CODEWITHSUNDEEP
javasquare-root
laish138
laish138

Reputation: 243

square numbers Java

"An array is used to store ten integer numbers. Write a Java program that determines and print the square numbers which are also odd numbers from the given array."

The problem I have is how to figure out if the number in the array is a square number. I tried this way but it's not correct!

import java.math.*;
public class JavaApplication43 {

    public static void main(String[] args) {

        int[] no = {22, 44, 25, 89, 81, 55, 23, 25, 55};

        for (int i = 0; i < no.length; i++) {

            int x = no[i];
            double y;
            if (x % 2 != 0) {
                y = Math.sqrt(x);
                if (x == (Math.pow(y, 2))) 
                    System.out.println(no[i]);
            }
       }
   }
}

This is the output it gives me

run:
25
81
55
25
55

55 is there too that means this method I used is not successful!

Upvotes: 4

Views: 18450

Answers (3)

Isuru
Isuru

Reputation: 129

hope this will be much easier,

    if((arr[i]%2 != 0) & (Math.sqrt(arr[i])%1 == 0)){
        System.out.println(arr[i]);
    }

In here inside if condition first I check whether number is an odd number by taking modulus division and the 2nd condition check whether number is a perfect square. First I get the square root of given number and then take the modulus division by 1 and check whether it's equal to 0. If number is a perfect square, square root of that number is an integer, when I take the modulus division of an integer answer should be equal to 0.

Upvotes: 2

Peter Lawrey
Peter Lawrey

Reputation: 533880

You can determine if a number is a square by checking if its square root is an integer.

double sqrt = Math.sqrt(x);
long sqrt2 = Math.round(sqrt);
if (Math.abs(sqrt - sqrt2) / sqrt < 1e-15)
   // we have a square.

If you know x is an int you shouldn't get a rounding error and you can do

int x = ...
double sqrt = Math.sqrt(x);
if ((int) sqrt == sqrt)
    // we have a square.

Upvotes: 5

Konstantin Yovkov
Konstantin Yovkov

Reputation: 62874

You could just do:

for (int i = 0; i < no.length; i++) {
    int x = no[i];
    if (x % 2 == 0) continue;
    int y = (int) Math.sqrt(x);
    if (x == y * y) { 
        System.out.println(x);
    }
}

Upvotes: 5

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