Amicable numbers | Project Euler | Problem #21

URL to the problem page: https://projecteuler.net/problem=21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).

If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284.
The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

#include <iostream>

using namespace std;

int main()

{

    long long i, j, b, a, result = 0, counter, counter2, sum, number;

    for (i = 1; i < 10000; i++) {

        counter = 0; a = 0; sum = 0; counter2 = 0; b = 0; number = 0;

        for (j = 1; j <= (i / 2); j++) {

            if (i % j == 0) {

                counter++;

            }

        }

        long long* factors = new long long[counter];

        for (j = 1; j <= (i / 2); j++) {

            if (i % j == 0) {

                factors[a] = j;

                a++;

            }

        }

        for (j = 0; j < a; j++) {

            sum += factors[j];

        }

        for (j = 1; j <= (sum / 2); j++) {

            if (sum % j == 0) {

                counter2++;

            }

        }

        long long* factors2 = new long long[counter2];

        for (j = 1; j <= (sum / 2); j++) {

            if (sum % j == 0) {

                factors2[b] = j;

                b++;

            }

        }

        for (j = 0; j < b; j++) {

            number += factors2[j];

        }

        if (i == number) {

            if (i != sum) {

                result += i;

            }

        }

    }

    cout << "Sum of all the amicable numbers under 10.000 is  =  " << result << endl;

    return 0;

}

CLICK TO SEE THE OUTPUT.