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我在朋友的帮助下做了这个，以便找到用户输入的数字的完美数字（除数之和 = 原始数字的数字）之和。有人可以帮我转换它，以便它在一行上打印每个数字，即使它不是一个完美的数字，并列出它的所有除数，然后如果它是完美的，它会在除数之后打印完美的数字？就像“6：1、2、3、6，完美数”一样不使用数组？

import java.util.*;
public class PerfNumbers {  
public static void main(String[] args){     
    System.out.println("Enter A Number:");      
    Scanner sc = new Scanner(System.in);    
    int input = sc.nextInt();       
     int sumOfDivisors = 0;

             for (int number = 1; number <= input; number++){       
                 for (int divisor = 1; divisor < number; divisor++)
                 {  

                     if (number % divisor == 0)

                         sumOfDivisors += divisor;
                     }

                         if (sumOfDivisors == number)
                         {

                     System.out.println(number);
                     }

                         sumOfDivisors = 0;  
             }

    sc.close();
}

}
4

2 回答 2

1 
import java.util.*;

/**
 *
 * @author Tyler Weaver
 */
public class Test {

    public static void main(String[] args) {
        System.out.print("Enter A Number: ");
        Scanner sc = new Scanner(System.in);
        int input = sc.nextInt();
        int sumOfDivisors = 0;
        Collection<Integer> col = new ArrayList<>();

        for (int number = 1; number <= input; number++) {
            for (int divisor = 1; divisor < number; divisor++) {

                if (number % divisor == 0) {
                    sumOfDivisors += divisor;
                    col.add(divisor);
                }
            }

            if (sumOfDivisors == number) {

                System.out.printf("Perfect Number: %,d%n[", number);
                for (Integer i : col) {
                    System.out.printf("%,d, ", i);
                }
                System.out.printf("\b\b]%n");
            }

            sumOfDivisors = 0;
            col.clear();
        }

        sc.close();
    }

}

只需添加一个集合来跟踪除数，然后添加一个 printf() 语句以正确格式化。

编辑：不使用集合的代码。

import java.util.*;

/**
 *
 * @author Tyler Weaver
 */
public class Test {

    public static void main(String[] args) {
        System.out.print("Enter A Number: ");
        Scanner sc = new Scanner(System.in);
        int input = sc.nextInt();
        int sumOfDivisors = 0;

        for (int number = 2; number <= input; number++) {
            System.out.printf("Number: %,d%n[", number);

            for (int divisor = 1; divisor < number; divisor++) {

                if (number % divisor == 0) {
                    sumOfDivisors += divisor;
                    System.out.printf("%,d, ", divisor);
                }
            }

            System.out.printf("\b\b]%n");

            if (sumOfDivisors == number) {
                System.out.printf("Therefore %,d is a perfect number.%n%n",
                        number);
            } else {
                System.out.printf("%n");
            }

            sumOfDivisors = 0;
        }

        sc.close();
    }
}
于 2014-10-17T02:15:19.987 回答
0 
public static void main(String[] args) 

{

    int i=6;
    int n=i;
    int r=0;
    while(i>0){
    i--;

if(i>0 && n%i==0){
    if (i!=0)
        r=r+i;
}

    }   
    System.out.println(r);
    if (r==n)
        System.out.println("its True");
    else
        System.out.println("its False");
}
于 2016-11-24T11:37:28.407 回答