检查是否存在带有X个因子的数字，其中正好是K为质数

给定两个整数X和K ，任务是确定是否存在正好有X个因数(其中K是素数)的数字。

例子：

Input: X = 8, K = 1
Output: Yes
Explanation:
The number is 128
Factors of 128 = {1, 2, 4, 8, 16, 32, 64, 128} which are 8 in count = X
Among these, only 2 is prime. Therefore count of prime factor = 1 = K

Input: X = 4, K = 2
Output: Yes
Explanation:
The number is 6
Factors of 6 = {1, 2, 3, 6} which are 4 in count = X
Among these, only 2 and 3 are prime. Therefore count of prime factor = 2 = K

方法：

• 假设数字N有X个因子，其中K是素数，例如$K_{1}, K_{2}, K_{3}, ...K_{M}$
• 因此，数字可以写成$N = K_{1}^{a}, K_{2}^{b}, K_{3}^{c}, ...K_{M}^{K}$其中，因子总数由下式计算： $X = (a+1) * (b+1) * (c+1) * ... *(K+1)$
• 可以看出，X是数的质数的“幂+1 ”的乘积。因此，如果我们能够将X划分为K个数的乘积，那么我们就可以形成一个正好有X个因子的数字，其中K是质数。

下面是上述方法的实现：

C++

// C++ program to check if there exists
// a number with X factors
// out of which exactly K are prime
  
#include 
using namespace std;
  
// Function to check if such number exists
bool check(int X, int K)
{
    int prime, temp, sqr, i;
  
    // To store the sum of powers
    // of prime factors of X which
    // determines the maximum count
    // of numbers whose product can form X
    prime = 0;
    temp = X;
    sqr = sqrt(X);
  
    // Determining the prime factors of X
    for (i = 2; i <= sqr; i++) {
  
        while (temp % i == 0) {
            temp = temp / i;
            prime++;
        }
    }
  
    // To check if the number is prime
    if (temp > 2)
        prime++;
  
    // If X is 1, then we cannot form
    // a number with 1 factor and K
    // prime factor (as K is atleast 1)
    if (X == 1)
        return false;
  
    // If X itself is prime then it
    // can be represented as a power
    // of only 1 prime factor which
    // is X itself so we return true
    if (prime == 1 && K == 1)
        return true;
  
    // If sum of the powers of prime factors
    // of X is greater than or equal to K,
    // which means X can be represented as a
    // product of K numbers, we return true
    else if (prime >= K)
        return true;
  
    // In any other case, we return false
    // as we cannot form a number with X
    // factors and K prime factors
    else
        return false;
}
  
// Driver code
int main()
{
    int X, K;
    X = 4;
    K = 2;
  
    if (check(X, K))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program to check if there exists
// a number with X factors
// out of which exactly K are prime
   
  
import java.util.*;
  
class GFG{
   
// Function to check if such number exists
static boolean check(int X, int K)
{
    int prime, temp, sqr, i;
   
    // To store the sum of powers
    // of prime factors of X which
    // determines the maximum count
    // of numbers whose product can form X
    prime = 0;
    temp = X;
    sqr = (int) Math.sqrt(X);
   
    // Determining the prime factors of X
    for (i = 2; i <= sqr; i++) {
   
        while (temp % i == 0) {
            temp = temp / i;
            prime++;
        }
    }
   
    // To check if the number is prime
    if (temp > 2)
        prime++;
   
    // If X is 1, then we cannot form
    // a number with 1 factor and K
    // prime factor (as K is atleast 1)
    if (X == 1)
        return false;
   
    // If X itself is prime then it
    // can be represented as a power
    // of only 1 prime factor which
    // is X itself so we return true
    if (prime == 1 && K == 1)
        return true;
   
    // If sum of the powers of prime factors
    // of X is greater than or equal to K,
    // which means X can be represented as a
    // product of K numbers, we return true
    else if (prime >= K)
        return true;
   
    // In any other case, we return false
    // as we cannot form a number with X
    // factors and K prime factors
    else
        return false;
}
   
// Driver code
public static void main(String[] args)
{
    int X, K;
    X = 4;
    K = 2;
   
    if (check(X, K))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 program to check if there exists
# a number with X factors
# out of which exactly K are prime
  
from math import sqrt
# Function to check if such number exists
def check(X,K):
  
    # To store the sum of powers
    # of prime factors of X which
    # determines the maximum count
    # of numbers whose product can form X
    prime = 0
    temp = X
    sqr = int(sqrt(X))
  
    # Determining the prime factors of X
    for i in range(2,sqr+1,1):
        while (temp % i == 0):
            temp = temp // i
            prime += 1
  
    # To check if the number is prime
    if (temp > 2):
        prime += 1
  
    # If X is 1, then we cannot form
    # a number with 1 factor and K
    # prime factor (as K is atleast 1)
    if (X == 1):
        return False
  
    # If X itself is prime then it
    # can be represented as a power
    # of only 1 prime factor w0hich
    # is X itself so we return true
    if (prime == 1 and K == 1):
        return True
  
    # If sum of the powers of prime factors
    # of X is greater than or equal to K,
    # which means X can be represented as a
    # product of K numbers, we return true
    elif(prime >= K):
        return True
  
    # In any other case, we return false
    # as we cannot form a number with X
    # factors and K prime factors
    else:
        return False
  
# Driver code
if __name__ == '__main__':
    X = 4
    K = 2
  
    if (check(X, K)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by Surendra_Gangwar

C#

// C# program to check if there exists
// a number with X factors
// out of which exactly K are prime
using System;
  
class GFG{
  
    // Function to check if such number exists
    static bool check(int X, int K)
    {
        int prime, temp, sqr, i;
      
        // To store the sum of powers
        // of prime factors of X which
        // determines the maximum count
        // of numbers whose product can form X
        prime = 0;
        temp = X;
        sqr = Convert.ToInt32(Math.Sqrt(X));
      
        // Determining the prime factors of X
        for (i = 2; i <= sqr; i++) {
      
            while (temp % i == 0) {
                temp = temp / i;
                prime++;
            }
        }
      
        // To check if the number is prime
        if (temp > 2)
            prime++;
      
        // If X is 1, then we cannot form
        // a number with 1 factor and K
        // prime factor (as K is atleast 1)
        if (X == 1)
            return false;
      
        // If X itself is prime then it
        // can be represented as a power
        // of only 1 prime factor which
        // is X itself so we return true
        if (prime == 1 && K == 1)
            return true;
      
        // If sum of the powers of prime factors
        // of X is greater than or equal to K,
        // which means X can be represented as a
        // product of K numbers, we return true
        else if (prime >= K)
            return true;
      
        // In any other case, we return false
        // as we cannot form a number with X
        // factors and K prime factors
        else
            return false;
    }
      
    // Driver code
    static public void Main ()
    {
        int X, K;
        X = 4;
        K = 2;
      
        if (check(X, K))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by shubhamsingh10

输出：

Yes