Peterson Number, Function overloading and Even Number with Start and end limit

Java program for Method Overloading Peterson Number and Even Number

Q. Write a program to compute the area of rectangle, square and circle using overloading function. get area(int, int),  get area(int) and get area(double a)

Solution:
Here we calculate Area of Rectangle, Area of Square and Area of Circle using Method Over Loading.

import java.io.*;
public class AreaCalc    // Class started
{
   int l,b,s,ar1;
   double r,ar2;
   int area(int a)            // Area of Square
   {
       s=a;
       ar1=s*s;
       return ar1;
   }
   int area(int x,int y)  // Area of Rectangle
   {
       l=x;
       b=y;
       ar1=l*b;
       return ar1;
    }
    double area(double x)  // Area of Circle
    {
        r=x;
        ar2=Math.PI*r*r;
        return ar2;
    }
    public static void main(String args[])throws IOException
    {
        int a;
        double ar;
        AreaCalc ob=new AreaCalc();
        a=ob.area(10);
        System.out.println("Area Of Square="+a);
        a=ob.area(10,5);
        System.out.println("Area Of Rectangle="+a);
        ar=ob.area(5.5);
        System.out.println("Area Of Circle="+ar);
    }
} // end of class

Output:
Area Of Square=100
Area Of Rectangle=50
Area Of Circle=95.03317777109123



Q2. Write a program to accept a number from user and check whether the number in Peterson or not, using a function (int n). A Peterson number is a number whose sum of factorial of all the digit is the number itself E.g., 145 = 1! + 4! + 5! = 1 + 24 +120 = 145

Solution:
import java.io.*;
/**
* class PetersonNumber
*/
public class PetersonNumber
{
    // instance variables
    private int n;
    private int f;
    public int peterson(int x)
    {
        f=1;
        n=x;
        for(int i=1; i<=n; i++)
        {
            f=f*i;
        }
        return f;
    }
    public static void main(String args[])throws IOException
    {
        // Local variable
        int a, b, n, m, s=0;
        PetersonNumber ob=new PetersonNumber ();
        InputStreamReader in=new InputStreamReader (System.in);
        BufferedReader br=new BufferedReader(in);
        System.out.print("Enter a number: ");
        n = Integer.parseInt(br.readLine());
        m=n;                // Store the n value to m
        while(n!=0)
        {
            a = n % 10;
            b = ob.peterson(a);
            s = s + b;
            n = n / 10;
        }
        if(m == s)
        {
            System.out.println ("PetersonNumber");
        }
        else
        {
            System.out.println ("Not PetersonNumber");
        }
    }   
} // end of class

Output:
Enter a number: 145
PetersonNumber
Enter a number: 273
Not PetersonNumber



Q3. Write a program to design a method called void sumEven(int start, int end) to take start and end limit from the main(). The function will display the number of even number and the sum of all the even number between start and end (both inclusive)
Solution:
import java.io.*;
public class SumEven
{
    int s,i,start,end;
    void sumEven(int s ,int e)
    {
        start=s;
        end=e;
        s=0;
        for(i=start;i<=end;i++)
        {
            if(i%2==0)
            {
                s=s+i;
                System.out.print(i+" ");
            }
        }
        System.out.println("\n Sum even="+s);
    }
    public static void main(String args[])throws IOException
    {
        // Local variable
        int a, b;
        SumEven ob=new SumEven ();
        InputStreamReader in=new InputStreamReader (System.in);
        BufferedReader br=new BufferedReader(in);
        System.out.print("Enter start and end: ");
        a = Integer.parseInt(br.readLine());
        b = Integer.parseInt(br.readLine());
        ob.sumEven(a,b);
    }
}

Output:
Enter start and end: 5
30
6 8 10 12 14 16 18 20 22 24 26 28 30 Sum even=234

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