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.

你的问题是判断一个特定的数字是否是质数吗?然后你需要一个质数测试(很简单)。或者你需要一个给定数字之前的所有质数吗?在这种情况下，素筛是很好的(简单，但需要内存)。或者你需要一个数的质因数?这将需要分解(如果你真的想要最有效的方法，对于较大的数字很难)。你看到的数字有多大?16位?32位?更大的吗?

一种聪明而有效的方法是预先计算质数表，并使用位级编码将它们保存在文件中。文件被认为是一个长位向量，而位n表示整数n。如果n是素数，则其位设置为1，否则为0。查找非常快(您可以计算字节偏移量和位掩码)，并且不需要在内存中加载文件。

一个非常快速的Atkin Sieve的实现是Dan Bernstein的primegen。这个筛子比埃拉托色尼的筛子更有效率。他的页面有一些基准测试信息。

i wrote it today in C,compiled with tcc, figured out during preparation of compititive exams several years back. don't know if anyone already have wrote it alredy. it really fast(but you should decide whether it is fast or not). took one or two minuts to findout about 1,00,004 prime numbers between 10 and 1,00,00,000 on i7 processor with average 32% CPU use. as you know, only those can be prime which have last digit either 1,3,7 or 9 and to check if that number is prime or not, you have to divide that number by previously found prime numbers only. so first take group of four number = {1,3,7,9}, test it by dividing by known prime numbers, if reminder is non zero then number is prime, add it to prime number array. then add 10 to group so it becomes {11,13,17,19} and repeat the process.

#include <stdio.h>
int main() {    
    int nums[4]={1,3,7,9};
    int primes[100000];
    primes[0]=2;
    primes[1]=3;
    primes[2]=5;
    primes[3]=7;
    int found = 4;
    int got = 1;
    int m=0;
    int upto = 1000000;
    for(int i=0;i<upto;i++){
        //printf("iteration number: %d\n",i);
        for(int j=0;j<4;j++){
            m = nums[j]+10;
            //printf("m = %d\n",m);
            nums[j] = m;
            got = 1;
            for(int k=0;k<found;k++){
                //printf("testing with %d\n",primes[k]);
                if(m%primes[k]==0){
                    got = 0;
                    //printf("%d failed for %d\n",m,primes[k]);
                    break;
                }
            }
            if(got==1){
                //printf("got new prime: %d\n",m);
                primes[found]= m;
                found++;
            }
        }
    }
    printf("found total %d prime numbers between 1 and %d",found,upto*10);
    return 0;
}

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。

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):