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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public void main(Scanner sc) {
        int n = sc.nextInt();
        int k = sc.nextInt();
        int a[] = new int[n + 1];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        sc.close();

        int b[] = new int[n + 1];
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int sum = 0;
        for (int i = 0; i <= n; i++) {
            b[i] = sum - k * i;
            map.putIfAbsent(b[i], null);
            sum += a[i];
        }
        int value = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            entry.setValue(value++);
        }

        int bn = Integer.highestOneBit((n + 1) * 2 - 1);
        long bit[] = new long[2 * bn];
        long ans = 0;
        for (int i = 0; i <= n; i++) {
            int c = map.get(b[i]);
            ans += query(c, bn, bit) + bit[c + bn];
            add(c, bn, bit);
        }

        System.out.println(ans);
    }

    private void add(int n, int off, long bit[]) {
        if (off == 0) {
            return;
        }

        bit[n + off]++;
        add(n / 2, off / 2, bit);
    }

    private long query(int n, int off, long bit[]) {
        if (off == 0) {
            return 0;
        }

        return (n % 2) * bit[n + off - 1] + query(n / 2, off / 2, bit);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        new Main().main(sc);
    }
}