無法在一個範圍內找到DISTINCT素因子的數量

淘氣數字是其不同素數因子的數目等於其十進制表示中的位數。 數字1被認爲是頑皮的號碼。以下是查找淘氣號碼的代碼。問題是方法的主要因素，它進入了一個無限循環。無法在一個範圍內找到DISTINCT素因子的數量

import java.util.ArrayList; 
import java.util.Scanner; 
import java.util.TreeSet; 
import java.util.Iterator; 

public class NaughtyNumber { 
    ArrayList <Integer> aldecrep = new ArrayList < >(); // use for decimal representation 
    TreeSet <Integer> tsprimefact = new TreeSet < >(); // use for store of prime factors  
    ArrayList <Integer> alsize1 = new ArrayList < >(); // use for storing number of prime factors 
    public static void main(String[] args) { 
      Scanner in = new Scanner(System.in); 
      int query, ul, dl; 
      System.out.println("Enter the nuber of queries"); 
      query = in .nextInt(); 
      System.out.println("Enter upperlimit and down limit"); 
      while (query != 0) { 
       dl = in .nextInt(); 
       ul = in .nextInt(); 
       NaughtyNumber n1 = new NaughtyNumber(); 
       //NaughtyNumber n2=new NaughtyNumber(); 
       //NaughtyNumber n3=new NaughtyNumber(); 
       n1.decal(dl, ul); 
       n1.primefactor(dl, ul); 
       n1.compare(dl); 
       query--; 
      } 
     } 
     //count number of digits   
    void decal(int dl, int ul) { 
      for (int i = dl; i <= ul; i++) { 
       int length = (int)(Math.log10(i) + 1); //calculation of length of a  number 
       aldecrep.add(length); 
      } 
     } 
     //count number of digits ends 
     //prime factorization starts 
    void primefactor(int dl, int ul) { 
      for (int i = dl; i <= ul; i++) { 
       for (int j = 2; j <= i; j++) { 
        if (i % j == 0) { 
         i = i/j; 
         tsprimefact.add(j); // add distinct prime factors 
         j--; 
        } 

        alsize1.add(tsprimefact.size()); //add treesize to set size1 
        tsprimefact.clear(); //empty ts 

       } 
      } 
      Iterator <Integer> itr1 = alsize1.iterator(); 
      while (itr1.hasNext()) { 
       Integer n1 = itr1.next(); 
       System.out.println(n1); 
      } 
     } 
     //prime factorization ends  
     // compare to find naughty number 
    void compare(int dl) { 
     Iterator <Integer> itr = aldecrep.iterator(); 
     Iterator <Integer> itr1 = alsize1.iterator(); 
     int count = 0; 
     if (dl == 1) { 
      count = count + 1; 
     } 
     while (itr.hasNext() && itr1.hasNext()) { 
      Integer n1 = itr.next(); 
      Integer n2 = itr1.next(); 
      if (n1 == n2) { 
       count++; 
      } 
     } 
     System.out.println(count); 
    } 
}

來源

2016-08-12 Fawkes

你能解決問題嗎？ – Mark

如果我想找到一個單一的質量因子，它的工作原理。但是，我試圖把它放入範圍它進入無限循環。所以我修改了代碼而不是傳遞（dl，ul）我開始傳遞單個值到主因子法。儘管我想知道爲什麼當我傳遞（dl，ul）參數時它會進入無限循環的原因。 – Fawkes

使用因式分解是這樣的：

public static Set<Integer> primeFactors(int numbers) { 
    int n = numbers; 
    Set<Integer> factors = new HashSet<Integer>(); 
    for (int i = 2; i <= n/i; i++) { 
     while (n % i == 0) { 
     factors.add(i); 
     n /= i; 
     } 
    } 
    if (n > 1) { 
     factors.add(n); 
    } 
    return factors; 
    }

我修改http://www.vogella.com/tutorials/JavaAlgorithmsPrimeFactorization/article.html

集合的源減少了在DISTINCT像這樣你的元素。

來源

2016-08-12 08:59:05 Mark

無法在一個範圍內找到DISTINCT素因子的數量

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