Codeforces Round #510 (Div. 2) D. Petya and Array

Codeforces Round #510 (Div. 2) D. Petya and Array

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Petya has an array a consisting of n integers. He has learned partial sums recently, and now he can calculate the sum of elements on any segment of the array really fast. The segment is a non-empty sequence of elements standing one next to another in the array.

Now he wonders what is the number of segments in his array with the sum less than t. Help Petya to calculate this number.

More formally, you are required to calculate the number of pairs l,r (l≤r) such that $a_l+a_{l+1}+\cdots +a_r<t$.

Input

The first line contains two integers n and t (1≤n≤200000,|t|≤2e14).

The second line contains a sequence of integers a1,a2,…,an (|ai|≤1e9) — the description of Petya’s array. Note that there might be negative, zero and positive elements.

Output

Print the number of segments in Petya’s array with the sum of elements less than t.

Examples

input

5 4
5 -1 3 4 -1

output

5

input

3 0
-1 2 -3

output

4

input

4 -1
-2 1 -2 3

output

3

The meaning of the question is very simple. Find how many intervals within 1-n whose sum is less than $t$ , we slightly transform:
$a_l+a_{l+1}+\cdots +a_r<t$
则:
$sum[r]-sum[l-1]<t$
$sum[r]-sum[l-1]\le t-1$
$sum\left[r\right]\le sum[l-1]+t-1\; The$
prefix sum can be preprocessed, and the title is converted to a certain number$x$ , find less than or equal to$x+t-$The number of elements of$1$ , it is easy to think of the weighted line segment tree to quickly find the number of less than or equal to a certain number in the interval. This problem must be discretized because of the amount of data. Use a$vector$ stores all the prefix sums, then the subscript of each prefix sum is the subscript in the segment tree after they are separated, then the title is the prefix and number of each prefix and query that meet the conditions. Pay attention tovector vectorThe elements in$v$$e$$c$$t$$o$$r$ must be de-duplicated, because the weighted line segment tree retains the state of the last insertion. If you do not de-duplicate, the answer will be repeated. The AC code is as follows:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=4e5+5;
int n,m,k;
ll t,a[N],sum[N],tree[N<<2];
vector<ll>v;
int getpos(ll x){
    
    
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
void pushup(int i)
{
    
    
    tree[i]=tree[i<<1]+tree[i<<1|1];
}

void build(int i,int l,int r)
{
    
    
    if(l==r)
    {
    
    
        tree[i]=0;
        return ;
    }
    int mid=l+r>>1;
    build(i<<1,l,mid);
    build(i<<1|1,mid+1,r);
    pushup(i);
}

void update(int i,int l,int r,int x,int k)
{
    
    
    if(l==r)
    {
    
    
        tree[i]+=k;
        return ;
    }
    int mid=l+r>>1;
    if(x<=mid) update(i<<1,l,mid,x,k);
    else update(i<<1|1,mid+1,r,x,k);
    pushup(i);
}

ll query(int i,int l,int r,int m,int n)
{
    
    
    if(m<=l&&r<=n) return tree[i];
    int mid=l+r>>1;
    ll ans=0;
    if(m<=mid) ans+=query(i<<1,l,mid,m,n);
    if(n>mid) ans+=query(i<<1|1,mid+1,r,m,n);
    return ans;
}

int main(){
    
    
    scanf("%d%lld",&n,&t);
    v.push_back(t-1);
    for(int i=1;i<=n;i++){
    
    
        scanf("%lld",&a[i]);
        sum[i]=sum[i-1]+a[i];
        v.push_back(sum[i]);
        v.push_back(sum[i]+t-1);
    }
    if(n==1){
    
    
        if(a[1]<t) printf("1");
        else printf("0");
    }else{
    
    
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        m=v.size();
        build(1,1,m);
        ll ans=0;
        for(int i=n;i>=0;i--){
    
    
            ans+=query(1,1,m,1,getpos(sum[i]+t-1));
            update(1,1,m,getpos(sum[i]),1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}