Divisor Algorithm

I have written a workable divisor algorithm that does this:

In order to get all divisors of a some number called "Num":

You search in the the range (1, num/2 + 1) and then do a for loop:

something like this:

for (int i =0; i < num; i ++){
if (num%i) == 0);
//add to the list of divisors

Someone else told me I need to make the range up to num/2 + 1 rather than num/2. Why the +1?

> Why the +1?
Because integer division truncates.

For example, if num == 20, then 20/2 is 10, and if your loop specifies < 10, then you won't find 10 as a divisor for 20. Therefore, you add 1 to 10, and now you will find 10 as a divisor.

You only need to loop up to sqrt(num). The other divisors can be found by then dividing num with the divisors you find within that range.

Code:

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int i, count = 0;

for( i = 0; i < sqrt(num); i++)
{
    if(num % i == 0)
    {
          // Add i to list of divisors.
          // Add num / i to list of divisors.
          count ++;
    }
}

count *= 2;

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This code will give you all divisors as well as the count, with twice as little work.

Quote:

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Originally Posted by IceDane

with twice as little work.

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as in half the work? ;)

QuantumPete

Quote:

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Originally Posted by QuantumPete

as in half the work? ;)

QuantumPete

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Pedantic semantics, but should have known - board full of programmers. Haha.

Hmmmmmmmmmmm

Fixed:

Code:

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int i, count = 0, square;

for( i = 0, square = sqrt(num); i < square; i++)
{
    if(num &#37; i == 0)
    {
          // Add i to list of divisors.
          // Add num / i to list of divisors.
          count ++;
    }
}
count *= 2;

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