Codeforces Round #632 (Div. 2)巧用小技巧

Codeforces Round #632 (Div. 2)点这
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
inputCopy
3
1 2 -3
outputCopy
5
inputCopy
3
41 -41 41
outputCopy
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.

思路 ：
记录前缀和数的位置，然后两个前缀和相等，那么中间这一段的数的和必定为0，巧用map记录位置，每一次循环，any加的是以i为尾的子串 符合条件的数量。

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
using namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
int dx[4]= {-1,1,0,0},dy[4]= {0,0,1,-1};
const ll mod=10001;
const ll N=1e6+10;
map<ll,ll> p;
int a[N];
int main()
{
    ios
    ll last=-1;
    ll any=0;
    ll sum=0;
    ll n;cin>>n;
    p[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        sum+=a[i];
        if(p.count(sum)) last=max(last,p[sum]);
        any+=(i-last-1);
        p[sum]=i;
    }
    cout<<any;
    return 0;
}