Dado un entero positivo N. La tarea es verificar si N es un número inusual o no. Escriba ‘SÍ’ si M es un número inusual, de lo contrario escriba ‘NO’.
Número inusual : En Matemáticas, un número inusual es un número natural cuyo mayor factor primo es estrictamente mayor que la raíz cuadrada de n.
Los primeros números inusuales son:
2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 26, 28, 29, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 51
Ejemplos :
Input : N = 14 Output : YES Explanation : 7 is largest prime factor of 14 and 7 is strictly greater than square root of 14 Input : N = 16 Output : NO Explanation : 2 is largest prime factor of 16 and 2 is less than square root of 16 ( i.e 4 ).
Acercarse :
- Encuentre el factor primo más grande del número N dado. Para encontrar el factor primo más grande de N, consulte este .
- Comprueba si el factor primo más grande de N es estrictamente mayor que la raíz cuadrada de N.
- Si ‘SÍ’, entonces N es un número inusual, de lo contrario, no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to check Unusual number #include <bits/stdc++.h> using namespace std; // Utility function to find largest // prime factor of a number int largestPrimeFactor(int n) { // Initialize the maximum prime factor // variable with the lowest one int max = -1; // Print the number of 2s that divide n while (n % 2 == 0) { max = 2; n >>= 1; // equivalent to n /= 2 } // n must be odd at this point, thus skip // the even numbers and iterate only for // odd integers for (int i = 3; i <= sqrt(n); i += 2) { while (n % i == 0) { max = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) max = n; return max; } // Function to check Unusual number bool checkUnusual(int n) { // Get the largest Prime Factor // of the number int factor = largestPrimeFactor(n); // Check if largest prime factor // is greater than sqrt(n) if (factor > sqrt(n)) { return true; } else { return false; } } // Driver Code int main() { int n = 14; if (checkUnusual(n)) { cout << "YES" << "\n"; } else { cout << "NO" << "\n"; } return 0; }
Java
// Java Program to check Unusual number class GFG { // Utility function to find largest // prime factor of a number static int largestPrimeFactor(int n) { // Initialize the maximum prime factor // variable with the lowest one int max = -1; // Print the number of 2s that divide n while (n % 2 == 0) { max = 2; n >>= 1; // equivalent to n /= 2 } // n must be odd at this point, thus skip // the even numbers and iterate only for // odd integers for (int i = 3; i <= Math.sqrt(n); i += 2) { while (n % i == 0) { max = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) max = n; return max; } // Function to check Unusual number static boolean checkUnusual(int n) { // Get the largest Prime Factor // of the number int factor = largestPrimeFactor(n); // Check if largest prime factor // is greater than sqrt(n) if (factor > Math.sqrt(n)) { return true; } else { return false; } } // Driver Code public static void main(String[] args) { int n = 14; if (checkUnusual(n)) { System.out.println("YES"); } else { System.out.println("NO"); } } }
Python3
# Python Program to check Unusual number from math import sqrt # Utility function to find largest # prime factor of a number def largestPrimeFactor(n): # Initialize the maximum prime factor # variable with the lowest one max = -1 # Print the number of 2s that divide n while n % 2 == 0: max = 2; n >>= 1; # equivalent to n /= 2 # n must be odd at this point, thus skip # the even numbers and iterate only for # odd integers for i in range(3,int(sqrt(n))+1,2): while n % i == 0: max = i; n = n / i; # This condition is to handle the case # when n is a prime number greater than 2 if n > 2: max = n return max # Function to check Unusual number def checkUnusual(n): # Get the largest Prime Factor # of the number factor = largestPrimeFactor(n) # Check if largest prime factor # is greater than sqrt(n) if factor > sqrt(n): return True else : return False # Driver Code if __name__ == '__main__': n = 14 if checkUnusual(n): print("YES") else: print("NO") # This code is contributed # by Harshit Saini
C#
// C# Program to check Unusual number using System; class GFG { // Utility function to find largest // prime factor of a number static int largestPrimeFactor(int n) { // Initialize the maximum prime factor // variable with the lowest one int max = -1; // Print the number of 2s that divide n while (n % 2 == 0) { max = 2; n >>= 1; // equivalent to n /= 2 } // n must be odd at this point, thus skip // the even numbers and iterate only for // odd integers for (int i = 3; i <= Math.Sqrt(n); i += 2) { while (n % i == 0) { max = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) max = n; return max; } // Function to check Unusual number static bool checkUnusual(int n) { // Get the largest Prime Factor // of the number int factor = largestPrimeFactor(n); // Check if largest prime factor // is greater than sqrt(n) if (factor > Math.Sqrt(n)) { return true; } else { return false; } } // Driver Code public static void Main() { int n = 14; if (checkUnusual(n)) { Console.WriteLine("YES"); } else { Console.WriteLine("NO"); } } }
PHP
<?php // PHP Program to check Unusual number // Utility function to find largest // prime factor of a number function largestPrimeFactor($n) { // Initialize the maximum prime factor // variable with the lowest one $max = -1; // Print the number of 2s that divide n while ($n % 2 == 0) { $max = 2; $n >>= 1; // equivalent to n /= 2 } // n must be odd at this point, thus skip // the even numbers and iterate only for // odd integers for ($i = 3; $i <= sqrt($n); $i += 2) { while ($n % $i == 0) { $max = $i; $n = $n / $i; } } // This condition is to handle the case // when n is a prime number greater than 2 if ($n > 2) $max = $n; return $max; } // Function to check Unusual number function checkUnusual($n) { // Get the largest Prime Factor // of the number $factor = largestPrimeFactor($n); // Check if largest prime factor // is greater than sqrt(n) if ($factor > sqrt($n)) { return true; } else { return false; } } // Driver Code $n = 14; if (checkUnusual($n)) { echo "YES"."\n"; } else { echo "NO"."\n"; } // This code is contributed // by Harshit Saini ?>
Javascript
<script> // javascript Program to check Unusual number // Utility function to find largest // prime factor of a number function largestPrimeFactor( n) { // Initialize the maximum prime factor // variable with the lowest one var max = -1; // Print the number of 2s that divide n while (n % 2 == 0) { max = 2; n >>= 1; // equivalent to n /= 2 } // n must be odd at this point, thus skip // the even numbers and iterate only for // odd integers for (var i = 3; i <= Math.sqrt(n); i += 2) { while (n % i == 0) { max = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) max = n; return max; } // Function to check Unusual number function checkUnusual(n) { // Get the largest Prime Factor // of the number var factor = largestPrimeFactor(n); // Check if largest prime factor // is greater than sqrt(n) if (factor > Math.sqrt(n)) { return true; } else { return false; } } // Driver Code var n = 14; if (checkUnusual(n)) { document.write("YES"); } else { document.write("NO"); } </script>
Producción:
YES
Complejidad temporal:
Espacio auxiliar: