I know it's somewhat later, but this could be useful to people arriving here from searches. Anyway, here's some JavaScript that relies on the fact that only prime factors need to be tested, so the earlier primes generated by the code are re-used as test factors for later ones. Of course, all even and mod 5 values are filtered out first. The result will be in the array P, and this code can crunch 10 million primes in under 1.5 seconds on an i7 PC (or 100 million in about 20). Rewritten in C it should be very fast.

var P = [1, 2], j, k, l = 3

for (k = 3 ; k < 10000000 ; k += 2)
{
  loop: if (++l < 5)
  {
    for (j = 2 ; P[j] <= Math.sqrt(k) ; ++j)
      if (k % P[j] == 0) break loop

    P[P.length] = k
  }
  else l = 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;
}

#include<iostream>
using namespace std;

void main()
{
    int num,i,j,prime;
    cout<<"Enter the upper limit :";
    cin>>num;

    cout<<"Prime numbers till "<<num<<" are :2, ";

    for(i=3;i<=num;i++)
    {
        prime=1;
        for(j=2;j<i;j++)
        {
            if(i%j==0)
            {
                prime=0;
                break;
            }
        }

        if(prime==1)
            cout<<i<<", ";

    }
}

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。