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 C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;


const int INF = 1e+9;

/* 数列 {a_n} に対して、
 * ・ a[i] += w
 * ・ sum(r,l) = a[r]+...+a[l] を求めたい
 * って時に使える
 * cf. http://hos.ac/slides/20140319_bit.pdf
 */
template <typename T>
class BinaryIndexTree
{
    private:
        T N;
        vector<T> bit;

    public:
        BinaryIndexTree(T n): N(n), bit(n+1, 0) {}

        // += w
        void add(T i, T w) {
            // x & -x は x の最も下の立っているbitを取り出す
            for( T x = i; x <= N; x += x & -x) bit[x] += w;
        }
        // [1, i] の和を計算.
        T sum(T i) {
            T ret = 0;
            for (T x = i; x > 0; x -= x & -x) ret += bit[x];
            return ret;
        }
        // [left+1, right] の和を計算.
        T sum(T left, T right) {
            return sum(right)-sum(left);
        }
};


// num より大きい数が a からできる数列にいくつあるかを求める
// 転倒数 (i<j かつ a[i]>a[j] となる組の数) の求め方とほぼ同じ
long long check( int num, vector<int>& a )
{
    int N = a.size()-1;

    vector<int> S(N+1, 0);
    for( int i = 1; i <= N; ++i ){
        if(a[i] > num) S[i] = -1;
        else S[i] = 1;
    }
    for( int i = 1; i <= N; ++i ) S[i] += S[i-1];

    long long ret = 0;
    BinaryIndexTree<long long> bit(INF);
    for( int i = 1; i <= N; ++i ){
        ret += bit.sum(num-1);
        bit.add(S[i], 1);
    }
    return ret;
}


int main()
{
    int N;
    cin >> N;
    vector<int> a(N+1);
    for( int i = 1; i <= N; ++i ) cin >> a[i], cout << i;

    cout << "a";
    // 中央値になる値を二分探索
    // 数列 M には、中央値より大きい数がちょうど [M/2] あることを利用
    int l = 0, r = INF;
    long long want = 1LL*(N+1)*N/4 + 1;  //　自分より大きい数がこれだけあればOK
    while(l < r){
        int c = (l+r)/2;
        if( check(c, a) >= want ) r = c;
        else l = c+1;
    }
    cout << l << endl;
    return 0;
}