POJ 3579- Median

The input consists of several test cases.
In each test case, N will be given in the first line. Then N numbers are given, representing X₁, X₂, ... , X_N, ( X_i ≤ 1,000,000,000  3 ≤ N ≤ 1,00,000 )

For each test case, output the median in a separate line.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;

public class Main {
	static int n;
	static int a[] = new int[100000];
	static int m;
	static int solve(int x,int y,int val,int flag){
		while(x<=y){
			int mid=(x+y)/2;
			if(val-a[mid]<=flag)
				y=mid-1;
			else
				x=mid+1;
			
		}
		return y;
	}
	static boolean ok(int val){
		int j,k=0;
		for(int i=2;i<=n;i++){
			j=solve(1,i-1,a[i],val);
			k+=i-j-1;
		}
		if(k<m)
			return false;
		else 
			return true;
	}
	public static void main(String[] args) throws IOException {
		//Scanner scan = new Scanner(System.in);
		StreamTokenizer st = new StreamTokenizer(new BufferedReader(
				new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		while (st.nextToken()!=StreamTokenizer.TT_EOF) {
			n =(int)st.nval;
			if(n*(n-1)/2%2!=0){
				m=(n*(n-1)/2+1)/2;
			}
			else
				m=n*(n-1)/2/2;
			
			for (int i = 1; i <=n; i++) {
				st.nextToken();
				a[i] = (int)st.nval;
			}
			Arrays.sort(a,1,n+1);
			int l = 0, r = a[n]-a[1];
			while(l<=r){
				int mid=(l+r)/2;
				if(ok(mid))
					r=mid-1;
				else
					l=mid+1;
			}
			out.println(l);
			out.flush();
		}
	}

}