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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
using namespace std;

typedef long long ll;
typedef uint64_t ull;
typedef pair<int, int> P;
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

template<typename T>
struct BinaryIndexedTree{
    vector<T> data;
    int sz;

    BinaryIndexedTree(int n){
        sz = n + 1;
        data.assign(sz, 0);
    }

    void add(int i, T x){
        ++i;
        while(i < sz){
            data[i] += x;
            i += i & -i;
        }
    }

    // [0, i] の区間和
    T sum(int i){
        T res = 0;
        ++i;
        while(i > 0){
            res += data[i];
            i -= i & -i;
        }
        return res;
    }

    T sum(int l, int r){
        return sum(r - 1) - sum(l - 1);
    }

    // val 以上を満たす 0-indexed の位置を返す
    int lower_bound(T val){
        // if(val <= 0)    return -1;
        int res = 0;
        int n = 1;
        while(n < sz)   n <<= 1;
        for(int k = n >> 1; k > 0; k >>= 1){
            if(res + k < sz && data[res + k] < val){
                val -= data[res + k];
                res += k;
            }
        }
        return res;     // 1-indexed であれば、res + 1 を返す
    }
};

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

    int n;
    cin >> n;
    vector<int> a(n);
    int maxe = 0;
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        chmax(maxe, a[i]);
    }
    int lb = 1, ub = maxe + 1;
    ll bound = (ll)n * (n + 1) >> 1;
    bound = (bound >> 1) + (bound % 2);
    while(ub - lb > 1){
        int x = (lb + ub) >> 1;
        int S = 0;
        vector<P> table;
        table.emplace_back(0, 0);
        for(int i = 0; i < n; ++i){
            S += (a[i] >= x) ? 1 : -1;
            table.emplace_back(S, i + 1);
        }
        sort(table.begin(), table.end(), greater<P>());
        BinaryIndexedTree<int> bit(n + 1);
        ll c = 0;
        for(auto p : table){
            c += bit.sum(p.second, n + 1);
            bit.add(p.second, 1);
        }
        if(c >= bound)  lb = x;
        else    ub = x;
    }
    cout << lb << endl;
}