Square Free Numbers - CodeVita 9 | Round 1

Square Free Numbers

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).

Problem Description

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

Other Test Case

Input
290990700

Output
255

Input
4491411836

Output
31

Program:

#include <stdio.h>

int isPerfectSquare(int n) 

{ 

    for (int i = 1; i * i <= n; i++) 

    { 

        if ((n % i == 0) && (n / i == i)) 

        { 

            return 1; 

        } 

    } 

    return 0; 

} 

int main() {

int x,i,cnt=0,j=0,y,k,a[10000];

scanf("%d",&x);

for(i=1;i<=x;i++)

{

    if(x%i==0)

    {

        a[j]=i;

        j++;

    }

}

for(i=0;i<j;i++)

{

    if((isPerfectSquare(a[i])==1)&&(a[i]!=0)&&(a[i]!=1))

    {

      y=a[i];

      for(k=0;k<j;k++)

      {

        if(a[k]!=0 && a[k]%y==0)

        a[k]=0;

       }

        a[i]=0;

     }

}

for(i=0;i<j;i++)

if(a[i]!=0)

cnt++;

printf("%d",cnt-1);

return 0;

}

You can also run it on an online IDE:    

https://ide.geeksforgeeks.org/tttw8dQwDN

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!  Happy Coding !! Cheers!!!

Related Links:

Grooving Monkeys: https://codepiggy.blogspot.com/2020/03/grooving-monkeys-code-vita-8-pre.html

Codu And Sum Love 2018

Codu And Sum Love

Problem Description

```

Scanner sc = new Scanner(System.in);

long sum = 0;

int N = sc.nextInt();

for (int i = 0; i < N; i++) {

final long x = sc.nextLong(); // read input

String str = Long.toString((long) Math.pow(1 << 1, x));

str = str.length() > 2 ? str.substring(str.length() - 2) : str;

sum += Integer.parseInt(str);

}

System.out.println(sum%100);

```

Given N number of x’'s, perform logic equivalent of the above Java code and print the output

Constraints

 1<=N<=10^7 0<=x<=10^18

Input Format

 First line contains an integer N

Second line will contain N numbers delimited by space

Output

 Number that is the output of the given code by taking inputs as specified above

Explanation

Example 1

Input

 4 8 6 7 4

 Output

 64

Example 2

Input

3

1 2 3

Output

14

Program

#include <stdio.h>
long power(long k)
{
    long i,j=1;
    for(i=1;i<=k;i++)
        j=j*2;
        return j;
}
long length(long k)
{
    long ct=0,n1;
    while(k)
    {
        ct++;
        k=k/10;
    }
    return ct;
}
long reduce(long p1,long r1)
{
    long ct=0,k,cnt,rev=0,a[1000],h=0,i,sum1=0;
    while(p1)
    {
      k=p1%10;
      a[h]=k;
      h++;
      p1=p1/10;
     }
      for(i=1;i>=0;i--)
      sum1=sum1*10+a[i];
    return sum1;
}
int main() {
 long x,n,sum=0,r[70],p,c,f=0,i;
 scanf("%ld",&n);
 for(i=0;i<n;i++)
 {
    scanf("%ld",&x);
    p=power(x);
    c=length(p);
    if(c>2)
    {
    r[f++]=reduce(p,c);
    }
    else
    r[f++]=p;
 }
 for(i=0;i<n;i++)
 
 sum=sum+r[i];
   printf("%ld",(sum%100));
 
 
 return 0;
}

Output:

4 8 6 7 4

64

You can also try running it on online IDE: https://ide.geeksforgeeks.org/G3gXokhTbW

Your doubts and feedback are welcomed! you can comment it below Cheers!

Related Link: Super Ascii