extended euclidean algorithm in java

1int gcd(int a, int b, int& x, int& y) {
2    if (b == 0) {
3        x = 1;
4        y = 0;
5        return a;
6    }
7    int x1, y1;
8    int d = gcd(b, a % b, x1, y1);
9    x = y1;
10    y = x1 - y1 * (a / b);
11    return d;
12}
13

1public class Main    
2{    
3public static void main (String args[])   
4{   
5    @SuppressWarnings("resource")    
6    System.out.println("How many times you would like to try ?")
7    Scanner read = new Scanner(System.in);    
8    int len = read.nextInt();    
9
10    for(int w = 0; w < len; w++)    
11    {
12        System.out.print("Please give the numbers seperated by space: ")
13        read.nextLine();
14        long tmp = read.nextLong();
15        long m = read.nextLong();
16        long n;
17        if (m < tmp) {      
18            n = m;
19            m = tmp;
20        }
21        else {
22            n = tmp;
23        }
24
25        long[] l1 = {m, 1, 0};
26        long[] l2 = {n, 0, 1};
27        long[] l3 = new long[3]; 
28
29        while (l1[0]-l2[0]*(l1[0]/l2[0]) > 0) {
30            for (int j=0;j<3;j++) l3[j] = l2[j]; 
31            long q = l1[0]/l2[0];        
32            for (int i = 0; i < 3; i++) {
33            l2[i] = (l1[i]-l2[i]*q);
34            }
35
36            for (int k=0;k<3;k++) l1[k] = l3[k];
37        }
38
39        System.out.printf("%d %d %d",l2[1],l2[2],l2[0]); // first two Bezouts identity Last One gcd
40    }
41}
42}    
43