c - Counting Positive Integers with a Given Number of Divisors -


Basically what I was trying to do involves an integer K which represents the number of divisors And then searches for all those numbers that are 1-100000 K separator 

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Although I do not seem to be able to do this, please help me because I am quite new to the programming scene and I have not received any response from anywhere else. 1 , * a 2 Sup> b 2 ... a n sub then the number of separators of this integer (B 1 + 1) * (b 2 + 1) ... (b n + 1).

Now that you have this theorem, you can modify the sieve of a little irretothenes so that all the integers can meet up to 100,000 canonical. Form

Here is some code that means me with the modified aerthostenes sieve.

const int size = 100000; Int devs [size + 1]; Zero compute_devs () {for (int i = 0; i

After calling compute_devs , the value of devs will be stored in the value of the largest original denominator of each number in size. I will leave you the rest of the work, but when this array is, it moves very straight forward.


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