# Euclid problem

Euclid problem

From Euclid, it is known that for any positive integers A and B there exist such integers X and Y that AX + BY = D, where D is the greatest common divisor of A and B. The problem is to find the corresponding X, Y, and D for a given A and B.

The input will consist of a set of lines with the integer numbers A and B, separated with space ( A, B < 1, 000, 000, 001).

For each input line the output line should consist of three integers X, Y, and D, separated with space. If there are several such X and Y, you should output that pair for which X<=Y and | X| + | Y| is minimal.

``````4 6
17 17

``````

``````-1 1 2
0 1 17

``````

``````#include"stdio.h"
#include"math.h"
long gcd(long p,long q,long *x,long *y)
{
long x1,y1;
long g;
if(q>p)
return gcd(q,p,y,x);
if(q==0)
{
*x=1;*y=0;
return p;
}
g=gcd(q,p%q,&x1,&y1);
*x=y1;
*y=(x1-floor(p/q)*y1);
return g;
}
int main()
{
long a,b,x,y,c;
while(scanf("%ld%ld",&a,&b)!=EOF)
{
c=gcd(a,b,&x,&y);
printf("%ld %ld %ld\n",x,y,c);
}
return 0;
}
``````

``````#include"stdio.h"
#include"math.h"
long gcd(long p,long q,long *x,long *y)
{
long x1,y1;
long g;
if(q>p)
return gcd(q,p,y,x);
if(q==0)
{
*x=1;*y=0;
return p;
}
g=gcd(q,p%q,&x1,&y1);
*x=y1;
*y=(x1-floor(p/q)*y1);
return g;
}
int main()
{
long a,b,x,y,c;
while(scanf("%ld%ld",&a,&b)!=EOF)
{
c=gcd(a,b,&x,&y);
printf("%ld %ld %ld\n",x,y,c);
}
return 0;
}
``````