Number of even divisors using square root

Question

I'm solving an problem where it need to get the difference between the number of even and odd divisors ,and i need to use sqrt() function because the limit of the number is 10^9 so looping on the whole number is not an option cause of the time limit exceed.

this a function I tried to make but it doesn't work perfectly on all number.

Ex. 4 & 48745.

Case 4 : should output 2 even divisors {2,4} and 1 odd divisor {1} -- the below function output 3 even 1 odd

Case 48745 :should output 0 even divisors and 4 odd divisors {1,5,9749,48745} -- the below function output 2 even 2 odd

int di(int x)
{
    int even=0,odd=0;
    for(int i=1;i<=sqrt(x);i++)
    {
        if(x%i==0)
        {
            if(i%2)
                odd++;
            else
                even++;
        if(x/i %2==0 && x/i!=i)
            even++;
        else if(x/i!=i)
            odd++;
        }
    }
    return even-odd;
}

Answer 1

Try more simple code:

#include <iostream>
#include <cmath>

int divdiff(int x)
{
    unsigned int even = 0;
    unsigned int odd  = 0;
    const unsigned int sqrtx = std::sqrt(x);

    for (int i = 1 ; i <= sqrtx ; ++i)
    {
        if (x % i == 0)
        {
            if (i % 2 == 0)
            {
                ++even;
            }
            else
            {
                ++odd;
            }
        }
    }

    even *= 2;
    odd  *= 2;

    if (x == sqrtx * sqrtx)
    {
        if (x % 2 == 0)
        {
            --even;
        }
        else
        {
            --odd;
        }
    }

    std::cerr << __func__ << '(' << x << "): even=" << even << ", odd=" << odd << std::endl;
    return even - odd;
}

int main()
{
    std::cout << divdiff(2*2) << std::endl;
    std::cout << divdiff(2*3) << std::endl;
    std::cout << divdiff(3*3) << std::endl;
    std::cout << divdiff(7*11*13*17*23) << std::endl;
}