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c - 埃拉托斯特尼筛法,素数 C

转载 作者:行者123 更新时间:2023-11-30 21:23:35 30 4
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我正在编写一个程序,使用埃拉托斯特尼筛法查找 1000 之前的素数...但是它不起作用...这是我的整个代码,但是查找素数的“计算”位于函数“markPrimes”内部(我确信其余代码都没有问题,所以我很确定问题出在这个函数中......

#include <stdio.h>
#include <stdlib.h>

typedef struct primal
{
int number; /* a number */
char mark; /* flag marking the number as active (1) or inactive (0) */
} primal;


void initialize(primal *s,int size)
{
int i;
for(i=0;i<1000;i++)
{
s[i].number=i+1;
s[i].mark=1; //1=prime number
}
}

void markPrimes(primal *s,int size)
{
/* add this function - it should mark all of the numbers in the passed primal array that are not prime numbers as inactive */
//Brute Sieve of Eratosthenes Approach (0=not prime number)

s[0].mark=0; //s[0].number=1 as on the function "initialize" I start from 1 not from 0
int i,j;
for (unsigned i = 2; i*i <size; i++)
{
if (s[i].mark == 1)
for (unsigned j = i<<1;j<size;j+=i)
s[j].mark = 0;
}
}

int main(void)
{
int i,j,prime_numbers[200];
primal source_numbers[1000]; /* an array of source values */

for(i=0;i<200;i++) prime_numbers[i]=0; /* initialize the prime numbers array to 0 */

initialize(source_numbers,1000); /* initialize the source numbers array to hold the numbers 1-1000 */

markPrimes(source_numbers,1000); /* identify the prime numbers in the source numbers array */

/* copy the primes from the source numbers to the prime numbers array */
for(i=0,j=0;i<1000;i++)
{
if(source_numbers[i].mark==1) /* if the current source number is a prime */
{
prime_numbers[j]=source_numbers[i].number; /* copy the number */
j++; /* increment the target index */
}
}

/* print the prime numbers */
for(i=0,j=0;prime_numbers[i]!=0;i++,j++)
{
printf("%3d ",prime_numbers[i]);
if(j==9) /* periodically print a newline and then reset j */
{
printf("\n");
j=-1;
}
}
return 0;
}

最佳答案

众所周知,C 允许程序员在他们的脚上打点(别担心,我们每个人都有这样的疤痕),这就是这里发生的事情。

在您的内部循环中,您从 1 开始,而不是 0,因此您需要将实际数字减一作为索引,并添加该数字而不是索引:

void markPrimes(primal *s,int size)
{
//Brute Sieve of Eratosthenes Approach (0=not prime number)

int i,j;
// 1 (one) is not prime per definition
s[0].mark = 0;
for (i = 1; i*i <size; i++)
{;
if (s[i].mark == 1) {
// you start at 1 instead of 0, so you need to take the actual number
// minus one as the index and add that number instead of the index.
for ( j = 2 * s[i].number - 1;j < size; j += s[i].number){
s[j].mark = 0;
}
}
}
}

关于c - 埃拉托斯特尼筛法,素数 C,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43217138/

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