另一个Python实现比死亡面具推销员的答案更直接，也更快:

import numpy as np

def prime_numbers(limit: int) -> list[int]:
    """Provide a list of all prime numbers <= the limit."""
    is_prime = np.full((limit + 1, ), True)
    is_prime[0:2] = False
    for n in range(2, limit + 1):
        if is_prime[n]:
            is_prime[n**2::n] = False
    return list(np.where(is_prime)[0])

你可以进一步优化，例如，排除2，或者硬编码更多质数，但我想保持简单。

*示例运行时比较(注意:我使用了其他实现的优化形式，见我的评论):

他，他我知道我是一个回答老问题的问题巫师，但我只是在网上搜索实现有效质数测试的方法时发现了这个问题。

到目前为止，我认为最快的质数测试算法是强或然质数(SPRP)。我引用Nvidia CUDA论坛:

One of the more practical niche problems in number theory has to do with identification of prime numbers. Given N, how can you efficiently determine if it is prime or not? This is not just a thoeretical problem, it may be a real one needed in code, perhaps when you need to dynamically find a prime hash table size within certain ranges. If N is something on the order of 2^30, do you really want to do 30000 division tests to search for any factors? Obviously not. The common practical solution to this problem is a simple test called an Euler probable prime test, and a more powerful generalization called a Strong Probable Prime (SPRP). This is a test that for an integer N can probabilistically classify it as prime or not, and repeated tests can increase the correctness probability. The slow part of the test itself mostly involves computing a value similar to A^(N-1) modulo N. Anyone implementing RSA public-key encryption variants has used this algorithm. It's useful both for huge integers (like 512 bits) as well as normal 32 or 64 bit ints. The test can be changed from a probabilistic rejection into a definitive proof of primality by precomputing certain test input parameters which are known to always succeed for ranges of N. Unfortunately the discovery of these "best known tests" is effectively a search of a huge (in fact infinite) domain. In 1980, a first list of useful tests was created by Carl Pomerance (famous for being the one to factor RSA-129 with his Quadratic Seive algorithm.) Later Jaeschke improved the results significantly in 1993. In 2004, Zhang and Tang improved the theory and limits of the search domain. Greathouse and Livingstone have released the most modern results until now on the web, at http://math.crg4.com/primes.html, the best results of a huge search domain.

更多信息请看这里: http://primes.utm.edu/prove/prove2_3.html和http://forums.nvidia.com/index.php?showtopic=70483

如果您只是需要一种方法来生成非常大的质数，而不关心生成所有<整数n的质数，您可以使用Lucas-Lehmer检验来验证梅森质数。梅森质数的形式是2^p -1。我认为卢卡斯-莱默检验是目前发现的梅森质数最快的算法。

如果你不仅想使用最快的算法，而且还想使用最快的硬件，那就尝试使用Nvidia CUDA来实现它，为CUDA写一个内核，然后在GPU上运行。

如果你发现足够大的质数，你甚至可以赚到一些钱，EFF提供的奖金从5万美元到25万美元不等: https://www.eff.org/awards/coop

这是找到从1到n的所有质数的最快算法(在我的电脑上，它只花了0.004秒就找到了从1到1000000的所有质数)。

#include <iostream>
#include <fstream>

using namespace std;

double FindPrime(bool* array, int size){
clock_t start;
double runtime;
for (int i = 2; i < size; i++)
    array[i] = true;
start = clock();
for (int i = 2; i <= size; i++)
    if (array[i])
        for (int j = 2 * i; j < size; j += i)
            array[j] = false;
runtime = (double)(clock() - start) / CLOCKS_PER_SEC;
return runtime;
}


int main() {
ofstream fout("prime.txt");
int n = 0;
cout << "Enter the upper limit of prime numbers searching algorithm:";
cin >> n;
bool* array = new bool[n + 1];
double duration = FindPrime(array, n + 1);
printf("\n%f seconds.\n", duration);
for (int i = 2; i <= n; i++)
    if (array[i])
        fout << i << endl;
fout.close();

return 0;
}

寻找因素的解决方案:

def divisors(integer):
    result = set()
    i = 2
    j = integer/2
    while(i <= j):
        if integer % i == 0:
            result.add(i)
            #it dont need to 
            result.add(integer//i)
        i += 1
        j = integer//i
    if len(result) > 0:
        return f"not  prime {sorted(result)}"
    else:
        return f"{integer} is prime"

—测试---- 导入的时间

start_time = time.time()
print(divisors(180180180180))
print("--- %s seconds ---" % (time.time() - start_time))

——0.06314539909362793秒——

start_time = time.time()
print(divs(180180180180180))
print("--- %s seconds ---" % (time.time() - start_time))

——1.5997519493103027秒——

start_time = time.time()
print(divisors(1827))
print("--- %s seconds ---" % (time.time() - start_time))

——0.0秒——

start_time = time.time()
print(divisors(104729))
print("--- %s seconds ---" % (time.time() - start_time))

——0.0秒——

下面的代码:

def divs(integer):
    result = set()
    i = 2
    j = integer / 2
    loops = 0
    while (i <= j):
        if integer % i == 0:
            print(f"loops:{loops}")
            return f"{integer} is not a prime"
        i += 1
        j = integer // i
        loops += 1
    print(f"loops:{loops}")
    
    return f"{integer} is prime"

——测试——

start_time = time.time()
print(divs(180180180180180180180180))
print("--- %s seconds ---" % (time.time() - start_time))

——0.0秒——

这取决于您的应用程序。这里有一些注意事项:

你需要的仅仅是一些数字是否是质数的信息，你需要所有的质数达到一定的限度，还是你需要(潜在的)所有的质数? 你要处理的数字有多大?

米勒-拉宾和模拟测试只比筛选超过一定规模的数字(我相信大约在几百万左右)的速度快。在这以下，使用试除法(如果你只有几个数字)或筛子会更快。

Rabin-Miller是一个标准的概率质数检验。(你运行K次，输入数字要么肯定是合数，要么可能是素数，误差概率为4-K。(经过几百次迭代，它几乎肯定会告诉你真相)

拉宾·米勒有一个非概率(确定性)的变体。

The Great Internet Mersenne Prime Search (GIMPS) which has found the world's record for largest proven prime (274,207,281 - 1 as of June 2017), uses several algorithms, but these are primes in special forms. However the GIMPS page above does include some general deterministic primality tests. They appear to indicate that which algorithm is "fastest" depends upon the size of the number to be tested. If your number fits in 64 bits then you probably shouldn't use a method intended to work on primes of several million digits.