Given a positive integer N, the task is to find the total number of distinct power of prime factor of the given number N.
Examples:
Input: N = 216
Output: 4
Explanation:
216 can be expressed as 2 * 22 * 3 * 32.
The factors satisfying the conditions are 2, 22, 3 and 32 as all of them are written as distinct positive powers of prime factors.
Input: N = 24
Output: 3
Explanation:
24 can be expressed as 2 * 22 * 3
Approach: The idea is to find all the prime factors of N and how many times each prime factor divides N.
Suppose the prime factor ‘p’ divides N ‘z’ times, then the required distinct prime factors are p, p2, …, pi.
To find the number of distinct primes factor for prime number p find the minimum value of i such that (1 + 2 + …. + i) ? z.
Therefore, for each prime number dividing N K number of times, find the minimum value of i such that (1 + 2 + …. + i) ? K.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number// of distinct positive power of// prime factor of integer Nint countFac(int n){ int m = n; int count = 0; // Iterate for all prime factor for (int i = 2; (i * i) <= m; ++i) { int total = 0; // If it is a prime factor, // count the total number // of times it divides n. while (n % i == 0) { n /= i; ++total; } int temp = 0; // Find the Number of distinct // possible positive numbers for (int j = 1; (temp + j) <= total; ++j) { temp += j; ++count; } } if (n != 1) ++count; // Return the final count return count;}// Driver Codeint main(){ // Given Number N int N = 24; // Function Call cout << countFac(N); return 0;} |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to count the number// of distinct positive power of// prime factor of integer Nstatic int countFac(int n){ int m = n; int count = 0; // Iterate for all prime factor for(int i = 2; (i * i) <= m; ++i) { int total = 0; // If it is a prime factor, // count the total number // of times it divides n. while (n % i == 0) { n /= i; ++total; } int temp = 0; // Find the Number of distinct // possible positive numbers for(int j = 1; (temp + j) <= total; ++j) { temp += j; ++count; } } if (n != 1) ++count; // Return the final count return count;}// Driver code public static void main(String[] args) { // Given Number N int N = 24; // Function Call System.out.println(countFac(N)); } } // This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approach# Function to count the number# of distinct positive power of# prime factor of integer Ndef countFac(n): m = n count = 0 # Iterate for all prime factor i = 2 while((i * i) <= m): total = 0 # If it is a prime factor, # count the total number # of times it divides n. while (n % i == 0): n /= i total += 1 temp = 0 # Find the Number of distinct # possible positive numbers j = 1 while((temp + j) <= total): temp += j count += 1 j += 1 i += 1 if (n != 1): count += 1 # Return the final count return count# Driver Code# Given number NN = 24# Function callprint(countFac(N))# This code is contributed by sanjoy_62 |
C#
// C# program for the above approach using System;class GFG{ // Function to count the number// of distinct positive power of// prime factor of integer Nstatic int countFac(int n){ int m = n; int count = 0; // Iterate for all prime factor for(int i = 2; (i * i) <= m; ++i) { int total = 0; // If it is a prime factor, // count the total number // of times it divides n. while (n % i == 0) { n /= i; ++total; } int temp = 0; // Find the Number of distinct // possible positive numbers for(int j = 1; (temp + j) <= total; ++j) { temp += j; ++count; } } if (n != 1) ++count; // Return the final count return count;}// Driver code public static void Main() { // Given Number N int N = 24; // Function Call Console.Write(countFac(N)); } } // This code is contributed by Code_Mech |
Javascript
<script>// Javascript program for the above approach// Function to count the number// of distinct positive power of// prime factor of integer Nfunction countFac(n){ var m = n; var count = 0; // Iterate for all prime factor for(var i = 2; (i * i) <= m; ++i) { var total = 0; // If it is a prime factor, // count the total number // of times it divides n. while (n % i == 0) { n /= i; ++total; } var temp = 0; // Find the Number of distinct // possible positive numbers for(var j = 1; (temp + j) <= total; ++j) { temp += j; ++count; } } if (n != 1) ++count; // Return the final count return count;}// Driver code// Given Number Nvar N = 24;// Function Calldocument.write(countFac(N)); // This code is contributed by Khushboogoyal499 </script> |
Output:
3
Time complexity: O(sqrt(N))
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