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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e5 + 5;
int a[MAXN] , binary[MAXN];
int n;
int s[MAXN] , pre[MAXN];
int c[MAXN * 2];
#define mid ((l+r)>>1)
long long ans;
inline void merge(int l,int r)
{
	if(l == r)	return ;
	merge(l,mid);
	merge(mid+1,r);
	int b[MAXN];
	int pos1 = l , pos2 = mid + 1;
	int tot = l; 
	while(pos1 <= mid && pos2 <= r)
	{
		if(pre[pos1] > pre[pos2])
		{
			b[tot++] = pre[pos2++];
			ans += mid - pos1 + 1; 
		}
		else
			b[tot++] = pre[pos1++];
	}
	while(pos1 <= mid)	b[tot++] = pre[pos1++];
	while(pos2 <= r)	b[tot++] = pre[pos2++];
	for(int i = l ; i <= r ; i++)
		pre[i] = b[i];
}
inline long long cnt(int x)//求新数列里有多少个数比x小 
{
	ans = 0;
	//a[i]小于等于x,s[i]就为-1 ,否则为1 
	for(int i = 1 ; i <= n ; i++)
		s[i] = (a[i] <= x) ? -1 : 1;
	//接下来求s数组有多少个子区间和小于0,若小于0,则代表比x小的数比比x大的数多，所以x>该区间中位数 
	pre[0] = 0;
	for(int i = 1 ; i <= n ; i++)
		pre[i] = pre[i-1] + s[i] , ans += pre[i] < 0 ? 1 : 0;
	//sum(s[l]~s[r]) = pre[r] - pre[l-1] < 0 
	//所以转为求pre数组逆序对
	merge(1,n);
	return ans; 
}
int main()
{
	scanf("%d",&n);
	for(int i = 1 ; i <= n ; i++)
		scanf("%d",a+i) , binary[i] = a[i];
	sort(binary+1,binary+n+1);
	int l = 1 , r = n;
	int m = (n + 1) * n / 2;
	while(l < r)
	{
		int x = cnt(binary[mid]);
		if(m >= x * 2)//猜的数大于等于中位数，尽量小 
			l = mid + 1;
		else
			r = mid;
	}
	printf("%d\n",binary[l]);
}