HackerEarth Count the array problem solution

1. N: Determine the count of distinct arrays of size <= N and >= 1 such that:
2. Each array element is a prime number
3. Product of the value of all the array elements is <= P
4. Array formed is palindromic

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int MAX_N = 1e6 + 14;
vector<int> primes;
bool is_not_prime[MAX_N];
int p;

int bt(int n, int z, int mul = 1) {
    if (n == 0)
        return 1;
    int ans = 0;
    for (auto pr : primes) {
        if (z * pr * mul > p)
            break;
        ans += bt(n - 1, 2, mul * z * pr);
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    for (int i = 2; i < MAX_N; ++i)
        if (!is_not_prime[i]) {
            primes.push_back(i);
            for (int j = i; j < MAX_N; j += i)
                is_not_prime[j] = true;
        }
    int q;
    cin >> p >> q;
    int ans[21] = {};
    for (int i = 1; i < 21; ++i)
        ans[i] = bt((i + 1) / 2, 2 - i % 2);
    while (q--) {
        int n;
        cin >> n;
        cout << accumulate(ans, ans + min(n + 1, 21), 0) << ' ';
    }
}