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我已经用 C++ 实现了自己的 Atkin 筛，它生成的素数很好，直到大约 860,000,000。在那里和更高的地方，程序开始返回几个复合材料，或者我认为。我在程序中有一个变量，用于计算找到的素数的数量，并且在 ~860,000,000 处，计数超过了应有的数量。我对照一个类似的埃拉托色尼筛子程序和几个互联网资源检查了我的计数。我是编程新手，所以这可能是一个愚蠢的错误。

无论如何，这里是：

#include <iostream>
#include <math.h>
#include <time.h>

int main(int argc, const char * argv[])
{
    long double limit;
    unsigned long long int term,term2,x,y,multiple,count=2;
    printf("Limit: ");
    scanf("%Lf",&limit);
    int root=sqrt(limit);
    int *numbers=(int*)calloc(limit+1, sizeof(int));
    clock_t time;


    //Starts Stopwatch
    time=clock();


    for (x=1; x<root; x++) {
        for (y=1; y<root; y++) {
            term2=4*x*x+y*y;
            if ((term2<=limit) && (term2%12==1 || term2%12==5)){
                numbers[term2]=!numbers[term2];
            }
            term2=3*x*x+y*y;
            if ((term2<=limit) && (term2%12==7)) {
                numbers[term2]=!numbers[term2];
            }
            term2=3*x*x-y*y;
            if ((term2<=limit) && (x>y) && (term2%12==11)) {
                numbers[term2]=!numbers[term2];
            }

        }
    }


    //Print 2,3
    printf("2 3 ");



    //Sieves Non-Primes That Managed to Get Through
    for (term=5; term<=root; term++) {
        if (numbers[term]==true) {
            multiple=1;
            while (term*term*multiple<limit){
                numbers[term*term*multiple]=false;
                multiple++;
            }
        }
    }

    time=clock()-time;

    for (term=5; term<limit; term++) {
        if (numbers[term]==true) {
            printf("%llu ",term);
            count++;
        }
    }


    printf("\nFound %llu Primes Between 1 & %Lf in %lu Nanoseconds\n",count,limit,time);



    return 0;
}
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1 回答 1

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来自 维基百科，

The following is pseudocode for a straightforward version of the algorithm:
// arbitrary search limit
limit ← 1000000         

// initialize the sieve
for i in [5, limit]: is_prime(i) ← false

// put in candidate primes: 
// integers which have an odd number of
// representations by certain quadratic forms
for (x, y) in [1, √limit] × [1, √limit]:
    n ← 4x²+y²
    if (n ≤ limit) and (n mod 12 = 1 or n mod 12 = 5):
        is_prime(n) ← ¬is_prime(n)
    n ← 3x²+y²
    if (n ≤ limit) and (n mod 12 = 7):
        is_prime(n) ← ¬is_prime(n)
    n ← 3x²-y²
    if (x > y) and (n ≤ limit) and (n mod 12 = 11):
        is_prime(n) ← ¬is_prime(n)

// eliminate composites by sieving
for n in [5, √limit]:
    if is_prime(n):
        // n is prime, omit multiples of its square; this is
        // sufficient because composites which managed to get
        // on the list cannot be square-free
        for k in {n², 2n², 3n², ..., limit}:
            is_prime(k) ← false

print 2, 3
for n in [5, limit]:
    if is_prime(n): print n

for (x, y) in [1, √limit] × [1, √limit]:是你的问题。

您使用过：

for (x=1; x<root; x++) 
        for (y=1; y<root; y++)

而是使用： 

for (x=1; x<=root; x++) 
        for (y=1; y<=root; y++)

希望这可以帮助 ！

于 2014-02-16T06:53:59.407 回答