Question

我已经在C ++中自己实现了Atkin的筛选，它可以生成质数，直到大约860,000,000。在那里和更高的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来计算找到的素数的数量，并且在~860,000,000时，计数超过它应该的数量。我查看了针对Eratosthenes筛选的类似程序和几个互联网资源的计数。我是编程的新手，所以这可能是一个愚蠢的错误。

无论如何，这是：

#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;
}

Answer 1

来自wikipedia，

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

对于[1，√limit]×[1，√limit]中的（x，y）

：是你的问题。

您使用过：

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

改为使用：

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

希望这有帮助！

Atkin的C ++筛选返回几种复合材料

1 个答案: