
    if(!b)
    {
        d=a;
        x=1;
        y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
int main()
{
    int x,y,a,b,d;
    while(~scanf("%d%d",&a,&b))
    {
        gcd(a,b,d,x,y);
        if(d!=1)
        {
            printf("sorry\n");
            continue;
        }
        else
        {
            while(x<=0)
            {
                x=x+b;
                y=y-a;
            }
            printf("%d %d\n",x,y);
        }
    }
    return 0;
}