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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

int n,m;
long long sum[300000];
int c[300000];
int d[300000];

int lowbit(int x){
	return x & -x;
}

void update(int x){
	while (x<=n+10){
		c[x]+=1;
		x+=lowbit(x);
	}
}

int query(int x){
	int s=0;
	while (x>0){
		s+=c[x];
		x-=lowbit(x);
	}
	return s;
}

struct o_O{
	long long b;
	int d;
}p[600000];

bool cmp(o_O a,o_O b){
	return a.b<b.b;
}

int main(){
	scanf("%d%d",&n,&m);
	int w;
	sum[0]=0;
	for (int i=1;i<=n;i++){
		scanf("%lld",&w);
		sum[i]=sum[i-1]+w;
	}
	for (int i=0;i<=n;i++){
		sum[i]=sum[i]-m*i;
		p[i].b=sum[i];
		p[i].d=i;
	}
	sort(p,p+n+1,cmp);
	sum[p[0].d]=1;
	int n1=1;
	for (int i=1;i<=n;i++){
		if (p[i].b==p[i-1].b){
			sum[p[i].d]=n1;
		}else {
			n1++;
			sum[p[i].d]=n1;
		}
	}
	memset(c,0,sizeof(c));
	long long ans=0;
	for (int i=0;i<=n;i++){
		ans+=query(sum[i]);
		update(sum[i]); 
	}
	printf("%lld\n",ans);
	return 0;
}