检查数字是否具有除数的质数

给定整数N ，任务是检查N的因数计数是否为质数。

例子：

Input: N = 13 
Output: Yes 
The divisor count is 2 (1 and 13) which is prime.

Input: N = 8 
Output: No 
The divisors are 1, 2, 4 and 8. 
 

方法：请阅读本文以找到数字的除数的数量。因此，求出每个质数p的i的最大值，使得N％(p ⁱ )= 0 。因此，除数的数量乘以(i + 1) 。除数的数量为(i ¹ +1)*(i ² +1)*…*(i ^k +1)。
现在可以看出，最大i只能有一个素数除数，如果N％p ⁱ = 0，则(i + 1)应该是素数。素数可以在sqrt(n)时间中检查，素数因子也可以在sqrt(n)时间中找到。因此，总体时间复杂度将为O(sqrt(n)) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true
// if n is prime
bool Prime(int n)
{
    // There is no prime
    // less than 2
    if (n < 2)
        return false;
 
    // Run a loop from 2 to sqrt(n)
    for (int i = 2; i <= sqrt(n); i++)
 
        // If there is any factor
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function that returns true if n
// has a prime count of divisors
bool primeCountDivisors(int n)
{
    if (n < 2)
        return false;
 
    // Find the prime factors
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0) {
 
            // Find the maximum value of i for every
            // prime divisor p such that n % (p^i) == 0
            long a = n, c = 0;
            while (a % i == 0) {
                a /= i;
                c++;
            }
 
            // If c+1 is a prime number and a = 1
            if (a == 1 && Prime(c + 1))
                return true;
 
            // The number cannot have two factors
            // to have count of divisors prime
            else
                return false;
        }
 
    // Else the number is prime so
    // it has only two divisors
    return true;
}
 
// Driver code
int main()
{
    int n = 13;
 
    if (primeCountDivisors(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function that returns true
    // if n is prime
    static boolean Prime(int n)
    {
        // There is no prime
        // less than 2
        if (n < 2)
            return false;
     
        // Run a loop from 2 to sqrt(n)
        for (int i = 2; i <= (int)Math.sqrt(n); i++)
     
            // If there is any factor
            if (n % i == 0)
                return false;
        return true;
    }
     
    // Function that returns true if n
    // has a prime count of divisors
    static boolean primeCountDivisors(int n)
    {
        if (n < 2)
            return false;
     
        // Find the prime factors
        for (int i = 2; i <= (int)Math.sqrt(n); i++)
            if (n % i == 0)
            {
     
                // Find the maximum value of i for every
                // prime divisor p such that n % (p^i) == 0
                long a = n, c = 0;
                while (a % i == 0)
                {
                    a /= i;
                    c++;
                }
     
                // If c+1 is a prime number and a = 1
                if (a == 1 && Prime((int)c + 1) == true)
                    return true;
     
                // The number cannot have two factors
                // to have count of divisors prime
                else
                    return false;
            }
     
        // Else the number is prime so
        // it has only two divisors
        return true;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 13;
     
        if (primeCountDivisors(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import sqrt
 
# Function that returns true
# if n is prime
def Prime(n) :
 
    # There is no prime
    # less than 2
    if (n < 2) :
        return False;
 
    # Run a loop from 2 to sqrt(n)
    for i in range(2, int(sqrt(n)) + 1) :
 
        # If there is any factor
        if (n % i == 0) :
            return False;
 
    return True;
 
# Function that returns true if n
# has a prime count of divisors
def primeCountDivisors(n) :
 
    if (n < 2) :
        return False;
 
    # Find the prime factors
    for i in range(2, int(sqrt(n)) + 1) :
        if (n % i == 0) :
 
            # Find the maximum value of i for every
            # prime divisor p such that n % (p^i) == 0
            a = n; c = 0;
            while (a % i == 0) :
                a //= i;
                c += 1;
 
            # If c + 1 is a prime number and a = 1
            if (a == 1 and Prime(c + 1)) :
                return True;
 
            # The number cannot have two factors
            # to have count of divisors prime
            else :
                return False;
         
    # Else the number is prime so
    # it has only two divisors
    return True;
 
# Driver code
if __name__ == "__main__" :
 
    n = 13;
 
    if (primeCountDivisors(n)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function that returns true
    // if n is prime
    static bool Prime(int n)
    {
         
        // There is no prime
        // less than 2
        if (n < 2)
            return false;
     
        // Run a loop from 2 to sqrt(n)
        for (int i = 2; i <= (int)Math.Sqrt(n); i++)
     
            // If there is any factor
            if (n % i == 0)
                return false;
        return true;
    }
     
    // Function that returns true if n
    // has a prime count of divisors
    static bool primeCountDivisors(int n)
    {
        if (n < 2)
            return false;
     
        // Find the prime factors
        for (int i = 2; i <= (int)Math.Sqrt(n); i++)
            if (n % i == 0)
            {
     
                // Find the maximum value of i for every
                // prime divisor p such that n % (p^i) == 0
                long a = n, c = 0;
                while (a % i == 0)
                {
                    a /= i;
                    c++;
                }
     
                // If c+1 is a prime number and a = 1
                if (a == 1 && Prime((int)c + 1) == true)
                    return true;
     
                // The number cannot have two factors
                // to have count of divisors prime
                else
                    return false;
            }
     
        // Else the number is prime so
        // it has only two divisors
        return true;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 13;
     
        if (primeCountDivisors(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript implementation of the approach
 
// Function that returns true
// if n is prime
function Prime(n)
{
     
    // There is no prime
    // less than 2
    if (n < 2)
        return false;
 
    // Run a loop from 2 to sqrt(n)
    for(var i = 2; i <= Math.sqrt(n); i++)
 
        // If there is any factor
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function that returns true if n
// has a prime count of divisors
function primeCountDivisors( n)
{
    if (n < 2)
        return false;
 
    // Find the prime factors
    for(var i = 2; i <= Math.sqrt(n); i++)
        if (n % i == 0)
        {
             
            // Find the maximum value of i for every
            // prime divisor p such that n % (p^i) == 0
            var a = n, c = 0;
             
            while (a % i == 0)
            {
                a /= i;
                c++;
            }
 
            // If c+1 is a prime number and a = 1
            if (a == 1 && Prime(c + 1))
                return true;
 
            // The number cannot have two factors
            // to have count of divisors prime
            else
                return false;
        }
 
    // Else the number is prime so
    // it has only two divisors
    return true;
}
 
// Driver rcode
n = 13;
 
if (primeCountDivisors(n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by SoumikMondal
 

输出：

Yes

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