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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input!(n: usize, a: [u32; n]);
    let mut l = a.iter().min().unwrap() - 1;
    let mut r = a.iter().max().unwrap() + 1;

    while 1 < r - l {
        let c = (l + r) / 2;
        let mut b = vec![n as i32; n + 1];
        (0..n).for_each(|i| b[i + 1] = b[i] + if a[i] <= c { 1 } else { -1 });

        let sz = 2 * n + 1;
        let mut seg = vec![0; 2 * sz];
        let mut cnt = 0;
        for x in b {
            let x = x as usize;

            let mut l = sz;
            let mut r = sz + x;
            while l < r {
                if l % 2 == 1 {
                    cnt += seg[l];
                    l += 1;
                }
                if r % 2 == 1 {
                    r -= 1;
                    cnt += seg[r];
                }
                l /= 2;
                r /= 2;
            }

            let mut i = x + sz;
            seg[i] += 1;
            while 0 < i {
                i /= 2;
                seg[i] = seg[2 * i] + seg[2 * i + 1];
            }
        }

        if (n * (n + 1)) / 4 < cnt {
            r = c;
        } else {
            l = c;
        }
    }
    println!("{}", r);
}

/*
 * 答えが X 以下であることの条件は、
 * 中央値が X 以下である区間が n(n+1)/4 個より多くあることです。
 *
 * 中央値が X 以下である区間を数えるために、X 以下 を 1、大きいものを -1 とします。
 * すると、中央値が X 以下である条件は、総和が 0 以上なことです。
 * そこで、この配列の累積和を取って、距離 1 以上の弱増加ペアを数え上げればよいです。
 *
 * うーん、セグツリーを書くしかなさそうですね。
 */