【原题】
Eugene likes working with arrays. And today he needs your help in solving one challenging task.
An array cc is a subarray of an array bb if cc can be obtained from bb 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
Help Eugene to calculate the number of nonempty good subarrays of a given array aa.
Input
The first line of the input contains a single integer nn (1≤n≤2×1051≤n≤2×105) — the length of array aa.
The second line of the input contains nn integers a1,a2,…,an (−109≤ai≤109−109≤ai≤109) — the elements of aa.
Output
Output a single integer — the number of good subarrays of a.
Examples
input
Copy
3
1 2 -3
output
5
input
3
41 -41 41
output
3
Note
In the first sample, the following subarrays are good:
In the second sample, three subarrays of size 1 are the only good subarrays. At the same time, the subarray
1 #include <algorithm> 2 #include <cmath> 3 #include <cstdio> 4 #include <cstring> 5 #include <list> 6 #include <map> 7 #include <iostream> 8 #include <queue> 9 #include <set> 10 #include <stack> 11 #include <string> 12 #include <vector> 13 #include <iomanip> 14 #define LL long long 15 #define inf 0x3f3f3f3f 16 #define INF 0x3f3f3f3f3f3f 17 #define PI 3.1415926535898 18 #define F first 19 #define S second 20 #define lson rt << 1 21 #define rson rt << 1 | 1 22 using namespace std; 23 24 const int maxn = 2e5 + 7; 25 const int maxm = 1e4 + 7; 26 int n, m; 27 LL a[maxn], sum[maxn]; 28 map<LL, int> mp; 29 30 LL read() 31 { 32 LL x = 0, f = 1;char ch = getchar(); 33 while (ch < '0' || ch>'9') { if (ch == '-')f = -1;ch = getchar(); } 34 while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0';ch = getchar(); } 35 return x * f; 36 } 37 38 int main() 39 { 40 ios::sync_with_stdio(false); 41 cin.tie(0); 42 n = read(); 43 for (int i = 1; i <= n; i++) 44 { 45 a[i] = read(); 46 sum[i] = sum[i - 1] + a[i]; 47 } 48 LL cnt = 0; 49 int num = 0; 50 mp[0] = 0; 51 for (int i = 1; i <= n; i++) 52 { 53 if (mp.count(sum[i])) num = max(mp[sum[i]] + 1, num); 54 mp[sum[i]] = i; 55 cnt += i - num; 56 } 57 cout << cnt << endl; 58 }