Question

我正在尝试使用Sierat of Eratosthenes方法来寻找大数的最大素数因子（Project Euler中的问题3）。

我的语法似乎是正确的，我使用Long（不是int），但我收到以下错误消息：

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 1, Size: 1
    at java.util.ArrayList.rangeCheck(Unknown Source)
    at java.util.ArrayList.get(Unknown Source)
    at problem3.ProblemThree.Factor(ProblemThree.java:49)
    at problem3.ProblemThree.Recursion(ProblemThree.java:37)
    at problem3.ProblemThree.main(ProblemThree.java:83)

我不知道为什么会这样。有人可以告诉我这里我做错了吗？

 package problem3;

 import java.util.List;
 import java.util.ArrayList;

 public class ProblemThree 
 {
    //initializing variables and lists
    long factorNo;
    long nowTesting;
    int i;  
    List<Long> allPrimeList = new ArrayList<Long>();
    List<Long> ourPrimes = new ArrayList<Long>();

    ProblemThree(long x)    //constructor; the input "x" is the number whose highest prime factor is being sought
    {
        factorNo = x;     
    }       

    void initialize()   //use the workaround initialization (add 2 to the allPrimesList, set nowTesting to 3). 
                        //If the factorNo is even, add 2 to the primes list
                        //TODO: need more elegant solution 
    {
        allPrimeList.add((long) 2);
        nowTesting=3;
        if(factorNo % 2 == 0) ourPrimes.add((long) 2);
        i = 0;
    }        

    void recursion()    //keep factoring the next nowTesting until the next nowTesting is greater than half of the factorNo
    {
        while (nowTesting <= (factorNo/2))
        {
            nowTesting = factor(nowTesting);
        }
        System.out.println(ourPrimes);
    }

    long factor(long t) //The factorization algorithm. Lists all the factors of long t
    {
        nowTesting = t;

 // Line 49:
     if ((nowTesting % allPrimeList.get(i)) == 0)
        {
            i = 0;
            return (nowTesting + 2);            
        }
        else
            if(i <= allPrimeList.size()) //if we have not yet reached the end of ourPrimeList
            {
                i++;
                return nowTesting;
            }
            else    //if the end of ourPrimeList has been reached without a single modulus==0, this number is a prime
            {
                allPrimeList.add(nowTesting);

                if(factorNo%nowTesting==0) //if the nowTesting is a prime factor of factorNo, it will be perfectly divisible
                {
                    ourPrimes.add(nowTesting);
                }                    
                i=0;
                return (nowTesting+2);   
            }            
    }

    public static void main (String[] args)
    {
        ProblemThree pt = new ProblemThree(600851475143L);
        pt.initialize();
        pt.recursion();
    }
 }

Answer 1

谢谢大家耐心地浏览我的代码，我意识到这一定是非常痛苦的事情：）

我刚刚解决了这个问题。回想起来，我以前的方法似乎非常复杂。这是我使用的最终解决方案，虽然仍有改进的余地，但仍然有点优雅：

//second attempt from the ground up!
package problem3;


public class BiggestPrime 
{
    long lInput;
    long factorTest;
    long currentHeight;
    boolean divided;

    public BiggestPrime(long n)
    {
        factorTest = 2;
        currentHeight = n;

        System.out.println("The prime factors of " + n + " are:"); 

        while (factorTest<currentHeight)
        {
            if (divided == true) {factorTest = 2; divided = false;}
            if (factorTest > currentHeight) {System.out.println("factorTest is greater than currentHeight; breaking"); break;}
            if (currentHeight%factorTest==0)
            {
                System.out.println(factorTest); 
                currentHeight /= factorTest; 
                divided = true;
            } 
            else { factorTest = factorTest + 1L; divided = false;}
        }
        if (factorTest == currentHeight)
        {
            System.out.println(factorTest);
        }
        System.out.println("The end"); 

    }


    public static void main (String[] args)
    {
        BiggestPrime bp = new BiggestPrime(600851475143L);
    }

}

Answer 2

一种有趣的方法。当然，没有人应该解决你的欧拉挑战。但是你知道第二次，你进入因素＆＃39; nowTesting是3？

// The factorization algorithm. Lists all the factors of long t
long factor (final long nowTesting) 
{
    System.out.println ("entering factor: " + nowTesting);

次要想法：

 allPrimeList.add ((long) 2);

可写：

 allPrimeList.add (2L);

并且你可能认出了＃34; final＆＃34;在“长”面前因子中的参数？如果您标记最终未更改的所有内容，它有助于推理代码。在实践中，结果是，你的Javacode混乱了最后的＆＃39;修饰语，但它是怎样的。它是良好代码的标志 - 也许不是好的设计。最终可能是默认值。

Answer 3

在第49行，你不应该检查nowTesting是否可以被i整除，而不是allPrimes的第i个元素？

项目欧拉概率。 3 IndexOutOfBoundsException

3 个答案: