P-Smooth Numbers or P-friable Number

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.

CPP

// 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 not
bool 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 function
int 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 not

import java.lang.*;

class GFG{

// function to check if number n
// is a P-smooth number or not

static 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 function
public 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 not

import math

# function to check if number n
# is a P-smooth number or not
def 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 function
n = 24
p = 7
if (check(n, p)):
    print( "yes" )
else:
    print( "no" )

# This code is contributed by
# Smitha Dinesh Semwal


Output:

yes

Reference:
http://oeis.org/wiki/P-smooth_numbers


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