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Dvinal [7]
3 years ago
12

Plz help me in this example

Computers and Technology
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer:

import java.util.*;

public class calcarms

{

public static void main(String [] args)

{

 Scanner input = new Scanner(System.in);

 //Declare Variable

 int num;

 //Initialize num to 0

 num = 0;

 //Print heading

 System.out.println("Armstrong Numbers\t\tDifference between Successive Armstrong Numbers");

 //Call Armstrong method

 boolean result = Armstrong(num);  

}

public static boolean Armstrong(int num)

{

 int nums;

 //Initialize old_Arms and new_Arms to 0

 int old_Arms =0; int new_Arms = 0;

 for(int i = 0;i<=500;i++) //Iterate from 0 to 500

 {

  nums = i;

 int temp,calc = 0;

 temp = nums;  

  //Check individual digit of each number

 while(nums > 0)

 {

  int a=nums%10;  

  nums/=10;

  calc+=(a*a*a);//Find cube of number

 }

 if(temp == calc) //Check if number is armstrong

 {

  new_Arms = calc;

  old_Arms = new_Arms - old_Arms;

  System.out.println(calc+"\t\t\t"+old_Arms);//Print

  old_Arms = calc;      

 }

 }

 return true;  

};

}

Explanation:

Comments are used to explain difficult lines

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The main difference between default constructor and overloaded constructor is that the default constructor does't have any parameters while the overloaded constructors have parameters.

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Consider the following Stack operations:
maxonik [38]

Answer:

Sequence of popped values: h,s,f.

State of stack (from top to bottom): m, d

Explanation:

Assuming that stack is  initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:

push(d) -> enters d to the Stack

Stack:  

d ->top

push(h) -> enters h to the Stack

Stack:

h ->top

d ->bottom

pop() -> removes h from the Stack:

Stack:

d ->top

p: Suppose p contains popped values so first popped value entered to p is h

p = h

push(f) -> enters f to the Stack

Stack:

f ->top

d ->bottom

push(s) -> enters s to the Stack

Stack:

s ->top

f

d ->bottom

pop() -> removes s from the Stack:

Stack:

f ->top

d -> bottom

p = h, s

pop() -> removes f from the Stack:

Stack:

d ->top

p = h, s, f

push(m) -> enters m to the Stack:

Stack:

m ->top

d ->bottom

So looking at p the sequence of popped values is:

h, s, f

the final state of the stack:

m, d

end that is the top of the stack:

m

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Ray owns a gaming cafe where users can play video games on computers. These games involve rich graphics and need high processing
tankabanditka [31]

Answer:

thick

Explanation:

i took the test and got it right

6 0
2 years ago
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