Problem:
Score : 700 points

Problem Statement

We will define the median of a sequence b of length M, as follows:

Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.

For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N.
For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a.
We will list m_{l, r} for all pairs (l, r) to create a new sequence m.
Find the median of m.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the median of m.

Sample Input 1

3
10 30 20

Sample Output 1

30

The median of each contiguous subsequence of a is as follows:

The median of (10) is 10.

The median of (30) is 30.

The median of (20) is 20.

The median of (10, 30) is 30.

The median of (30, 20) is 30.

The median of (10, 30, 20) is 20.

Thus, m = (10, 30, 20, 30, 30, 20) and the median of m is 30.

Sample Input 2

1
10

Sample Output 2

10

Sample Input 3

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3

8

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a[] = new int[n];
        for(int i=0;i<n;i++) {
            a[i] = sc.nextInt();
        }
        sc.close();
        int maxim = 1000000001;

        int ans = bSearch(0, maxim, a);
        
        System.out.println(ans);        
    }
    
    /*
    条件を満たすxの最大値をにぶたんで求める
    */
    public static int bSearch(int ok,int ng,int[] numbers){
        if(ng == ok + 1){
            return(ok);
        } else {
            int mid = (ok+ng)/2;
            if(judge(mid, numbers)) {
                return(bSearch(mid, ng, numbers));
            } else {
                return(bSearch(ok, mid, numbers));
            }
        }
    }

    public static boolean judge(int x, int[] numbers){
        /*
        int xが条件を満たすかどうか判定する
        条件：x以上のmedian yが半数以上存在する
        */
        int n = numbers.length;
        int[] cums = new int[n+1];
        cums[0] = 0;
        for(int i=0;i<n;i++){
            if(numbers[i] >= x) {
                cums[i+1] = cums[i] + 1;
            } else {
                cums[i+1] = cums[i] - 1;
            }
        }
        //転倒数が全区間数の半分以下
        int half = ((n+1)*n/2)/2;
        int invNumber = 0;
        MyPair msort = invNum(cums);
        invNumber = msort.value;
        if(invNumber <= half) {
            return(true);
        } else {
            return(false);
        }
    }

    //left, rightはそれぞれソート済み
    private static MyPair merge(int[] left, int[] right){
        int l = left.length;
        int r = right.length;
        int i = 0;
        int j = 0;
        int ansValue = 0;
        int[] ansList = new int[l+r];
        int index = 0;
        while(i < l && j < r){
            if(left[i] <= right[j]) {
                ansList[index] = left[i];
                i++;
            } else {
                ansList[index] = right[j];
                j++;
                ansValue += l-i;
            }
            index++;
        }
        while(i < l){
            ansList[index] = left[i];
            index++;
            i++;
        }
        while(j < r){
            ansList[index] = right[j];
            index++;
            j++;
        }
        MyPair ans = new MyPair();
        ans.list = ansList;
        ans.value = ansValue;
        return(ans);
    }

    private static MyPair invNum(int[] numbers) {
        int l = numbers.length;
        MyPair ans = new MyPair();
        
        //基底部
        if(l <= 1){
            ans.value = 0;
            ans.list = numbers;
            return(ans);
        }
        // l>=2の場合は分割統治
        int half = l/2;
        int[] left = new int[half];
        int[] right = new int[l-half];
        for(int i=0;i<l;i++) {
            if(i < half){
                left[i]=numbers[i];
            } else {
                right[i-half]=numbers[i];
            }
        }
        MyPair leftP = invNum(left);
        MyPair rightP = invNum(right);
        MyPair mergeP = merge(leftP.list,rightP.list);
        ans.list = mergeP.list;
        ans.value = leftP.value + rightP.value + mergeP.value;
        return(ans);
    }
  }

class MyPair{
    public int value;
    public int[] list;
}