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

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 su

const int N=2e5+7;
ll t,n,m,k=0;
ll a[N],b[N],c[N];
map<ll,ll> mp;
ll ans=0,j=0;
int main() {
    mp[0]=1;
    scanf("%lld",&n);
    for(ll i=1; i<=n; i++) {
        scanf("%lld",&a[i]);
        b[i]=a[i]+b[i-1];
        if(mp[b[i]]==0)
            ans+=i-j;
        else {
            if(j>mp[b[i]]) {
                ans+=i-j;
            } else {
                ans+=i-mp[b[i]];
            }
            j=max(j,mp[b[i]]);
        }
        mp[b[i]]=i+1;
    }
    printf("%lld\n",ans);
    return 0;
}

map里的count操作更加简化

int main() {
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        scanf("%d",&a[i]);
    }
    vis[0]=0;
    for(int i=1; i<=n; i++) {
        sum+=a[i];
        if(vis.count(sum)) {
            now=max(now,vis[sum]+1);
        }
        vis[sum]=i;
        ans+=i-now;
    }
    cout<<ans<<endl;
    return 0;
}