For this Question: You are given an array of integers, You have to print the total number of non-empty subarray count of an array.The first line of the input contains a single integer N denoting the size of an array.
The second line of the input contains space-separated integers A<sub>1</sub>, A<sub>2</sub>,. , A<sub>n</sub>.

<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A[i] ≤ 10<sup>5</sup>
<b>Note</b>: Use long for big multiplications.Print the count of the total number of subarrays in an array.Sample Input 
3
1 2 3

Sample Output
6

Explanation
[1], [2], [3], [1, 2], [2, 3], [1, 2, 3]
Hence the total number of non empty subarrays are 6., I have written this Solution Code: #include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif

#define int long long

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
    uniform_int_distribution<int64_t> generator(l, r);
    return generator(rng);
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    auto start = std::chrono::high_resolution_clock::now();

    int tt = 1;
    // cin >> tt;
    while (tt--) {
        long n;
        cin >> n;
        for (int i = 0; i < n; i++) {
            int temp;
            cin >> temp;
        }
        long ans = (n * (n + 1)) / 2;
        cout << ans << "\n";
    }

    auto stop = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
    cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s\n";

    return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?