Project Euler #23 - Non-abundant sums

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

출처: https://projecteuler.net/problem=23

자신을 제외한 약수를 모두 더한 값이 자신보다 클 경우 abundant 라고 합니다. 28123을 넘는 모든 정수는 두 abundant의 합으로 표현 가능함을 보일 수가 있습니다. 두 abundant 의 합으로 나타낼 수 없는 모든 양의 정수의 합은 얼마입니까?

my solving

c++

#include <fstream>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
 
 
int sumProperDivisors(int number)
{
    int ret = 1;
    int sqrtn = sqrt(number);
    for (int i = 2; i <= sqrtn; i++)
    {
        if (number % i == 0)
        {
            ret += i;
            int pairNum = number / i;
            if (pairNum != i)
                ret += pairNum;
        }
    }
    return ret;
}
 
int main()
{
    int ret = 0;
    vector<int> abundants;
 
    for (int i = 1; i <= 28123 ; i++)
    {
        int sum = sumProperDivisors(i);
 
        if (i < sum)
            abundants.push_back(i);
    }
 
    bool isSumAbundant[28124];
    memset(isSumAbundant, 0, sizeof(isSumAbundant));
 
    for (int i = 0; i < abundants.size(); i++)
    {
        for (int j = i; j < abundants.size(); j++)
        {
            if (abundants[i] + abundants[j] <= 28123)
                isSumAbundant[abundants[i] + abundants[j]] = true;
            else
                break;
        }
    }
 
 
    for (int i = 1; i <= 28123; i++)
    {
        if (isSumAbundant[i] == false)
            ret += i;
    }
 
    cout << ret << endl;
    system("pause");
    return 0;
}