Petya and Array (权值线段树+逆序对)

Petya and Array

http://codeforces.com/problemset/problem/1042/D

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya has an array aa consisting of nn 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 tt. Help Petya to calculate this number.

More formally, you are required to calculate the number of pairs l,rl,r (l≤rl≤r) such that al+al+1+⋯+ar−1+ar<tal+al+1+⋯+ar−1+ar<t.

Input

The first line contains two integers nn and tt (1≤n≤200000,|t|≤2⋅10141≤n≤200000,|t|≤2⋅1014).

The second line contains a sequence of integers a1,a2,…,ana1,a2,…,an (|ai|≤109|ai|≤109) — 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 tt.

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

Note

In the first example the following segments have sum less than 44:

• [2,2][2,2], sum of elements is −1−1
• [2,3][2,3], sum of elements is 22
• [3,3][3,3], sum of elements is 33
• [4,5][4,5], sum of elements is 33
• [5,5][5,5], sum of elements is −1

　　参考博客 http://mamicode.com/info-detail-2452129.html

　　找区间和小于t的个数，区间和的问题，一般用前缀和来做

　　可以看成sum[i]>t+sum[k]的个数，i<k<=n。这样就变成了一个逆序对的问题

　　

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<cstdio>
 #include<vector>
 #define maxn 500005
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 typedef long long ll;
 using namespace std;

 vector<ll>v;
 ll n;
 ll a[maxn];
 ll sum[maxn];

 int tree[maxn<<];

 int getid(ll x){
     return lower_bound(v.begin(),v.end(),x)-v.begin()+;
 }

 void pushup(int rt){
     tree[rt]=tree[rt<<]+tree[rt<<|];
 }

 void build(int l,int r,int rt){
     if(l==r){
         tree[rt]=;
         return;
     }
     int mid=(l+r)/;
     build(lson);
     build(rson);
     pushup(rt);
 }

 void add(int L,int k,int l,int r,int rt){
     if(l==r){
         tree[rt]+=k;
         return;
     }
     int mid=(l+r)/;
     if(L<=mid) add(L,k,lson);
     else add(L,k,rson);
     pushup(rt);
 }

 ll query(int L,int R,int l,int r,int rt){
     if(L<=l&&R>=r){
         return tree[rt];
     }
     int mid=(l+r)/;
     ll ans=;
     if(L<=mid) ans+=query(L,R,lson);
     if(R>mid) ans+=query(L,R,rson);
     return ans;
 }

 int main(){

     std::ios::sync_with_stdio(false);
     ll t;
     cin>>n>>t;
     for(int i=;i<=n;i++){
         cin>>a[i];
     }
     v.push_back(t-);
     for(int i=;i<=n;i++){
         sum[i]=a[i]+sum[i-];
         v.push_back(sum[i]);
         v.push_back(sum[i]+t-);
     }
     if(n==){
         if(a[]<t){
             cout<<<<endl;
         }
         else{
             cout<<<<endl;
         }
     }
     else{
         sort(v.begin(),v.end());
         v.erase(unique(v.begin(),v.end()),v.end());
         int Size=v.size();
         build(,Size,);
         add(getid(sum[n]),,,Size,);
         ll ans=;
         for(int i=n-;i>=;i--){
             ans+=query(,getid(sum[i]+t-),,Size,);
             add(getid(sum[i]),,,Size,);
         }
         ans+=query(,getid(t-),,Size,);

         cout<<ans<<endl;
     }

 }