Problem:
Score : 600 points

Problem Statement

You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.

a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints

All input values are integers.

1 ≤ N ≤ 2 \times 10^5

1 ≤ K ≤ 10^9

1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N K
a_1
a_2
:
a_N

Output

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

Sample Input 1

3 6
7
5
7

Sample Output 1

5

All the non-empty contiguous subsequences of a are listed below:

{a_1} = {7}

{a_1, a_2} = {7, 5}

{a_1, a_2, a_3} = {7, 5, 7}

{a_2} = {5}

{a_2, a_3} = {5, 7}

{a_3} = {7}

Their means are 7, 6, 19/3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a_1} and {a_3} are indistinguishable by the values of their elements, but we count them individually.

Sample Input 2

1 2
1

Sample Output 2

0

Sample Input 3

7 26
10
20
30
40
30
20
10

Sample Output 3

13

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <cstdio>

using namespace std;

int sum(vector<int> &tree, int i) {
    int ans = 0;
    while (i > 0) {
        ans += tree[i];
        i -= (i & -i);
    }
    return ans;
}
void update(vector<int> &tree, int i, int diff) {
    while (i < tree.size()) {
        tree[i] += diff;
        i += (i & -i);
    }
}

int main(){
    int n, k;
    cin >> n >> k;
    
    vector<int> a(n);
    for (int i=0;i<n;i++){cin >> a[i];}
    
    vector<int> b(n+1,0);
    vector<int> c(n+1,0);
    vector<int> tree(n+1,0);
    vector<pair<int,int>> p(n+1);

    for (int i=1;i<n+1;i++){b[i] = b[i-1] + a[i-1];}
    for (int i=1;i<n+1;i++){b[i] -= i*k;}
    for (int i=0;i<n+1;i++){p[i] = make_pair(b[i],i);}
    
    sort(p.begin(),p.end());
    int num = 1;
    c[p[0].second] = 1;
    for (int i=1;i<n+1;i++){
        if(p[i].first == p[i-1].first) c[p[i].second] = num;
        else c[p[i].second] = ++num;
    }

    int cnt = 0;
    for (int i=1;i<n+1;i++){
        update(tree, c[i], 1);
        if ((c[i]-1)>0) cnt += sum(tree, c[i]-1);
    }
    
    cout << cnt << '\n';
    return 0;
}