CODEWITHSUNDEEP
javabooleanperfect-numbers
FAT
FAT

Reputation: 11

Check for perfect number using boolean

I have problem with my Java coding to find the perfect number using boolean method. I want to print out like this: Example : ‘6 is a perfect number. 6 is the sum of 1, 2, 3’ Otherwise : ‘9 is not a perfect number’

But I don't know how to make the coding for "6 is the sum of 1, 2, 3". Can anyone help me? Here is my coding :

import java.util.Scanner;

public class trial
{
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub

        Scanner perfect = new Scanner(System.in);
        System.out.print("Enter any integer number : ");
        int n = perfect.nextInt();

        if(isPerfectNumber(n))
        {
            System.out.println(n+" is a perfect number");

        }
        else
        {
            System.out.println(n+" is not a perfect number");
        }
    }

    public static boolean isPerfectNumber(int n)
    {
        int sum = 0;
        for (int i=1; i<n; i++)
        {
            if (n%i == 0)
            {
                sum = sum + i;
            }
        }
        if (sum == n)
        {
            return true;
        }
        else
        {
            return false;
        }
    }
}

Upvotes: 0

Views: 3499

Answers (3)

Mureinik
Mureinik

Reputation: 311448

I'd change isPerfectNumber to return a List of the factors that make n or null if it's not a perfect number. Then you have a single result that both contains factors and can be used to determine if n is perfect:

public static void main(String[] args) {
    Scanner perfect = new Scanner(System.in);
    System.out.print("Enter any integer number : ");
    int n = perfect.nextInt();

    List<Integer> factors = getPerfectFactors(n);
    if (factors != null) {
        System.out.println
            (n + " is a perfect number. It's the sum of" +
             factors.stream()
                    .map(String::valueOf)
                    .collect(Collectors.joining(", "));
    } else {
        System.out.println(n+" is not a perfect number");
    }
}

public static List<Ingeger> getPerfectFactors(int n) {
    int sum = 0;
    List<Ingeger> factors = new LinkedList<>();
    for (int i = 1; i < n; i++) {
        if (n % i == 0)  {
            sum += i;
            factors.add(i);
        }
        if (sum > n) { // Early return optimization, not material to the solution
            return null;
        }
    }
    if (sum == n) {
        return factors;
    } else {
        return null;
    }
}

Upvotes: 1

Navoneel Talukdar
Navoneel Talukdar

Reputation: 4598

Well in the for loop if condition if modulus is true then you know the i has to be kept in track somehow.

So I would declare an arraylist of integers to keep track of summation numbers.

ArrayList<Integer> numbs = new ArrayList<Integer>();

Then 

public static boolean isPerfectNumber(int n)
{
        int sum = 0;            
        for (int i=1; i<n; i++)
        {
            if (n%i == 0)
            {
                numbs.add(i);  //here keep track of those summation numbers
                sum = sum + i;
            }
        }
        if (sum == n)
        {
            return true;
        }
        else
        {
            return false;
        }
}

Now while printing you can do like

System.out.println(n + " is a perfect number. " + n + " is sum of " + StringUtils.join(numbs, ","));

Don't forget to import StringUtils: import org.apache.commons.lang3.StringUtils commons-lang3 library.

Upvotes: 0

paxdiablo
paxdiablo

Reputation: 881573

Well, you already know how to get the factors of the number, as per the if (n%i == 0) line.

So, hint only since this is almost certainly class work.

At the same point you add the factor to the sum, you should add it to a list of some description and have that made available to the calling function.

One possibility would be to return a list, the second and subsequent elements being all the factors of the given number, and the first element being the sum of those.

So, for 6, you would get the list {6, 1, 2, 3}, 12 would give you {16, 1, 2, 3, 4, 6}, and 7 would give you {7, 1}.

That way, you simply have to check the original number against the first element and, if they're equal, print out the other elements. In other words, pseudo-code such as:

input num
factorList = getFactorList(num)
if factorList[0] == num:
    print num, " is perfect, factors are:"
    for idx = 1 to factorList.size() - 1 inclusive:
        print " ", factorList[idx]
    println "."
else:
    println num, " is not perfect."

Upvotes: 1

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