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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using pdd = pair<double, double>;
using uint = unsigned int;
template <typename T>
using Graph = vector<vector<T>>;
const int MOD = 1e9 + 7;
const ld PI = acosl(-1);

template <typename T>
struct BIT {
    int siz;
    vector<T> bit;

    BIT(int N) : siz(N + 1), bit(siz) {}

    void add(int k, T x) {
        for (int i = k; i < siz; i += (i & -i))
            bit[i] += x;
    }

    T sum(int k) {
        T res(0);
        for (int i = k; i > 0; i -= (i & -i))
            res += bit[i];
        return res;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N;
    cin >> N;
    vector<int> a(N);
    for (int i = 0; i < N; ++i)
        cin >> a[i];
    ll mi = (ll)N * (N + 1) / 4 + 1;

    auto acpy = a;
    sort(acpy.begin(), acpy.end());
    auto end = unique(acpy.begin(), acpy.end());
    int siz = end - acpy.begin();

    auto isok = [&](int x) -> bool {
        vector<int> cnt(N + 1);
        for (int i = 0; i < N; ++i)
            cnt[i + 1] = cnt[i] + (a[i] <= x ? 1 : -1);

        BIT<int> bit(2 * N + 1);
        ll res = 0;
        for (int i = 0; i <= N; ++i) {
            res += bit.sum(cnt[i] + N);
            bit.add(cnt[i] + N + 1, 1);
        }

        return mi <= res;
    };

    auto search = [&]() -> int {
        int ng = -1, ok = siz - 1;
        while (ok - ng > 1) {
            int mid = (ok + ng) / 2;
            (isok(acpy[mid]) ? ok : ng) = mid;
        }
        return acpy[ok];
    };

    cout << search() << endl;
    return 0;
}