检查一个数字是否可以表示为正K个除数的乘积

给定整数N ，任务是检查它是否可以表示为正好K个质数的乘积。
例子：

Input: N = 12, K = 3
Output: Yes
Explanation:
12 can be expressed as product of 2×2×3.

Input: N = 14, K = 3
Output:  No
Explanation:
14 can be only expressed as product of 2×7.

方法：
为了解决上述问题，我们给定值N，我们将找到可以将N分解为的最大数量的值。我们可以将N的素因式分解表示为$\prod_{i=1}^{K} {p_{i}}^{a_{i}}$其中p _i是N的素因子，而a _i是指数。我们知道N的除数的总数是$\prod_{i=1}^{K} (a_{i}+1)$ 。因此，我们可以观察到必须检查是否有可能将N表示为K个数的乘积。如果最大分割小于K，则不可能精确地用K个素数来表达，否则总是可能的。

C++

// CPP implementation to Check if a
// number can be expressed as a
// product of exactly K prime divisors
 
#include 
using namespace std;
 
// function to find K prime divisors
void KPrimeDivisors(int N, int K)
{
    int maximum_split = 0;
 
    // count number of 2s that divide N
    while (N % 2 == 0) {
        maximum_split++;
        N /= 2;
    }
 
    // N must be odd at this point.
    // So we can skip one element
    for (int i = 3; i * i <= N; i = i + 2) {
 
        while (N % i == 0) {
            // divide the value of N
            N = N / i;
 
            // increment count
            maximum_split++;
        }
    }
 
    // Condition to handle the case when n
    // is a prime number greater than 2
    if (N > 2)
        maximum_split++;
 
    // check if maximum_split is less than K
    // then it not possible
    if (maximum_split < K) {
        printf("No\n");
        return;
    }
 
    printf("Yes\n");
}
 
/* Driver code */
int main()
{
    // initialise N and K
    int N = 12;
    int K = 3;
 
    KPrimeDivisors(N, K);
 
    return 0;
}

Java

// Java implementation to Check if a
// number can be expressed as a
// product of exactly K prime divisors
class GFG {
     
    // function to find K prime divisors
    static void KPrimeDivisors(int N, int K)
    {
        int maximum_split = 0;
     
        // count number of 2s that divide N
        while (N % 2 == 0) {
            maximum_split++;
            N /= 2;
        }
     
        // N must be odd at this point.
        // So we can skip one element
        for (int i = 3; i * i <= N; i = i + 2) {
     
            while (N % i == 0) {
                // divide the value of N
                N = N / i;
     
                // increment count
                maximum_split++;
            }
        }
     
        // Condition to handle the case when n
        // is a prime number greater than 2
        if (N > 2)
            maximum_split++;
     
        // check if maximum_split is less than K
        // then it not possible
        if (maximum_split < K) {
            System.out.println("No");
            return;
        }
     
        System.out.println("Yes");
    }
     
    /* Driver code */
    public static void main (String[] args)
    {
        // initialise N and K
        int N = 12;
        int K = 3;
     
        KPrimeDivisors(N, K);
    }
}
 
// This code is contributed by Yash_R

Python3

# Python implementation to Check if a
# number can be expressed as a
# product of exactly K prime divisors
 
import math as mt
 
# function to find K prime divisors
def KPrimeDivisors(n, k):
     
    # To count maximum split of N
    maximum_split = 0
     
    # count number of 2s that divide N
    while n % 2 == 0:
        maximum_split+= 1
        n = n // 2
         
    # n must be odd at this point
    # so we skip one element
    for i in range(3, mt.ceil(mt.sqrt(n)), 2):
        while n % i == 0:
            n = n / i;
            maximum_split+= 1
             
    # Condition to handle the case when n
    # is a prime number greater than 2
    if n > 2:
        maximum_split+= 1
         
    # check if maximum_split is less than K
    # then it not possible
    if maximum_split < k:
        print("No")
        return
 
    print("Yes")
         
     
 
# Driver code
N = 12
K = 3
KPrimeDivisors(N, K)

C#

// C# implementation to Check if a
// number can be expressed as a
// product of exactly K prime divisors
using System;
 
class GFG {
      
    // function to find K prime divisors
    static void KPrimeDivisors(int N, int K)
    {
        int maximum_split = 0;
      
        // count number of 2s that divide N
        while (N % 2 == 0) {
            maximum_split++;
            N /= 2;
        }
      
        // N must be odd at this point.
        // So we can skip one element
        for (int i = 3; i * i <= N; i = i + 2) {
      
            while (N % i == 0) {
 
                // divide the value of N
                N = N / i;
      
                // increment count
                maximum_split++;
            }
        }
      
        // Condition to handle the case when n
        // is a prime number greater than 2
        if (N > 2)
            maximum_split++;
      
        // check if maximum_split is less than K
        // then it not possible
        if (maximum_split < K) {
            Console.WriteLine("No");
            return;
        }
      
        Console.WriteLine("Yes");
    }
      
    /* Driver code */
    public static void Main(String[] args)
    {
        // initialise N and K
        int N = 12;
        int K = 3;
      
        KPrimeDivisors(N, K);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript implementation to Check if a
// number can be expressed as a
// product of exactly K prime divisors    
 
// function to find K prime divisors
    function KPrimeDivisors(N , K)
    {
        var maximum_split = 0;
 
        // count number of 2s that divide N
        while (N % 2 == 0)
        {
            maximum_split++;
            N /= 2;
        }
 
        // N must be odd at this point.
        // So we can skip one element
        for (i = 3; i * i <= N; i = i + 2)
        {
 
            while (N % i == 0)
            {
             
                // divide the value of N
                N = N / i;
 
                // increment count
                maximum_split++;
            }
        }
 
        // Condition to handle the case when n
        // is a prime number greater than 2
        if (N > 2)
            maximum_split++;
 
        // check if maximum_split is less than K
        // then it not possible
        if (maximum_split < K)
        {
            document.write("No");
            return;
        }
 
        document.write("Yes");
    }
 
    /* Driver code */
     
        // initialise N and K
        var N = 12;
        var K = 3;
        KPrimeDivisors(N, K);
 
// This code is contributed by gauravrajput1.

输出：

Yes

时间复杂度： O(sqrt(N))