Petya and Array CodeForces

D. Petya and Array

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 $\mathrm{44:}$

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

先补上树状数组的解法  

sum[i] - sum[j] < t , 1 <= j < i,所以sum[i] - t < sum[j]

因为j是i之前的前缀和，那么我们可以找当前sum[j],小于等于sum[i]-t的，然后用ans+=（i-query），然后一直wa，看了网上题解，加入一个0的虚拟节点后（相当于每个当前前缀和），树状数组记录当前时刻1到n+1,而不是1到n；

这样就可以知道当前前缀和之前的前缀和出现情况。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 2e5+5;

ll tree[maxn];
ll sum[maxn];
ll num[maxn];
int n;
ll t;

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

void add(int x)
{
    for(int i=x;i<=n+1;i+=lowbit(i))
    {
        tree[i]++;
    }
}

ll query(int x)
{
    ll ans = 0;
    for(int i=x;i>0;i-=lowbit(i))
    {
        ans += tree[i];
    }
    return ans;
}

int main()
{
     scanf("%d%lld",&n,&t);
     for(int i=1;i<=n;i++)
     {
         scanf("%lld",&sum[i]);
         sum[i] += sum[i-1];
         num[i] = sum[i];
     }
     sort(num,num+n+1);
     ll ans = 0;
     for(int i=1;i<=n;i++)
     {
         add(lower_bound(num,num+n+1,sum[i-1])-num+1);
         ans += i - query(upper_bound(num,num+n+1,sum[i]-t)-num);
     }
    printf("%lld
",ans);
}