质数更好的算法

Question

I'm working on a program which will found the n th. prime number. For example, By listing the first six prime numbers: 2, 3, 5, 7, 11 and 13 , we can see that the 6th prime is 13 . I'm trying to making an algorithm, like, if I want to see 50th prime, I will add 1 to ends of the range() function. I'm using this algorithm to find primes at the moment;

cnt = 1
print (2)
for x in range(3,40,2):
    div = False
    for y in range(2,round(x**0.5)+1):
        if x%y == 0:
            div = True
    if div == False:
        print (x)
        cnt += 1

print ("\nThere is {} prime numbers.".format(cnt))

You see that, I put 40 . But I want to put n there, so for example, untill reaching the 50th prime, add +1 to n . But it's not going to like that, if I tried something like;

cnt = 1
n = 40 #example
while cnt<50:
    for x in range(3,n,2):
        #codes
    if div == False:
        n += 1

I thought when the program finds a prime, it will add +1 to n and while loop will process untill find the 50th prime. But it didn't, primes are wrong if I use this one also, nothing relevant what I want to do.

• How to make this algorithm, obviously changing the last element of range() function does not working.

• Is there better/elegant algorithm/way? If I want to find 200.000th prime, I need faster codes.

Edit: I was working with lists first but, I got MemoryError all the time when working with big numbers. So I pass that and using a variable that counting how much primes are there cnt .

Answer 1

Here is a much faster version

primes = []
primecount = 0

candidate = 2
while primecount<50:
    is_prime = True
    for prime in primes:
        if candidate%prime == 0:
            is_prime = False
            break
        elif candidate < prime**2:
            break
    if is_prime:
        primes.append(candidate)
        primecount += 1

    candidate += 1
print primes[-1]

note small edit adding the candidate<prime**2 test that OP included but I neglected initially.

Your code is going to be very slow for several reasons. If 2 divides a number you know it's not prime, but you're still checking whether 3 or 4 or 5... divides it. So you can break out as soon as you know it's not prime. Another major issue is that if 2 does not divide a number, there's no reason to check if 4 divides it as well. So you can restrict your attention to just checking if the primes coming before it divide it.

In terms of run time:

Answer 2

First off, for backwards compatibility with python 2, I added an int() around your rounding of root x.

From what I understand of your question, you are looking for something like this:

cnt = 1
maximum=50 #Specify the number you don't want the primes to exceed
print (2)
n=20 #highest number 
while cnt<maximum: #
    for x in range(3,n,2): #using "n" as you hoped
        div = False
        for y in range(2,int(round(x**0.5)+1)):
            if x%y == 0:
                div = True
        if div == False:
            if x<maximum: #we only want to print those up to the maximum
                print str(x)
            else: #if it is greater than the maximum, break the for loop
                break
            cnt += 1
    break #break the while loop too.
print ("\nThere are {} prime numbers.".format(cnt))

This gave me the correct primes. As for better/more elegant solutions. If by better you want speed, use a library that has an optimized, or look at this for an example of a fast program. Elegance is subjective, so I will leave that to the rest of the internet.