尝试仅将素数打印到输出文件中

Question

我必须创建一个包含一串数字的输入文件。我需要我的程序来创建一个输出文件，其中只有输入文件的质数。我完全不知道如何创建一个循环来检查我的输入文件中的素数并创建一个只有素数的输出文件。

import java.util.Scanner;
import java.io.*;


public class ClassWork5_3 
{


    public static void main(String[] args) throws IOException 
    {
        Scanner keyboard = new Scanner(System.in);
        System.out.println("Enter file name: ");
        String filename = keyboard.nextLine();
        PrintWriter pw = new PrintWriter("output.txt");

        File file = new File(filename);
        Scanner inputFile = new Scanner(file);
        int line = inputFile.nextInt();


       while(inputFile.hasNext())
       {
         isPrime(line);

       }

    }

    public static boolean isPrime(int num)
    {
        boolean status;
        for(int i = 2; i < num/2; i++)
        {
            if (num%i==0)
            {
             status = false;
            }

        }
        return true;
    }   
}

Answer 1

首先，你的方法isPrime()不正确。
这个应该可以工作，尽管它可能不是最有效的：

public static boolean isPrime(int n)
{
  // Manage easy cases
  if (n <= 1)
    return (false);
  else if (n == 2)
    return (true);
  else if ((n % 2) == 0)
    return (false);

  // Check if (odd) number can be divided by something
  for (int i = 3; i <= n / 2; i += 2)
  {
    if ((num % i) == 0)
      return (false);
  }

  // If we get here, we got a prime number
  return (true);

} // isPrime

那么，你的while 循环应该是这样的：

while(inputFile.hasNext())
{
  line = inputFile.nextInt();
  if (isPrime(line))
    pw.println(line);
}

Answer 2

此方法遍历n 到Math.sqrt(n) 的可能除数：

public static boolean isPrime(int n) {
    if (n < 2)
        return false;
    else if (n <= 3 )
        return true;

    boolean notPrime = true;

    for (int divisor = 2; divisor <= Math.sqrt(n); divisor++) {
        notPrime = (n % divisor == 0);
        if (notPrime)
            break;
    }

    return !notPrime;
}

Answer 3

这是使用 while 循环和 Math.sqrt() 的另一种可能的解决方案：

public class X1 {

public static void main(String[] args) {
    File inputFile = null;
    Scanner fileInput = null;
    PrintStream fileOutput = null;

    try {
        inputFile = new File("input.txt");
        fileInput = new Scanner(inputFile, "UTF-8");
        fileOutput = new PrintStream("output.txt", "UTF-8");

        while (fileInput.hasNext()) {
            int number = fileInput.nextInt();
            if (isPrime(number))
                fileOutput.println(number);
        }

    } catch (FileNotFoundException fnfe) {
        System.out.println("File not found!");
    } catch (UnsupportedEncodingException uee) {
        System.out.println("Unsupported encoding!");
    } finally {
        if (fileInput != null) {
            fileInput.close();
        }
        if (fileOutput != null) {
            fileOutput.close();
        }
    }

}

private static boolean isPrime(int n) {
    int divider = 2;
    boolean prime = true;
    while (prime && (divider <= (int) Math.sqrt(n))) {
        if (n % divider == 0) {
            prime = false;
        }
        divider++;
    }
    return prime;
}

}

Answer 4

代码是针对C++的，但是算法是一样的。

bool IsPrime(unsigned int num)
{
    // 2 is the first prime number; 
    if(num < 2) return false;

    // We check dividers up to the root of the given number,
    //because after that the multipliers are the same, but with switched places. 
    //Example: 12 = 1*12 = 2*6 = 3*4 = 4*3 = 6*2 = 12*1
    unsigned int limit = sqrt(num);

    //We calculated the limit outside the loop so it's calculated only once
    //instead of being calculated at every iteration of the loop.
    for(unsigned int i=2; i < limit; ++i)
    {
        //If we divide without a remainder, then it's not prime.
        if(num%i==0)
            return false;
    }

    //We have tried with all the possible multipliers and have found no dividers.
    return true;
}