Which is the 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.

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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: YES Input: 32

Output: YES Input: 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++

#include using namespace std; bool isPowerful( int n) { while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } if (power == 1) return false ; } for ( int factor = 3; factor <= sqrt (n); factor += 2) { int power = 0; while (n % factor == 0) { n = n / factor; power++; } if (power == 1) return false ; } return (n == 1); } int main() { isPowerful(20) ? cout << "YES

" : cout << "NO

" ; isPowerful(27) ? cout << "YES

" : cout << "NO

" ; return 0; } Java

class GFG { static boolean isPowerful( int n) { while (n % 2 == 0 ) { int power = 0 ; while (n % 2 == 0 ) { n /= 2 ; power++; } if (power == 1 ) return false ; } for ( int factor = 3 ; factor <= Math.sqrt(n); factor += 2 ) { int power = 0 ; while (n % factor == 0 ) { n = n / factor; power++; } if (power == 1 ) return false ; } return (n == 1 ); } public static void main(String[] args) { if (isPowerful( 20 )) System.out.print( "YES

" ); else System.out.print( "NO

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" ); if (isPowerful( 27 )) System.out.print( "YES

" ); else System.out.print( "NO

" ); } } Python3

import math def isPowerful(n): while (n % 2 = = 0 ): power = 0 while (n % 2 = = 0 ): n = n / / 2 power = power + 1 if (power = = 1 ): return False for factor in range ( 3 , int (math.sqrt(n)) + 1 , 2 ): power = 0 while (n % factor = = 0 ): n = n / / factor power = power + 1 if (power = = 1 ): return false return (n = = 1 ) print ( "YES" if isPowerful( 20 ) else "NO" ) print ( "YES" if isPowerful( 27 ) else "NO" ) C# using System; class GFG { static bool isPowerful( int n) { while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } if (power == 1) return false ; } for ( int factor = 3; factor <= Math.Sqrt(n); factor += 2) { int power = 0; while (n % factor == 0) { n = n / factor; power++; } if (power == 1) return false ; } return (n == 1); } public static void Main() { if (isPowerful(20)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); if (isPowerful(27)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } } PHP " : "NO

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" ; echo $d ; $d = isPowerful(27) ? "YES

" : "NO

" ; echo $d ; ?> Javascript

Output NO YES

Time complexity: O(sqrt(n))

Auxiliary space: O(1)

References:

https://en.wikipedia.org/wiki/Powerful_number

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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