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javaperfect-square
Anonymous
Anonymous

Reputation: 17

How do I find perfect square from a set of numbers in Java?

So I'm trying to find out the perfect squares within a set of numbers. I declared the necessary variables, added a for loop, added a sqroot = Math.sqrt(num) and a print method to list the numbers. What I can't figure out is how can I make the program pick out the perfect squares within the range of numbers, and find the average of them?

This is for an assignment I'm working on for a class and I've been stuck on this for a while now. I'm also rather new to Java so sorry if this is a stupid question. The code is below:


public class Test {

    public static void main(String[] args) {
        int num;
        double sqroot = 0;
        int sumPsq = 0; //sum variable
        int psq = 0; //counter for perfect squares
        double avg = 0;

        for(num = 101; num <= 149; num += 2){
           sqroot = Math.sqrt(num);

            if(sqroot*sqroot == num){ //Condition to find perfect squares
                psq += 1; //counting perfect squares
                sumPsq = sumPsq + num; //Find the sum of all the perfect squares

                System.out.println(num); //Print out the perfect squares
            }
        }
        avg = 1.0 * sumPsq/psq;
        System.out.println(avg);
    }

}

This is just a piece of the code from the entire assignment, so if you need more of it then I'm more than willing to provide it. Thanks!

Upvotes: 0

Views: 3857

Answers (3)

Jakha
Jakha

Reputation: 11

I used the following mathematical formula to find the perfect square:

1 + 3 + 5 + 7 .... = n ^ 2

for example: 1 + 3 + 5 = 9 = 3 ^ 2

and sample code:

   int i = 1;

    while (num > 0) {
        num = num - i;
        i = i + 2;
    }

    return num == 0;

Upvotes: 1

Liviu Stirb
Liviu Stirb

Reputation: 6085

A perfect square is a number that can be expressed as the product of two equal integers so it must be an int after the sqrt. If you do a sqroot*sqroot == num you are just checking that sqrt is working correctly. Some number won't pass the check because of this but usually, you will get way more numbers than you want.

So what you need to do is just checking that after the sqrt the result is an int:

if (sqrootd % 1 == 0) { ... }

An optimization you can do is to check the number is an integer before sqrt.   Other than that, your code looks fine to me

Upvotes: 3

Jignesh
Jignesh

Reputation: 471

The best way to check if square root is an integer you will need below condition 

if ((sqroot - Math.floor(sqroot)) == 0){

instead of 

if(sqroot*sqroot == num){

The Math.sqrt() method finds the square root of the given number and the floor() method finds the largest (closest to positive infinity) floating-point value that less than or equal to the argument (here square root value returned by sqrt() method) and is equal to a mathematical integer.

Then we calculate the difference between these two to check if the difference is zero. For a perfect square number this difference is always zero. The reason is: square root of perfect square number is integer.

reference

Upvotes: 0

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