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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef pair<double,double> pdd;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define SZ(i) ll(i.size())
#define FOR(i, j, k, in) for (ll i=j ; i<k ; i+=in)
#define RFOR(i, j, k, in) for (ll i=j ; i>=k ; i-=in)
#define REP(i, j) FOR(i, 0, j, 1)
#define REP1(i,j) FOR(i, 1, j+1, 1)
#define RREP(i, j) RFOR(i, j, 0, 1)
#define ALL(_a) _a.begin(),_a.end()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define X first
#define Y second
#ifdef tmd
#define debug(...) do{\
    fprintf(stderr,"%s - %d (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<" ,";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
#else
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0)
#endif

template<class T> inline bool cmax(T &a, const T &b) { return b > a ? a = b, true : false; }
template<class T> inline bool cmin(T &a, const T &b) { return b < a ? a = b, true : false; }
template<class T> using MaxHeap = priority_queue<T>;
template<class T> using MinHeap = priority_queue<T, vector<T>, greater<T>>;

const ll MOD=1000000007;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll MAXN=1e5+5;
const ll MAXLG=__lg(MAXN)+2;



const ll MAXB = MAXN*5;
ll bit[MAXB];
ll n,a[MAXN];

void init() {
    REP (i,MAXB) {
        bit[i] = 0;
    }
}

void add (ll pos,ll val) {
    pos += MAXN;
    for(;pos<=MAXB-1;pos+=-pos&pos) {
        bit[pos] += val;
    }
}

ll query(ll pos) {
    pos += MAXN;
    ll ret = 0;
    for (;pos>=1;pos-=pos&-pos) {
        ret += bit[pos];
    }
    return ret;
}


ll check(ll num) {
    ll pre = 0;
    ll tmp[MAXN];
    REP (i,n) {
        if (num >= a[i]) pre--;
        else pre++;
        tmp[i] = pre;
        add(pre,1);
    }
    pary(tmp,tmp+n);

    ll ret = 0;
    ll sum = -1;
    pre = 0;
    REP (i,n) {
        debug(i,query(sum),sum);
        ret += query(sum);
        if (num >= a[i]) {
            pre--;
            sum--;
        } else {
            pre++;
            sum++;
        }
        add(pre,-1);
    }

    return ret;
}
/********** Good Luck :) **********/
int main()
{
    IOS();
    cin >> n;
    vector<ll> val;
    REP (i,n) {
        cin >> a[i];
        val.eb(a[i]);
    }
    if (n == 1) {
        cout << a[0] << endl;
        return 0;
    }
    sort(ALL(val));
    val.resize(unique(ALL(val))-val.begin());

    // debug(check(20));

    ll cb = n * (n-1) / 2;
    cb = (cb + 1) / 2;

    ll L = 0,R = SZ(val);
    while (L < R - 1) {
        ll mid = L+R>>1;
        ll ch = check(val[mid]);
        debug(val[mid],ch);
        if (ch >= cb) {
            L = mid;
        } else {
            R = mid;
        }
    }

    debug(L,R);
    cout << val[L] << endl;
    return 0;
}