If a number n is not a prime, it can be factored into two factors a and b: n = a * b Now a and b can’t be both greater than the square root of n, since then the product a * b would be greater than sqrt(n) * sqrt(n) = n. So in … Read more
Fastest way to list all primes below N
Just declare notAPrime inside for(int i = n; i > 2; i–) to make it local for each i – candidate to be prime. Now you increase notAPrime and never reset it to zero.
The range function creates an array of numbers between its parameters. For example: >>> range(1, 4) => [1, 2, 3] In your code you’re saying a%b which is finding the remainder between a number the user inputted and a range object. Think of the range object as a python list. So what you’re doing is … Read more
The only way to do this in O(1) would be an array of all prime numbers. So, you’d only be able to support a certain number depending on your computer’s memory. Edit: There might be some way to compute this involving a bunch of calculus, but that’s beyond me 🙂
The are several problems with your program. The first one is that the loop starting with statement while (number1 == 5) is infinite because number1 is not changed within the loop. The second one is that you must always initialize variable a to zero within the loop. And it should be defined also within the … Read more
First for future reference you should probably post that on code review, unless there is a specific problem then you should create a Minimal, Complete, and Verifiable post. There is nothing inherently wrong other than you do not check that N>0 which could lead to an infinite loop, j should be going to i not … Read more