You are given an integer sequence of length N, a= {a1,a2,…,aN}, and an integer K.
a has N(N+1)⁄2 non-empty contiguous subsequences, {al,al+1,…,ar} (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≤2×105
1≤K≤109
1≤ai≤109

Input is given from Standard Input in the following format:
N K
a1
a2
:
aN

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

3 6

7

5

7

5

All the non-empty contiguous subsequences of a are listed below:
{a1} = {7}
{a1,a2} = {7,5}
{a1,a2,a3} = {7,5,7}
{a2} = {5}
{a2,a3} = {5,7}
{a3} = {7}
Their means are 7, 6, 19⁄3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a1} and {a3} are indistinguishable by the values of their elements, but we count them individually.

数组要从0开始算，不然会少算长度为1的连续子序列。

AC代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
struct poi{ll sum,pos;}a[200010];
ll n,k,ans,cnt,tree[200010],lisan[200010];
void read(ll &k)
{
k=0;int f=1;char c=getchar();
while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
while(c<='9'&&c>='0')k=k*10+c-'0',c=getchar();
k*=f;
}
bool cmp(poi a,poi b){return a.sum<b.sum;}
int lowbit(int x){return x&-x;}
void add(int x,int delta)
{
for(int i=x;i<=cnt;i+=lowbit(i))
tree[i]+=delta;
}
int sum(int x)
{
int s=0;
for(int i=x;i>=1;i-=lowbit(i))
s+=tree[i];
return s;
}
int main()
{
read(n);read(k);
for(int i=1;i<=n;i++)read(a[i].sum),a[i].sum-=k;
for(int i=1;i<=n;i++)a[i].sum+=a[i-1].sum,a[i].pos=i;
sort(a,a+1+n,cmp);
for(int i=0;i<=n;i++)
{
if(a[i].sum!=a[i-1].sum||i==0)cnt++;
lisan[a[i].pos]=cnt;
}
for(int i=n;i>=0;i--)
{
ans+=sum(cnt)-sum(lisan[i]-1);
add(lisan[i],1);
}
printf("%lld\n",ans);
}