C++阿特金的筛子返回了几种复合材料

C++ Sieve of Atkin Returning Several Composites

本文关键字:复合材料 几种 返回 C++      更新时间:2023-10-16

>我已经在C++年实现了我自己的阿特金筛子,它生成素数很好,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物,至少我认为是这样。我在程序内有一个变量,用于计算找到的素数的数量,并且在 ~860,000,000 处,计数比应有的多。我根据埃拉托色尼筛子的类似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 Nanosecondsn",count,limit,time);

    return 0;
}

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
对于 [1, √限制]

× [1, √限制] 中的 (x, y):是你的问题。

您使用了 :

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++)

希望这有帮助!