Codeforces Round #632 (Div. 2) C.Eugene and an array

Codeforces Round #632 (Div. 2) C.Eugene and an array

题目链接
Eugene likes working with arrays. And today he needs your help in solving one challenging task.

An array c is a subarray of an array b if c can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Let’s call a nonempty array good if for every nonempty subarray of this array, sum of the elements of this subarray is nonzero. For example, array [−1,2,−3] is good, as all arrays [−1], [−1,2], [−1,2,−3], [2], [2,−3], [−3] have nonzero sums of elements. However, array [−1,2,−1,−3] isn’t good, as his subarray [−1,2,−1] has sum of elements equal to 0.

Help Eugene to calculate the number of nonempty good subarrays of a given array a.

Input

The first line of the input contains a single integer n (1≤n≤2×105) — the length of array a.

The second line of the input contains n integers a1,a2,…,an (−109≤ai≤109) — the elements of a.

Output

Output a single integer — the number of good subarrays of a.

Examples

input

3
1 2 -3

output

5

input

3
41 -41 41

output

3

一开始考虑用总情况去减，后来发现比较麻烦，因为会存在前缀和与数都为0的情况，不好判断，改成累加答案~
用 $map$ 更新前缀和 $sum$ 的位置，对当前前缀和 $sum$，$ans=ans+i-m[sum]$，注意前缀和为 $0$ 时，$ans$ 要少加一个1，即 $m[0]=1$，AC代码如下：

#include 
typedef long long ll;
using namespace std;
ll n,k,sum=0,pos=0,ans=0;
map<ll,ll> m;
int main()
{
	cin>>n;
	m[0]=1;
	for(ll i=1;i<=n;i++)
	{
		cin>>k;
		sum+=k;
		pos=max(pos,m[sum]);
		ans+=i-pos;
		m[sum]=i+1;
	}
	cout<<ans;
}