The brute force approach to find all the divisors of a number is to iterate through every number from 1 to N and check whether it is a divisor or not. We can store all the divisors and return the list of divisors after iteration.

Code

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Function to find all divisors
    vector<int> getDivisors(int N) {
        // Create a vector to store divisors
        vector<int> res;

        // Loop from 1 to N
        for (int i = 1; i <= N; i++) {
            // Check if i is a divisor of N
            if (N % i == 0) {
                // Add i to the result
                res.push_back(i);
            }
        }
        // Return the list of divisors
        return res;
    }
};

int main() {
    // Create object of Solution class
    Solution sol;

    // Input number
    int N = 36;

    // Call the function to get divisors
    vector<int> result = sol.getDivisors(N);

    // Print the result
    cout << "Divisors of " << N << ": ";
    for (int val : result) {
        cout << val << " ";
    }
    cout << endl;

    return 0;
}
import java.util.*;

class Solution {
    // Function to find all divisors
 