N的素数的异次幂计数

给定正整数N ，任务是找到给定数N的质数的幂次方的总数。
例子：

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 
 

方法：想法是找到N的所有素数以及每个素数除N的次数。
假设素数因子“ p”除以N个“ z”倍，则所需的不同素数因子为p，p ² ，…，p ⁱ 。
为了找到素数p的不同素数因子的数量，找到i的最小值，使得(1 + 2 +…。+ i)≤z 。
因此，对于每个除以N K次的质数，求出i的最小值，使得(1 + 2 +…。+ i)≤K 。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
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
int 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 N
static 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 N
def 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 N
N = 24
 
# Function call
print(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 N
static 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

 
// Javascript program for the above approach
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
function 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 N
var N = 24;
 
// Function Call
document.write(countFac(N));
 
// This code is contributed by Khushboogoyal499
    

输出：

3

时间复杂度： O(sqrt(N))