Powerful Number
A number n is said to be Powerful Number if for every prime factor p of it, p2 also divides it. For example:- 36 is a powerful number. It is divisible by both 3 and square of 3 i.e, 9.
The first few Powerful Numbers are:
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64 ….
Given a number n, our task is to check if this is powerful or not.
Examples :
Input: 27
Output: YESInput: 32
Output: YESInput: 12
Output: NO
Approach:
The idea is based on the fact that if a number n is powerful, then all prime factors of it and their squares should be divisible by n. We find all prime factors of given number. And for every prime factor, we find the highest power of it that divides n. If we find a prime factor whose highest dividing power is 1, we return false. If highest dividing power of all prime factors is more than 1, we return true.
Below is implementation of above idea.
C++
// C++ program to find if a number is powerful or not. #include <bits/stdc++.h> using namespace std; // function to check if the number is powerful bool isPowerful( int n) { // First divide the number repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // If only 2^1 divides n (not higher powers), // then return false if (power == 1) return false ; } // if n is not a power of 2 then this loop will execute // repeat above process for ( int factor = 3; factor <= sqrt (n); factor += 2) { // Find highest power of "factor" that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // If only factor^1 divides n (not higher powers), // then return false if (power == 1) return false ; } // n must be 1 now if it is not a prime numenr. // Since prime numbers are not powerful, we return // false if n is not 1. return (n == 1); } // Driver program to test above function int main() { isPowerful(20) ? cout << "YES\n" : cout << "NO\n" ; isPowerful(27) ? cout << "YES\n" : cout << "NO\n" ; return 0; } |
Java
// Java program to find if a // number is powerful or not. class GFG { // function to check if the // number is powerful static boolean isPowerful( int n) { // First divide the number // repeatedly by 2 while (n % 2 == 0 ) { int power = 0 ; while (n % 2 == 0 ) { n /= 2 ; power++; } // If only 2^1 divides n (not higher powers), // then return false if (power == 1 ) return false ; } // if n is not a power of 2 then this loop will execute // repeat above process for ( int factor = 3 ; factor <= Math.sqrt(n); factor += 2 ) { // Find highest power of "factor" that divides n int power = 0 ; while (n % factor == 0 ) { n = n / factor; power++; } // If only factor^1 divides n (not higher powers), // then return false if (power == 1 ) return false ; } // n must be 1 now if it is not a prime numenr. // Since prime numbers are not powerful, we return // false if n is not 1. return (n == 1 ); } // Driver code public static void main(String[] args) { if (isPowerful( 20 )) System.out.print( "YES\n" ); else System.out.print( "NO\n" ); if (isPowerful( 27 )) System.out.print( "YES\n" ); else System.out.print( "NO\n" ); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to find # if a number is powerful or not. import math # function to check if # the number is powerful def isPowerful(n): # First divide the number repeatedly by 2 while (n % 2 = = 0 ): power = 0 while (n % 2 = = 0 ): n = n / / 2 power = power + 1 # If only 2 ^ 1 divides # n (not higher powers), # then return false if (power = = 1 ): return False # if n is not a power of 2 # then this loop will execute # repeat above process for factor in range ( 3 , int (math.sqrt(n)) + 1 , 2 ): # Find highest power of # "factor" that divides n power = 0 while (n % factor = = 0 ): n = n / / factor power = power + 1 # If only factor ^ 1 divides # n (not higher powers), # then return false if (power = = 1 ): return false # n must be 1 now if it # is not a prime number. # Since prime numbers are # not powerful, we return # false if n is not 1. return (n = = 1 ) # Driver code print ( "YES" if isPowerful( 20 ) else "NO" ) print ( "YES" if isPowerful( 27 ) else "NO" ) # This code is contributed # by Anant Agarwal. |
C#
// C# program to find if a // number is powerful or not. using System; class GFG { // function to check if the // number is powerful static bool isPowerful( int n) { // First divide the number // repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // If only 2^1 divides n // (not higher powers), // then return false if (power == 1) return false ; } // if n is not a power of 2 then // this loop will execute repeat // above process for ( int factor = 3; factor <= Math.Sqrt(n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // If only factor^1 divides n // (not higher powers), then // return false if (power == 1) return false ; } // n must be 1 now if it is not // a prime numenr. // Since prime numbers are not // powerful, we return false if // n is not 1. return (n == 1); } // Driver code public static void Main() { if (isPowerful(20)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); if (isPowerful(27)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find if a // number is powerful or not // function to check if // the number is powerful function isPowerful( $n ) { // First divide the // number repeatedly by 2 while ( $n % 2 == 0) { $power = 0; while ( $n % 2 == 0) { $n /= 2; $power ++; } // If only 2^1 divides // n (not higher powers), // then return false if ( $power == 1) return false; } // if n is not a power of 2 // then this loop will execute // repeat above process for ( $factor = 3; $factor <= sqrt( $n ); $factor += 2) { // Find highest power of // "factor" that divides n $power = 0; while ( $n % $factor == 0) { $n = $n / $factor ; $power ++; } // If only factor^1 divides // n (not higher powers), // then return false if ( $power == 1) return false; } // n must be 1 now if it is // not a prime number. Since // prime numbers are not powerful, // we return false if n is not 1. return ( $n == 1); } // Driver Code $d = isPowerful(20) ? "YES\n" : "NO\n" ; echo $d ; $d = isPowerful(27) ? "YES\n" : "NO\n" ; echo $d ; // This code is contributed by ajit. ?> |
Javascript
<script> // Javascript program to find if a // number is powerful or not. // function to check if the // number is powerful function isPowerful(n) { // First divide the number // repeatedly by 2 while (n % 2 == 0) { let power = 0; while (n % 2 == 0) { n /= 2; power++; } // If only 2^1 divides n (not higher powers), // then return false if (power == 1) return false ; } // if n is not a power of 2 then this loop will execute // repeat above process for (let factor = 3; factor <= Math.sqrt(n); factor += 2) { // Find highest power of "factor" that divides n let power = 0; while (n % factor == 0) { n = n / factor; power++; } // If only factor^1 divides n (not higher powers), // then return false if (power == 1) return false ; } // n must be 1 now if it is not a prime numenr. // Since prime numbers are not powerful, we return // false if n is not 1. return (n == 1); } // Driver code to test above methods if (isPowerful(20)) document.write( "YES" + "<br>" ); else document.write( "NO" + "<br>" ); if (isPowerful(27)) document.write( "YES" + "<br>" ); else document.write( "NO" + "<br>" ); // This code is contributed by avijitmondal1998. </script> |
NO YES
Time complexity: O(sqrt(n))
Auxiliary space: O(1)
References:
https://en.wikipedia.org/wiki/Powerful_number
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