Norm of a Number in Java

Norm of a Number in Java

Question : Write a program to calculate Norm of a Number.

Norm of a number is square root of sum of squares of all digits of the number.

Example :-

Input : 68

            6 X 6 + 8 X 8 = 100 ;  Square root of 100 = 10

Output : Norm of 68 is 10.

Logic :-

• First input number.
• Copy the input number in another variable
• extract each digit of the number
• Store the sum of squares of the digit of the number in a variable say 's'
• Print the square root of the result. (i.e. square root of 's')
• Compile and run the program

Note : To understand the program clearly watch this video : https://youtu.be/tXDl56akraE

Program to calculate Norm of a Number:-

Program :-

import java.io.*;

import java.util.*;

class Norm

{

    public static void main(String args[])

    {

        Scanner in = new Scanner(System.in);

        int n,p,dig,s=0;

        System.out.println("Enter Number");

        n= in.nextInt();//input number

        p=n;//copying input number

        while(p>0)

        {

            dig= p%10;

            s= s+(dig*dig);//storing square of each digit

            p=p/10;

        }

        System.out.println("Norm of "+n+" is : "+Math.sqrt(s));//print norm of a number

    }//end of main method

}//end of class

For Proper Understanding Watch the Video :-

Watch this video : Norm of a Number in Java.

All the best :)

Keep Learning :)