Question

我目前正在学习C并试图解决问题。我需要使用数组和循环来查找2到100之间的所有素数。

我已经知道解决此问题的方法，但是我在查找代码错误时遇到了麻烦。这是我第一次使用StackOverflow，所以希望我对所有内容进行了正确的评论：）

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

int main(){

  int prime_numbers[50] = {2,3};                //initializes the array which will be printed at the end
  int counter = 1;                              //initializes the index of the last prime element in array
  int checker = 0;                              //initializes a checker used to determine if the number is prime

  for(int i = 5; i <= 100; i++) {               //goes through numbers 5 to 100 as the first two primes are hard coded
    for(int j = 0; j <= counter; j++){          //goes through array untill it reaches last prime using the before initialized counter
        if(i % prime_numbers[j] != 0) {         //check to see if a number that is being checked is not divisible by j'th element in array
            checker++;                          //if so, checker is incremented
        }
        if(checker == counter + 1) {            //check to see if number was not divisible by any prime in our array
            checker = 0;                        //if so checker is reset to 0 for the next iteration
            ++counter;                          //counter is incremented as there is one more prime in our array
            prime_numbers[counter] = i;         //add inside array the found prime number
            break;                              //break should not be necessary, however for some reason, it yields a different result when I don't put it in
        }                                       //most likely the error in the code. Need to find out why loop does not stop after second if is done
    }
  }

  for(int g = 0; g <= 50; g++) {                //prints out all the prime numbers in array
    if(prime_numbers[g] != 0) {
        printf("%d ", prime_numbers[g]);
    }
  }

  return 0;
}

我希望程序打印所有从0到100的质数，之间要有空格。

Answer 1

程序还会查找非质数。它的逻辑是一个相当奇怪的反转。候选对象仅需被数组中的一个素数整除。报告还打破了数组界限。

更正的代码：

#include <stdio.h>

int main(void) {                            // correct function signature

    int prime_numbers[50] = {2,3};
    int counter = 1;
    int checker;                            // initialise within each loop

    for(int i = 5; i <= 100; i++){
        checker = 0;                        // inintialise here
        for(int j = 0; j <= counter; j++) {
            if(i % prime_numbers[j] == 0) { // opposite test to yours
                checker++;                  // flag a non-prime
                break;
            }
        }

        if(checker == 0) {                  // moved outside the loop
            ++counter;
            prime_numbers[counter] = i;
        }
    }

    for(int g = 0; g < 50; g++) {           // corrected the bounds error
        if(prime_numbers[g] != 0){
            printf("%d ", prime_numbers[g]);
        }
    }
    printf("\n");                           // flush the output buffer
    return 0;
}

程序输出：

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

报告循环依赖于初始化为0的数组，并且更好的方法是

for(int j = 0; j <= counter; j++) {
    printf("%d ", prime_numbers[j]);
}
printf("\n");

Answer 2

以下代码效果很好。

int main()
{
    int k=1,temp,j;
    const int N_prime = 10000;       // number of primes to be generated
    int primes[N_prime];             // array to save primes
    primes[0] = 2;
    primes[1] = 3;
    temp = 5;
    while (k!=N_prime-1){     // generating only N_prime prime numbers
        for (j = 0; j <= k && primes[j] * primes[j] <= temp; j++){
            if (temp%primes[j] == 0){    // if temp%primes[j] == 0 then temp is divisible by
                temp += 2;               // a prime and is not a prime itself, therefore 
                break;                   // immediately break
            }
            else if (primes[j+1] * primes[j+1]>temp){             // if no such primes found, temp is prime,
                primes[k + 1] = temp;                             // save it and increase the value for
                k++;                                              // next check
                temp += 2;
            }
        }
    }
    for (int ind = 0; ind < N_prime; ind++) printf("%d\n",primes[ind]);
    getch();
    return 0;
}

输出：

正如我注意到的那样，您的代码会生成所有最大为100的奇数。

使用C查找素数不起作用

2 个答案: