Codevita - Square Free Number

 In the theory of numbers, square free numbers have a special place.  A square free number is one that is not divisible by a perfect square (other than 1).  Thus 72 is divisible by 36 (a perfect square), and is not a square free number, but 70 has factors 1, 2, 5, 7, 10, 14, 35 and 70.  As none of these are perfect squares (other than 1), 70 is a square free number.

For some algorithms, it is important to find out the square free numbers that divide a number.  Note that 1 is not considered a square free number. 

In this problem, you are asked to write a program to find the number of square free numbers that divide a given number.

Input:-

  • The only line of the input is a single integer N which is divisible by no prime number larger than 19.

Output:-

  • One line containing an integer that gives the number of square free numbers (not including 1).

Constraints:-

  • N   < 10^9

Complexity:-

  • Simple

Time Limit:-

  • 1

Examples


Example 1

  • Input:-
    20
  • Output:-
    3

Explanation

N=20

If we list the numbers that divide 20, they are

1, 2, 4, 5, 10, 20

1 is not a square free number, 4 is a perfect square, and 20 is divisible by 4, a perfect square.  2 and 5, being prime, are square free, and 10 is divisible by 1,2,5 and 10, none of which are perfect squares.  Hence the square free numbers that divide 20 are 2, 5, 10.  Hence the result is 3.

Example 2

  • Input:-
    72
  • Output:-
    3

Explanation

N=72.  The numbers that divide 72 are

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

1 is not considered square free.   4, 9 and 36 are perfect squares, and 8,12,18,24 and 72 are divisible by one of the.  Hence only 2, 3 and 6 are square free.  (It is easily seen that none of them are divisible by a perfect square).  The result is 3.

Method - 1

import javax.swing.plaf.synth.SynthOptionPaneUI;
import java.util.Scanner;

public class SquareFree {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long num;
        double sqnum;
        int i,j=0, flag, chksqr;
        Integer[] temp = new Integer[10000];
        int count = 0, k;

        System.out.println("Enter the number : ");
        num = sc.nextInt();
        for (i = 2; i < num/2; i++) {
            // if i is a factor of the number
            if(num % i == 0){
                count++;
                sqnum = Math.sqrt(i);
                chksqr = (int) sqnum;

                // if i is a non square free factor of the number
                if (chksqr == sqnum){
                    count--;

                    // Adding to the temp list to check future factor
                    // has a non square free sub-factor
                    temp[j] = i;
                    j++;
                }
                // if i is not a perfect square number
                else{
                    for (k = 0; k < j; k++) {
                        if(i > temp[k] && j!=0 ){
                            if(i%temp[k] == 0){
                                count--;
                                k = j+1;
                            }
                        }else
                            break;
                    }
                }
            }
        }

        System.out.println(count);
    }
}

Method - 2

import java.util.Scanner;

public class SquareFree2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        Integer[] primes = {2,3,5,7,11,13,17,19};
        int nonSqrFreeCount = 0;

        for (int i = 0; i < primes.length; i++) {
            int count = 0;
            while (num % primes[i] == 0){
                count++;
                num = num / primes[i];
            }
            if (count > 0){
                nonSqrFreeCount++;
            }
        }
        System.out.println((int) Math.pow(2, nonSqrFreeCount) - 1);

    }
}

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