solve-play with math # ENIGMATH on spoj

ENIGMATH - PLAY WITH MATH

You would have been fed up with competitive programming questions so far, now it is time to solve little math.

Assume you have a equation A * x - B * y = 0

For a given value of A and B, find the minimum positive integer value of x and y that satisfies this equation.

Input

First line contains T, number of test cases 0 <= T <=1000 followed by T lines.

First line of each test case contains two space seperated integers A and B. 1 <= A, B <=1 000 000 000.

Output

For each test case, output a single line containing two integers x and y (seperated by a single space)

. 

SOLUTION-

#include<stdio.h>

int main()

{ long long int t,a,b,c,d,m,n,l,x;

scanf("%lld",&t);

while(t--)

{

scanf("%lld%lld",&a,&b);

m=a;

n=b;

while(b!=0)

{

x=a%b;

a=b;

b=x;

}

l=(m*n)/a;

c=l/m;

d=l/n;

printf("%lld %lld",c,d);

printf("\n");

}

}