A P-smooth number or P-friable number is an integer whose largest prime factor is less than or equal to P. Given N and P, we need to write a program to check whether it is P-friable or not.
Examples:
Input : N = 24 , P = 7
Output : YES
Explanation : The prime divisors of 24 are 2 and 3 only.
Hence its largest prime factor is 3 which
is less than or equal to 7, it is P-friable.
Input : N = 22 , P = 5
Output : NO
Explanation : The prime divisors are 11 and 2, hence 11>5,
so it is not a P-friable number.
The approach will be to prime factorize the number and store the maximum of all the prime factors. We first divide the number by 2 if it is divisible, then we iterate from 3 to Sqrt(n) to get the number of times a prime number divides a particular number which reduces every time by n/i and store the prime factor i if its divides N. We divide our number n (by prime factors) by its corresponding smallest prime factor till n becomes 1. And if at the end n > 2, it means its a prime number, so we store that as a prime factor as well. At the end the largest factor is compared with p to check if it is p-smooth number or not.
C++
// CPP program to check if a number is// a p-smooth number or not#include <iostream>#include<math.h>using namespace std;
// function to check if number n// is a P-smooth number or notbool check(int n, int p)
{ int maximum = -1;
// prime factorise it by 2
while (!(n % 2))
{
// if the number is divisible by 2
maximum = max(maximum, 2);
n = n/2;
}
// check for all the possible numbers
// that can divide it
for (int i = 3; i <= sqrt(n); i += 2)
{
// prime factorize it by i
while (n % i == 0)
{
// stores the maximum if maximum
// and i, if i divides the number
maximum = max(maximum,i);
n = n / i;
}
}
// if n at the end is a prime number,
// then it a divisor itself
if (n > 2)
maximum = max(maximum, n);
return (maximum <= p);
}// Driver program to test above functionint main()
{ int n = 24, p = 7;
if (check(n, p))
cout << "yes";
else
cout << "no";
return 0;
} |
Java
// Java program to check if a number is// a p-smooth number or notimport java.lang.*;
class GFG{
// function to check if number n// is a P-smooth number or notstatic boolean check(int n, int p)
{ int maximum = -1;
// prime factorise it by 2
while ((n % 2) == 0)
{
// if the number is divisible by 2
maximum = Math.max(maximum, 2);
n = n/2;
}
// check for all the possible numbers
// that can divide it
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
// prime factorize it by i
while (n % i == 0)
{
// stores the maximum if maximum
// and i, if i divides the number
maximum = Math.max(maximum,i);
n = n / i;
}
}
// if n at the end is a prime number,
// then it a divisor itself
if (n > 2)
maximum = Math.max(maximum, n);
return (maximum <= p);
}// Driver program to test// above functionpublic static void main(String[] args)
{ int n = 24, p = 7;
if (check(n, p))
System.out.println("yes");
else
System.out.println("no");
}}// This code is contributed by// Smitha Dinesh Semwal |
Python3
# Python 3 program to# check if a number is# a p-smooth number or notimport math
# function to check if number n# is a P-smooth number or notdef check(n, p) :
maximum = -1
# prime factorise it by 2
while (not(n % 2)):
# if the number is divisible by 2
maximum = max(maximum, 2)
n = int(n/2)
# check for all the possible numbers
# that can divide it
for i in range(3,int(math.sqrt(n)), 2):
# prime factorize it by i
while (n % i == 0) :
# stores the maximum if maximum
# and i, if i divides the number
maximum = max(maximum,i)
n = int(n / i)
# if n at the end is a prime number,
# then it a divisor itself
if (n > 2):
maximum = max(maximum, n)
return (maximum <= p)
# Driver program to test above functionn = 24
p = 7
if (check(n, p)):
print( "yes" )
else:
print( "no" )
# This code is contributed by# Smitha Dinesh Semwal |
C#
// C# program to check if a number is// a p-smooth number or notusing System;
class GFG {
// function to check if number n
// is a P-smooth number or not
static bool check(int n, int p)
{
int maximum = -1;
// prime factorise it by 2
while ((n % 2) == 0)
{
// if the number is divisible by 2
maximum = Math.Max(maximum, 2);
n = n / 2;
}
// check for all the possible numbers
// that can divide it
for (int i = 3; i <= Math.Sqrt(n); i += 2)
{
// prime factorize it by i
while (n % i == 0)
{
// stores the maximum if maximum
// and i, if i divides the number
maximum = Math.Max(maximum, i);
n = n / i;
}
}
// if n at the end is a prime number,
// then it a divisor itself
if (n > 2)
maximum = Math.Max(maximum, n);
return (maximum <= p);
}
// Driver program to test
// above function
public static void Main()
{
int n = 24, p = 7;
if (check(n, p))
Console.Write("yes");
else
Console.Write("no");
}
}// This code is contributed by vt_m. |
PHP
<?php// PHP program to check if a number is// a p-smooth number or not// function to check if number n// is a P-smooth number or notfunction check($n, $p)
{ $maximum = -1;
// prime factorise it by 2
while (!($n % 2))
{
// if the number is
// divisible by 2
$maximum = max($maximum, 2);
$n = $n / 2;
}
// check for all the possible
// numbers that can divide it
for ($i = 3; $i <= sqrt($n); $i += 2)
{
// prime factorize it by i
while ($n % $i == 0)
{
// stores the maximum if maximum
// and i, if i divides the number
$maximum = max($maximum, $i);
$n = $n / $i;
}
}
// if n at the end is a prime number,
// then it a divisor itself
if ($n > 2)
$maximum = max($maximum, $n);
return ($maximum <= $p);
}// Driver Code$n = 24; $p = 7;
if (check($n, $p))
echo("yes");
else echo("no");
// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript program to check if a number is// a p-smooth number or not // function to check if number n
// is a P-smooth number or not
function check(n, p)
{
let maximum = -1;
// prime factorise it by 2
while (!(n % 2))
{
// if the number is divisible by 2
maximum = Math.max(maximum, 2)
n = n/2;
}
// check for all the possible numbers
// that can divide it
var i;
for (i = 3; i*i <= n; i += 2)
{
// prime factorize it by i
while (n % i == 0)
{
// stores the maximum if maximum
// and i, if i divides the number
maximum = Math.max(maximum, i);
n = n / i;
}
}
// if n at the end is a prime number,
// then it a divisor itself
if (n > 2)
maximum = Math.max(maximum, n);
if (maximum <= p)
return true;
else
return false;
}
// Driver Code
let n = 24, p = 7;
if (check(n, p))
document.write("yes");
else
document.write("no");
// This code is contributed by ajaykrsharma132.
</script> |
Output:
yes
Time Complexity: O(sqrt(N)*logN), as we are using nested loops to traverse sqrt(N)*logN times.
Auxiliary Space: O(1), as we are not using any extra space.
Reference:
http://oeis.org/wiki/P-smooth_numbers