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: Accepted)

#include <bits/stdc++.h>
 
#define forn(i, n) for (llong i = 0ll; i < (llong) n; ++i)
#define fornn(i, l, r) for (llong i = (llong) l; i < (llong) r; ++i)
#define size(x) ((int) (x.size()))
 
using namespace std;
 
typedef long long llong;
const llong inf = (llong) 1e+9 + 7ll;
const llong linf = (llong) 1e+18 + 7ll;
const long double eps = (long double) 1e-9;
const long double pi = acosl((long double) -1.0);
const int alph = 26;

mt19937 mrand(random_device{} ()); 

template<typename T, typename U> inline llong umin(const T& a, const U& b) { return a < b ? a : b; }
template<typename T, typename U> inline llong umax(const T& a, const U& b) { return a > b ? a : b; }
 
static char buff[(int) 2e6 + 17]; // reads std::string
const int maxn = (int) 2e5 + 17;

int n, k;
llong ans = 0;
llong a[maxn];

bool read()
{
	if (scanf("%d %d", &n, &k) != 2)
		return false;

	forn (i, n)
		scanf("%lld", a + i);

	return true;
}

void calc(int l, int r)
{
	if (l == r)
	{
		ans += a[l] >= 0;
		return ;
	}

	int m = (l + r) / 2;
	calc(l, m);
	calc(m + 1, r);
	vector<llong> ps;
	llong cur = 0;

	for (int i = m; i >= l; --i)
	{
		cur += a[i];
		ps.push_back(cur);
	}

	cur = 0;
	sort(ps.begin(), ps.end());

	for (int i = m + 1; i <= r; ++i)
	{
		cur += a[i];
		int cnt = lower_bound(ps.begin(), ps.end(), -cur) - ps.begin();
		ans += size(ps) - cnt;
	}
}

void solve()
{
	forn (i, n)
		a[i] -= k;

	calc(0, n - 1);
	printf("%lld\n", ans);
}

int main()
{
#if SEREZHKA
	freopen("file.in", "r", stdin);
#endif

	while (read())
		solve();

	return 0;
}