CodeForces#632Div.2C. Eugene and an array

网址：https://codeforces.ml/contest/1333/problem/C
题目描述:
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

Note
In the first sample, the following subarrays are good: [1], [1,2], [2], [2,−3], [−3]. However, the subarray [1,2,−3] isn't good, as its subarray [1,2,−3] has sum of elements equal to 0.

In the second sample, three subarrays of size 1 are the only good subarrays. At the same time, the subarray [41,−41,41] isn't good, as its subarray [41,−41] has sum of elements equal to 0.

题解：对于一个数组a要想找到他所有子数组，只需要从数组a第一个元素开始一直遍历到最后一个元素，每次遍历是从当前元素向前（此元素之前的元素）看。
这道题就是利用sum(前i个元素之和)，使用了map数据结构，map[sum]=i。准确的说是每此遍历找到以a[i]为最后一个元素的good的子数组。
这里比较重要的就是变量pos(在后面的代码中),理解了pos就可以。
AC代码：

    #include<iostream>
    #include<cstdio>
    #include<map>
    #define ll long long int
    using namespace std;

int main(){
	ll n,sum=0,a,ans=0,pos=-1;
	cin>>n;
	map<ll,ll>mp;
	mp[0]=0;
	for(int i=1;i<=n;i++){
		cin>>a;
		sum+=a;
		if(mp.find(sum)!=mp.end()){//存在sum 
			pos=max(pos,mp[sum]);
		}
		mp[sum]=i;
		ans=ans+i-pos-1;
	}
	cout<<ans<<endl;
	return 0;
}

注：碰到了许多的算法题基本都是数学题，但有时数学逻辑是真的想不到，即便是别人稍微一提及自己就能明白，但做的时候自己想着实没有想到。应该还是不熟练所致。