# 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.

## HackerEarth In an array problem solution.

`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());    }}`