# Time limit exceeded in hackerearth exercise

I’m trying to solve this exercise in HackerEarth. But I have an error of time limit exceeded. This is the code that I wrote :

```import java.io.BufferedReader;
import java.util.*;

class TestClass {
//gcd
public static long gcd(long num1, long num2) {
if (num2 != 0) {
return gcd(num2, num1 % num2);
} else {
return num1;
}
}

public static void main(String args[] ) throws Exception {
int T = Integer.parseInt(br.readLine()); // Reading input from STDIN
while (T-- > 0) {
StringTokenizer st1 = new StringTokenizer(br.readLine());
long a = Long.parseLong(st1.nextToken());
long b = Long.parseLong(st1.nextToken());
long A = a/gcd(a,b);
long B = b/gcd(a,b);
System.out.printf("%d%1s%d%n",B,"",A);
}
}
}
```

Your solution is a little bit slow because of the bad implementation. I rewrote your solution in better implementation with the same logic and time complexity of your solution and get Accepted and none of the test cases exceeded .8 second

```import java.util.*;

class TestClass {
// Same gcd function but it's better code :)
public static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}

public static void main(String args[] ) throws Exception {
Scanner s = new Scanner(System.in);
int t = s.nextInt();  // this way of reading input is faster alot.
while(t-- > 0) {
int a = s.nextInt();
int b = s.nextInt();  // No need to use long it's  just 1e9

int tmp = gcd(a, b);  // It's better to save the value of the gcd(a, b) instead of calculate it twice.

int A = a/tmp;
int B = b/tmp;
System.out.println(B+" "+A);
}
}
}
```