Codeforces Round #632 (Div. 2)點這
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 ：
記錄前綴和數的位置，然后兩個前綴和相等，那么中間這一段的數的和必定為0，巧用map記錄位置，每一次循環，any加的是以i為尾的子串 符合條件的數量。

代碼：

#include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <stack> #include <vector> #include <set> #include <map> #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() {iosll 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; }

總結

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