Java 将主生成器存储到阵列中

我现在正在学习阵列

我做了一个简短的小节目。这是一个测试然后打印素数的循环。 特别是1到100之间的数字

我想知道我是否可以进行循环并将每个素数存储到一个数组中。这是我的密码：

public class QC3PrimeNumbers
{

    public static void main (String[] args)
   {
      System.out.println ("Here are the prime numbers: ");

       for (int index = 2; index < 100; index++)
      {
         if (index%2 != 0 && index%3 !=0)
         System.out.print (index + " ");         
      }

   }
}

---

如果使用此条件：如果索引%2！=0&索引%3=0，那么它永远不会考虑2个或3个素数。如果您将其更改为：

 if (index%2 != 0 && index%3 !=0 && index%5 !=0 && index%7 !=0)

它不会发现2、3、5或7是素数

我修正了你的算法，并决定使用一个存储素数列表，因为它的大小是动态的。否则，我们首先要找到你所在区域的素数来确定数组的大小，然后再做另一个同样的循环，除了这次将这些数添加到数组中

一旦你找到一个素数。使用ArrayList.addprimeNumberFound；的名称

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使用ArrayList执行此操作 _________________________________________________________________________________

使用常规数组执行此操作 以下是如何使用不推荐使用的常规阵列执行此操作，但是：

public class QC3PrimeNumbers
{

    public static void main (String[] args)
    {
        System.out.println ("Here are the prime numbers: ");

        int numOfPrimes = 0; // counts the # of prime nums in your region

        // this loop will count up the number of prime numbers
        for (int index = 2; index < 100; index++)
        {
            boolean isPrime = true; // initially true, reset this every loop

            // for every number that can factor into the index number
            for (int i = 2; i < index; i++)
                // if a number is found that factors into it, it's not prime
                if (index % i == 0) isPrime = false;

            if (isPrime) // if this number is prime
                numOfPrimes++; // add it to the count
        }

        // array to hold the prime nums w/ size=how many prime nums we counted
        int[] primeNums = new int[numOfPrimes];

        int count = 0; // keep track of the position in the primeNums array

        for (int index = 2; index < 100; index++)
        {
            boolean isPrime = true; // initially true, reset this every loop

            // for every number that can factor into the index number
            for (int i = 2; i < index; i++)
                // if a number is found that factors into it, it's not prime
                if (index % i == 0) isPrime = false;

            if (isPrime) // if this number is prime
            {
                primeNums[count] = index; // add this prime num to the array
                count++; // change the array position tracker for next number
            }
        }

        for (int n : primeNums) // for every integer in primeNums array
           System.out.print(n + " "); // display that integer to the console
    }
}

---

这不会生成素数。修复您的算法并回答您的问题，您可以使用ArrayList存储找到的素数。注意：数组的大小是固定的。您确定对素数的测试是正确的吗？素数是一个大于1的自然数，除1和它本身外，没有其他正因子。大于1的非素数的自然数称为复合数。索引%5、索引%7以及2和3的所有非倍数如何？herpite derp。你们说得对。所以我必须加上5和7？还有其他可除数吗？数组是固定的，但您不需要填充它们，不是吗？编译器给了我两个无法找到指向列表和数组列表的符号错误。我正在使用第一段代码。@munchshair尝试添加import java.util.ArrayList；并导入java.util.List；到你们班的第一名。哇！现在它跑了！它还活着

public class QC3PrimeNumbers
{

    public static void main (String[] args)
    {
        System.out.println ("Here are the prime numbers: ");

        int numOfPrimes = 0; // counts the # of prime nums in your region

        // this loop will count up the number of prime numbers
        for (int index = 2; index < 100; index++)
        {
            boolean isPrime = true; // initially true, reset this every loop

            // for every number that can factor into the index number
            for (int i = 2; i < index; i++)
                // if a number is found that factors into it, it's not prime
                if (index % i == 0) isPrime = false;

            if (isPrime) // if this number is prime
                numOfPrimes++; // add it to the count
        }

        // array to hold the prime nums w/ size=how many prime nums we counted
        int[] primeNums = new int[numOfPrimes];

        int count = 0; // keep track of the position in the primeNums array

        for (int index = 2; index < 100; index++)
        {
            boolean isPrime = true; // initially true, reset this every loop

            // for every number that can factor into the index number
            for (int i = 2; i < index; i++)
                // if a number is found that factors into it, it's not prime
                if (index % i == 0) isPrime = false;

            if (isPrime) // if this number is prime
            {
                primeNums[count] = index; // add this prime num to the array
                count++; // change the array position tracker for next number
            }
        }

        for (int n : primeNums) // for every integer in primeNums array
           System.out.print(n + " "); // display that integer to the console
    }
}

import java.io.*;
import java.util.*;              
class numbers
{
public static void main(String args[])
{
    int[] prime=new int[20];
    int[] nonprime=new int[20];
    int count=0;
    int count1=0;`enter code here`
    try
        {
        InputStreamReader r=new InputStreamReader(System.in);  
        BufferedReader br=new BufferedReader(r);  
        System.out.println("enter n value");
        int n=Integer.parseInt(br.readLine());
        for(int i=1;i<n;i++)
        {   
            if(i%2!=0)
            {
                if(i%3!=0)
                    {
                    prime[count]=i;
                    count++;
                    }
            }
            else    
            {
            nonprime[count1]=i;
            count1++;
            }
        }
        System.out.println("prime");
        for(int m=0;m<count;m++)
        {
            System.out.println(prime[m]);
        }
        System.out.println("Non-prime");
        for(int k=0;k<count1;k++)
        {
            System.out.println(nonprime[k]);
        }
    }
    catch(Exception e)
    {
        System.out.println(e);
    }
}
}