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 $\mathrm{aa consisting\; of nn integers.\; He\; has\; learned\; partial\; sums\; recently,\; and\; now\; he\; can\; calculate\; the\; sum\; of\; elements\; on\; any\; segment\; of\; the\; array\; really\; fast.\; The\; segment\; is\; a\; non-empty\; sequence\; of\; elements\; standing\; one\; next\; to\; another\; in\; the\; array.}$

Now he wonders what is the number of segments in his array with the sum less than $\mathrm{tt.\; Help\; Petya\; to\; calculate\; this\; number.}$

More formally, you are required to calculate the number of pairs $\mathrm{l,rl,r (l\le rl\le r)\; such\; that al+al+1+\cdots +ar-1+ar<tal+al+1+\cdots +ar-1+ar<t.}$

Input

The first line contains two integers $\mathrm{nn and tt (1\le n\le 200000,|t|\le 2\cdot 10141\le n\le 200000,|t|\le 2\cdot 1014).}$

The second line contains a sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (|ai|\le 109|ai|\le 109)\; —\; the\; description\; of\; Petya\text{'}s\; array.\; Note\; that\; there\; might\; be\; negative,\; zero\; and\; positive\; elements.}}^{}$

Output

Print the number of segments in Petya's array with the sum of elements less than $\mathrm{tt.}$

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 $\mathrm{44:}$

• $[2,2][2,2],\; sum\; of\; elements\; is -1-1$
• $[2,3][2,3],\; sum\; of\; elements\; is 22$
• $[3,3][3,3],\; sum\; of\; elements\; is 33$
• $[4,5][4,5],\; sum\; of\; elements\; is 33$
• $[5,5][5,5],\; sum\; of\; elements\; is -1$

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

　　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<<3];

int getid(ll x){
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}

void pushup(int rt){
    tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}

void build(int l,int r,int rt){
    if(l==r){
        tree[rt]=0;
        return;
    }
    int mid=(l+r)/2;
    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)/2;
    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)/2;
    ll ans=0;
    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=1;i<=n;i++){
        cin>>a[i];
    }
    v.push_back(t-1);
    for(int i=1;i<=n;i++){
        sum[i]=a[i]+sum[i-1];
        v.push_back(sum[i]);
        v.push_back(sum[i]+t-1);
    }
    if(n==1){
        if(a[1]<t){
            cout<<1<<endl;
        }
        else{
            cout<<0<<endl;
        }
    }
    else{
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        int Size=v.size();
        build(1,Size,1);
        add(getid(sum[n]),1,1,Size,1);
        ll ans=0;
        for(int i=n-1;i>=1;i--){
            ans+=query(1,getid(sum[i]+t-1),1,Size,1);
            add(getid(sum[i]),1,1,Size,1);
        }
        ans+=query(1,getid(t-1),1,Size,1);

        cout<<ans<<endl;
    }

}