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;
}

#include<stdio.h>
main()
{
    long long unsigned x,y,b,z,e,r,c;
    scanf("%llu",&x);
    if(x<2)return 0;
    scanf("%llu",&y);
    if(y<x)return 0;
    if(x==2)printf("|2");
    if(x%2==0)x+=1;
    if(y%2==0)y-=1;
    for(b=x;b<=y;b+=2)
    {
        z=b;e=0;
        for(c=2;c*c<=z;c++)
        {
            if(z%c==0)e++;
            if(e>0)z=3;
        }
        if(e==0)
        {
            printf("|%llu",z);
            r+=1;
        }
    }
    printf("|\n%llu outputs...\n",r);
    scanf("%llu",&r);
}    

我最近写了这段代码来求数字的和。它可以很容易地修改，以确定一个数字是否是质数。基准测试在代码之上。

// built on core-i2 e8400
// Benchmark from PowerShell
// Measure-Command { ExeName.exe }
// Days              : 0
// Hours             : 0
// Minutes           : 0
// Seconds           : 23
// Milliseconds      : 516
// Ticks             : 235162598
// TotalDays         : 0.00027217893287037
// TotalHours        : 0.00653229438888889
// TotalMinutes      : 0.391937663333333
// TotalSeconds      : 23.5162598
// TotalMilliseconds : 23516.2598
// built with latest MSVC
// cl /EHsc /std:c++latest main.cpp /O2 /fp:fast /Qpar

#include <cmath>
#include <iostream>
#include <vector>

inline auto prime = [](std::uint64_t I, std::vector<std::uint64_t> &cache) -> std::uint64_t {
    std::uint64_t root{static_cast<std::uint64_t>(std::sqrtl(I))};
    for (std::size_t i{}; cache[i] <= root; ++i)
        if (I % cache[i] == 0)
            return 0;

    cache.push_back(I);
    return I;
};

inline auto prime_sum = [](std::uint64_t S) -> std::uint64_t {
    std::uint64_t R{5};
    std::vector<std::uint64_t> cache;
    cache.reserve(S / 16);
    cache.push_back(3);

    for (std::uint64_t I{5}; I <= S; I += 8)
    {
        std::uint64_t U{I % 3};
        if (U != 0)
            R += prime(I, cache);
        if (U != 1)
            R += prime(I + 2, cache);
        if (U != 2)
            R += prime(I + 4, cache);
        R += prime(I + 6, cache);
    }
    return R;
};

int main()
{
    std::cout << prime_sum(63210123);
}

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。

如果它必须非常快，你可以包括一个质数列表: http://www.bigprimes.net/archive/prime/

如果你只想知道某个数是不是质数，维基百科上列出了各种质数判别法。它们可能是确定大数是否为质数的最快方法，特别是因为它们可以告诉你一个数是否为质数。

#include <iostream>

using namespace std;

int set [1000000];

int main (){

    for (int i=0; i<1000000; i++){
        set [i] = 0;
    }
    int set_size= 1000;
    set [set_size];
    set [0] = 2;
    set [1] = 3;
    int Ps = 0;
    int last = 2;

    cout << 2 << " " << 3 << " ";

    for (int n=1; n<10000; n++){
        int t = 0;
        Ps = (n%2)+1+(3*n);
        for (int i=0; i==i; i++){
            if (set [i] == 0) break;
            if (Ps%set[i]==0){
                t=1;
                break;
            }
        }
        if (t==0){
            cout << Ps << " ";
            set [last] = Ps;
            last++;
        }
    }
    //cout << last << endl;


    cout << endl;

    system ("pause");
    return 0;
}