如何从区间[100]中返回具有强数的元素列表999]

Question

我是java的新手，我被分配了阿姆斯特朗号码。我已经创建了一个新的类ArmstrongNumber.java，我从这个网站初始化方法:http://www.programmingsimplified.com/java/source-code/java-program-armstrong-number

现在，在一个类的主方法中，我创建了另一个方法，我调用armstrong number类，现在我必须返回interval从[100到999]的armstrong number。

这就是我现在被困的地方。

public static void armtrongNumbs()
    {
        ArmstrongNumber returnObj = new ArmstrongNumber(); // here i m calling class.
        int start = 100;
        int end = 999;
        for(int i = start; i<= end; i++)
        {
           number = i + number;
           returnObj.Armstrong(number);   
        }
        //returnObj.Armstrong();
    }

为什么我的循环只返回阿姆斯特朗数?

编辑:ArmstrongNumber类

class ArmstrongNumber
{
   public void Armstrong(int number)
   {
      int n, sum = 0, temp, remainder, digits = 0;
      Scanner in = new Scanner(System.in);
      System.out.println("Input a number to check if it is an Armstrong number");      
      n = in.nextInt();     
      temp = n;     
      // Count number of digits 
      while (temp != 0) {
         digits++;
         temp = temp/10;
      }
      temp = n;
      while (temp != 0) {
         remainder = temp%10;
         sum = sum + power(remainder, digits);
         temp = temp/10;
      }
      if (n == sum)
         System.out.println(n + " is an Armstrong number.");
      else
         System.out.println(n + " is not an Armstrong number.");         
   }
   static int power(int n, int r) {
      int c, p = 1;
      for (c = 1; c <= r; c++) 
         p = p*n;
      return p;   
   }
}

Answer 1

根据您的需求，您需要ArmstrongNumber.java的逻辑，并根据您的需求对其进行建模。

您只需要使用以下代码，就可以停止担心使用ArmstrongNumber.java

package hello;
public class Abc {
public static void main(String[] args) {
    int n, sum, temp, remainder, digits;
    int start = 100;
    int end = 999;
    for (int i = start; i <= end; i++) {
        sum = 0;
        digits = 0;
        temp = i;
        // Count number of digits
        while (temp != 0) {
            digits++;
            temp = temp / 10;
        }
        temp = i;
        while (temp != 0) {
            remainder = temp % 10;
            sum = sum + power(remainder, digits);
            temp = temp / 10;
        }
        if (i == sum)
            System.out.println(i + " is an Armstrong number.");
    }
}
static int power(int n, int r) {
    int c, p = 1;
    for (c = 1; c <= r; c++)
        p = p * n;
    return p;
}
}

在这里你可以看到，如何将每个数字的和和数字初始化为零，然后其余的逻辑是相同的。您可以验证153、370、371,407被打印为阿姆斯特朗编号。

希望能有所帮助

Answer 2

try like

public int[] Armstrong(int start ,int end){
int a[],i=0;
for(int i = start; i<= end; i++)
{
      number = i + number;
      int n, sum = 0, temp, remainder, digits = 0;
      Scanner in = new Scanner(System.in);
      System.out.println("Input a number to check if it is an Armstrong number");      
      n = in.nextInt();     
      temp = n;     
      // Count number of digits 
      while (temp != 0) {
         digits++;
         temp = temp/10;
      }
      temp = n;
      while (temp != 0) {
         remainder = temp%10;
         sum = sum + power(remainder, digits);
         temp = temp/10;
      }
      if (n == sum)
         a[i++]=n;
      else
         System.out.println(n + " is not an Armstrong number.");  
}
return a;       
   }
static int power(int n, int r) {
    int c, p = 1;
    for (c = 1; c <= r; c++)
        p = p * n;
    return p;
}
}