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: Accepted)

#include <bits/stdc++.h>
using namespace std;

template <class T>
struct BIT {
  vector<T> lst;
  int treesize;
  // constructor
  BIT(int newn = 0) {
    treesize = newn;
    lst.resize(treesize + 1);
    for(int i = 0; i <= treesize; ++i) lst[i] = 0;
  }
  // a_place += num
  bool add(int place, T num) {
    while(place <= treesize) {
      lst[place] += num;
      place += place & -place;
    }
    return 1;
  }
  // add wide ver.(add num between l and r)
  bool addw(int left, int right, T num) {
    while(left <= right && left >= 1 && right <= treesize)
      add(left++, num);
    return 1;
  }
  // sum between l and r
  T calcsum(int place) {
    T answer = 0;
    while(place > 0) {
      answer += lst[place];
      place -= place & -place;
    }
    return answer;
  }
  // sum between l and r(calcsum wide ver.)
  T calcsumw(int left, int right) {
    return calcsum(right) - calcsum(left - 1);
  }
};

long long n, req;
vector<long long> a, ls;

long long solve();
bool isans(long long now);

int main() {
  cin >> n;
  for(int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    a.push_back(x);
    ls.push_back(x);
  }
  req = (n * (n + 1) + 2) / 4;
  sort(ls.begin(), ls.end());
  ls.erase(unique(ls.begin(), ls.end()), ls.end());
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long l = 0, r = ls.size() - 1, now = ls[0];
  while(r - l > 0) {
    if(r - l == 1) {
      if(isans(ls[r]))
        now = ls[r];
      else
        now = ls[l];
      break;
    }
    else {
      now = ls[(r + l) / 2];
      if(isans(now))
        l = (r + l) / 2;
      else
        r = (r + l) / 2;
    }
  }
  return now;
}
// yes 1,no 0
bool isans(long long now) {
  BIT<long long> bits = BIT<long long>(1 + n),
                 inv = BIT<long long>(3 * n);
  long long ans = 0;
  // calculate cumlative number(1 or -1 progression)
  for(int i = 0; i <= n; ++i) {
    if(i == 0 || a[i - 1] >= now)
      bits.add(i + 1, 1);
    else
      bits.add(i + 1, -1);
  }
  // avoid runtime error
  bits.add(1, n);
  // calculate inversion number
  for(int i = 1; i <= n + 1; ++i) {
    ans += inv.calcsum(bits.calcsum(i));
    inv.add(bits.calcsum(i), 1);
  }
  // cout << now << " " << ans << " " << req << endl;
  if(ans >= req)
    return 1;
  else
    return 0;
}