加速大数的素数查找算法

Question

My code for Project Euler problem 7 seems correct because it works for small numbers, but it takes forever for large numbers. Where did I go wrong

prime_numbers = []
for number in (range(2,2000000000)):
    if number > 1:
        for i in range(2, number):
            if number % i == 0:
                break
        else:
            prime_numbers.append(number)
print(prime_numbers[10001])

Answer 1

Look at comments and other answers for inspiration on using a faster algorithm. This answer suggests some small speed optimizations to your algorithm, which may be enough to get the result in time.

1. Stop as soon as you have your result:

    prime_numbers = [] number = 1 while True: number += 1 for i in range(2, number): if number % i == 0: break else: prime_numbers.append(number) if len(prime_numbers) > 10001: break print(prime_numbers[10001])

2. No need to look at even numbers or divisors:

    prime_numbers = [2] number = 1 while True: number += 2 for i in range(2, number, 2): if number % i == 0: break else: prime_numbers.append(number) if len(prime_numbers) > 10001: break print(prime_numbers[10001])

3. The upper limit of i can be lower: the square root of number:

    import math prime_numbers = [2] number = 1 while True: number += 2 for i in range(2, math.isqrt(number) + 1, 2): if number % i == 0: break else: prime_numbers.append(number) if len(prime_numbers) > 10001: break print(prime_numbers[10001])

4. Only look at prime divisors:

    import math prime_numbers = [2] number = 1 while True: number += 2 divisor_limit = math.isqrt(number) no_divisor = True for d in prime_numbers: if d > divisor_limit: break if number % d == 0: no_divisor = False break if no_divisor: prime_numbers.append(number) if len(prime_numbers) > 10001: break print(prime_numbers[10001])

Answer 2

Your code is valid, but not very fast - and therefore for large numbers, your program is just going to take far too long. You could try to make some changes, skip even numbers, don't range from 2 to the number but rather 2 to the square root and so on, but you could also look at a different, faster approach called the Sieve of Eratosthenes, you can read about it here .