Euler Project No. 14

I am currently solving a Problem with Project Euler 14 :

The following iterative sequence is defined for a set of positive integers:

n → n/2 (n is even)
n → 3n + 1 (n is odd)  

Using the rule above and starting with 13, we generate the following sequence:
               13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1  

Which starting number, under one million, produces the longest chain?

To solve the problem, I developed the following algorithm:

• Instead of alternately finding a series for each number that will contain a lot of redundant calculations, I try to expand the series back from 1. That is, start with a number and predict the element in front of it.
• Since multiple episodes can be generated, save the last number of all episodes using the linked list. (The idea is to store only those items that have longer rows.)
• I will loop this until I find all the numbers under the given limit; the last number under the limit will have the longest row.

Here is my code:

void series_generate(long num)
{
    long n = 1;
    addhead(n);             //add 1 to list.
    while (n < num) 
    {
            series *p;
            for (p = head; p != NULL; p = p->next)
            {
                    long bf = p->num - 1;
                    if (p->num%2 == 0 && bf != 0 && bf%3 == 0) {
                            bf /= 3;
                            if (bf != 1)
                                    addhead(bf);
                            if (bf < num)
                                    n++; 
                    }
                    p->num *= 2;
                    if ( p->num < num)
                            n++;

            }       
    }       
}

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