Javaプログラミングを使用した完全な数字

Question

Hellooo 1とユーザーとその要素によって入力された整数の間の完全な数字をすべて表示しようとしています。私は

に...コードそれは私の出力はかなり私が所望の出力と一致していないだけということです細かい作業を持ってここに私のコードです：

import java.util.Scanner;  

public class PerfectNumbers {  
    public static boolean isPerfect(int a) {    
     int n = a; 
     int sum = 0; 
     boolean perfect; 

     while (n-- >1) {     
      if(a%n==0) 
       sum = sum + n; 
      } 
      if (sum == a) { 
       perfect = true; 
      } else { 
       perfect = false; 
      } 
      return perfect; 
     } 

    public static void main(String[] args) { 
     Scanner input = new Scanner(System.in); 
     System.out.print("Enter the number up to which you would like to look for perfect numbers:"); 
     int i = input.nextInt(); 
     String factors = "";   
     System.out.printf("Looking for perfect numbers from 1 to %d%n", i); 

     while (i > 0) { 
      if (PerfectNumbers.isPerfect(i) == true) { 
       int w = i-1; 
       while (w-- > 1) { 
        if(i % w == 0) 
         factors = factors + " " + Integer.toString(w); 
       } 
       System.out.println(i + " is a perfect number it's factors are:" + factors); 
      } 
      i = i - 1; 
     } 
    } 
} 

これは何を私の出力を返す

ですしかし、私は、それは逆の順序で、この代わりに

Answer 1

あなたのループを表示したい、FRを言います1000から0まで。要因

EDIT

ソリューション提案のループのためにあなたの

while (i>0) 

で

for (int j=1;j<i+1;j++) 

同じことを変更し、変更のみが主な方法である

public static void main(String[] args) { 

    Scanner input = new Scanner(System.in); 

    System.out.print("Enter the number up to which you would like to look for perfect numbers:"); 
    int inp = input.nextInt(); 


    System.out.printf("Looking for perfect numbers from 1 to %d%n", inp); 

    for(int j=1;j<inp+1;j++) { 

     if (PerfectNumbers.isPerfect(j) == true) { 


     String factors = ""; 
     for (int w=1; w<j;w++) { 

     if(j % w == 0) 
      factors = factors + " " + Integer.toString(w); 

     } 
     System.out.println(j + " is a perfect number it's factors are:" + factors); 
    } 

    } 
} 

Answer 2

それはbこのように書かれたe？

import java.util.Scanner; 

public class PerfectNumbers { 

    public static boolean isPerfect(int a) { 

     int n = a; 
     int sum = 0; 
     boolean perfect; 

     while (n-- > 1) { 

      if (a % n == 0) 
       sum = sum + n; 

     } 
     if (sum == a) { 

      perfect = true; 

     } else { 

      perfect = false; 

     } 

     return perfect; 

    } 

    public static void main(String[] args) { 

     Scanner input = new Scanner(System.in); 

     System.out.print("Enter the number up to which you would like to look for perfect numbers:"); 
     int i = input.nextInt(); 
     String factors = ""; 

     System.out.printf("Looking for perfect numbers from 1 to %d%n", i); 

     for (int j = 1; j < i + 1; j++) { 

      if (PerfectNumbers.isPerfect(i) == true) { 

       int w = i - 1; 

       for (j = 1; j < i + 1; j++) { 

        if (i % w == 0) 
         factors = factors + " " + Integer.toString(w); 

       } 
       System.out.println(i + " is a perfect number it's factors are:" + factors); 
      } 
      i = i - 1; 
     } 
    } 
} 

Answer 3

このコードは正常に動作します。それは{

public static boolean isPerfect(int a) { 

    int n = a; 
    int sum = 0; 
    boolean perfect; 

    while (n-- >1) { 

     if(a%n==0) 
      sum = sum + n; 

    } 
    if (sum == a) 
    { 

     perfect = true; 

    } 
    else 
    { 

     perfect = false; 

    } 

     return perfect; 

    } 

public static void main(String[] args) { 
    // TODO Auto-generated method stub 

    Scanner input = new Scanner(System.in); 

    System.out.print("Enter the number up to which you would like to look for perfect numbers:"); 
    int i = input.nextInt(); 


    System.out.println("Looking for perfect numbers from 1 to "+i); 
    int inc = 1; 
    while (inc < i) { 

     if (PerfectNumbers.isPerfect(inc) == true) { 
      String factors = ""; 
      int w = inc-1; 
      int j = 1; 
      while (j < w) { 

       if(inc % j == 0) 
        factors = factors + " " + Integer.toString(j); 

       j++; 
      } 
      System.out.println(inc + " is a perfect number it's factors are:" + factors); 

     } 
     inc = inc + 1; 
    } 
    input.close(); 
} 

}

Answer 4

1ループここ

コード

を働いている間に は factors = "";でこれを追加するには、コードに思います..あなたが期待通りに出力として パブリッククラスPerfectNumbersを与えます

public class PerfectNumbers { 

    public static boolean isPerfect(int a) { 
     int n = a; 
     int sum = 0; 
     boolean perfect = false; 

     while (n-- >1) { 
      if(a%n==0) { 
       sum = sum + n; 
      } 
      if (sum == a) { 
       perfect = true; 
      } else { 
       perfect = false; 
      } 
     } 
     return perfect; 
    } 

    public static void main(String[] args) { 
     Scanner input = new Scanner(System.in); 
     System.out.print("Enter the number up to which you would like to look for perfect numbers:"); 
     int i = input.nextInt(); 
     String factors = ""; 
     System.out.printf("Looking for perfect numbers from 1 to %d%n", i); 
     while (i > 0) { 
      if (PerfectNumbers.isPerfect(i) == true) { 
       int w = i-1; 
       while (w-- > 1) { 
        if(i % w == 0) 
        factors = factors + " " + Integer.toString(w);     
       } 
       System.out.println(i + " is a perfect number it's factors are:" + factors); 
       factors = ""; 
      } 
      i = i - 1; 
     } 
    } 
} 

Answer 5

ここでは、Java 8ストリームベースのソリューションをご紹介します。

public class PerfectNumbers { 
    public static void main(String[] args) { 
     IntStream.range(1, 500) 
       .filter(n -> factors(n).sum() == n) 
       .forEach(n -> System.out.println(n + " is a prefect numbers its factors are " 
        + factors(n).mapToObj(Integer::toString).collect(Collectors.joining(" ")))); 
    } 

    private static IntStream factors(int n) { 
     return IntStream.range(1, n).filter(d -> n % d == 0); 
    } 
}