HackerEarth In an array problem solution

In this HackerEarth In an array problem solution, You are given an array A of size N, where the ith integer of the array is A[i] and its value ranges between 1 and 1000 inclusive. You are required to complete the following task:

Assume that you are provided with 3 additional numbers K, X, and Y. Your task is to report the number of unordered pairs of elements (i,j) from this array, such that (1 <= i < j <= N), (A[i] + A[j])%K = X, and (A[i] x A[j])%K = Y.

import java.io.*;
import java.util.*;
public final class checkpoint_b
{
    static BufferedReader br;
    static FastScanner sc;
    static PrintWriter out;
    static Random rnd=new Random();
    static long[] cnt;
    static int maxn=(int)(1005);
    
    static void init(String curr) throws Exception
    {
        br=new BufferedReader(new FileReader(new File("in"+curr+".txt")));
        sc=new FastScanner(br);
        out=new PrintWriter(new FileWriter("out"+curr+".txt"));
    }
    
    public static void main(String args[]) throws Exception
    {
        init(args[0]);int n=sc.nextInt(),k=sc.nextInt(),x=sc.nextInt(),y=sc.nextInt();int[] a=new int[n];cnt=new long[maxn];
        for(int i=0;i<n;i++)
        {
            a[i]=sc.nextInt();cnt[a[i]]++;
        }
        long res=0;
        for(int i=1;i<maxn;i++)
        {
            for(int j=i;j<maxn;j++)
            {
                int val1=(i+j)%k,val2=(i*j)%k;
                if(val1==x && val2==y)
                {
                    if(i==j)
                    {
                        long curr=(cnt[i]*(cnt[i]-1L))/2L;res=res+curr;
                    }
                    else
                    {
                        long curr=(cnt[i]*cnt[j]);res=res+curr;
                    }
                }
            }
        }
        out.println(res);out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
    
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
    
    public String next() throws Exception {
        return nextToken().toString();
    }
    
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}