给定数N的可被K整除的除数的数量

给定一个数字N和一个数字K。任务是找到可被K整除的N个除数。这里K是一个始终小于或等于√(N)的数字。

例子：

Input: N = 12, K = 3
Output: 3

Input: N = 8, K = 2
Output: 3

简单方法：一种简单方法是检查从1到N的所有数字，并检查是否有任何数字是N的除数并且可以被K整除。计算满足两个条件的小于N的数字。
下面是上述方法的实现：

C++

// C++ program to count number of divisors
// of N which are divisible by K
  
#include 
using namespace std;
  
// Function to count number of divisors
// of N which are divisible by K
int countDivisors(int n, int k)
{
    // Variable to store
    // count of divisors
    int count = 0, i;
  
    // Traverse from 1 to n
    for (i = 1; i <= n; i++) {
  
        // increase the count if both
        // the conditions are satisfied
        if (n % i == 0 && i % k == 0) {
  
            count++;
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    int n = 12, k = 3;
  
    cout << countDivisors(n, k);
  
    return 0;
}

Java

// Java program to count number of divisors
// of N which are divisible by K
  
import java.io.*;
  
class GFG {
     
// Function to count number of divisors
// of N which are divisible by K
 static int countDivisors(int n, int k)
{
    // Variable to store
    // count of divisors
    int count = 0, i;
  
    // Traverse from 1 to n
    for (i = 1; i <= n; i++) {
  
        // increase the count if both
        // the conditions are satisfied
        if (n % i == 0 && i % k == 0) {
  
            count++;
        }
    }
  
    return count;
}
  
// Driver code
    public static void main (String[] args) {
      int n = 12, k = 3;
  
    System.out.println(countDivisors(n, k));
    }
}
// This code is contributed by shashank..

Python3

# Python program to count number 
# of divisors of N which are 
# divisible by K 
  
# Function to count number of divisors 
# of N which are divisible by K 
def countDivisors(n, k) :
  
    # Variable to store 
    # count of divisors 
    count = 0
  
    # Traverse from 1 to n 
    for i in range(1, n + 1) :
  
        # increase the count if both 
        # the conditions are satisfied 
        if (n % i == 0 and i % k == 0) :
  
            count += 1
              
    return count
  
# Driver code     
if __name__ == "__main__" :
  
    n, k = 12, 3
    print(countDivisors(n, k))
  
# This code is contributed by ANKITRAI1

C#

// C# program to count number 
// of divisors of N which are
// divisible by K
using System;
  
class GFG
{
      
// Function to count number 
// of divisors of N which
// are divisible by K
static int countDivisors(int n, int k)
{
    // Variable to store
    // count of divisors
    int count = 0, i;
  
    // Traverse from 1 to n
    for (i = 1; i <= n; i++) 
    {
  
        // increase the count if both
        // the conditions are satisfied
        if (n % i == 0 && i % k == 0) 
        {
            count++;
        }
    }
  
    return count;
}
  
// Driver code
public static void Main () 
{
    int n = 12, k = 3;
      
    Console.WriteLine(countDivisors(n, k));
}
}
  
// This code is contributed by Shashank

---

输出

3

时间复杂度：O(N)
高效的方法：想法是从1到<√(N)循环运行，并检查数字是否为N的因数并且可以被K整除，我们还将检查(N / i)是否可以被K整除。 。如果i是N的因数，则(N / i)也将是N的因数。
下面是上述方法的实现：

C++

// C++ program to count number of divisors
// of N which are divisible by K
#include 
using namespace std;
  
// Function to count number of divisors
// of N which are divisible by K
int countDivisors(int n, int k)
{
    // integer to count the divisors
    int count = 0, i;
  
    // Traverse from 1 to sqrt(N)
    for (i = 1; i <= sqrt(n); i++) 
    {
  
        // Check if i is a factor
        if (n % i == 0) 
        {
            // increase the count if i
            // is divisible by k
            if (i % k == 0) 
            {
                count++;
            }
  
            // (n/i) is also a factor
            // check whether it is divisible by k
            if ((n / i) % k == 0) 
            {
                count++;
            }
        }
    }
      
    i--;
  
    // If the number is a perfect square
    // and it is divisible by k
    if ((i * i == n) && (i % k == 0)) 
    {
        count--;
    }
  
    return count;
}
  
// Driver code
int main()
{
    int n = 16, k = 4;
  
    // Function Call
    cout << countDivisors(n, k);
    return 0;
}

Java

// Java  program to count number of divisors 
// of N which are divisible by K
import java.io.*;
  
class GFG {
      
// Function to count number of divisors 
// of N which are divisible by K 
static int countDivisors(int n, int k) 
{ 
    // integer to count the divisors 
    int count = 0, i; 
  
    // Traverse from 1 to sqrt(N) 
    for (i = 1; i <= Math.sqrt(n); i++) 
    { 
  
        // Check if i is a factor 
        if (n % i == 0) 
        { 
            // increase the count if i 
            // is divisible by k 
            if (i % k == 0) 
            { 
                count++; 
            } 
  
            // (n/i) is also a factor 
            // check whether it is divisible by k 
            if ((n / i) % k == 0) 
            { 
                count++; 
            } 
        } 
    } 
    
    i--;
  
    // If the number is a perfect square 
    // and it is divisible by k 
    if ((i * i == n) && (i % k == 0)) 
    { 
        count--; 
    } 
  
    return count; 
} 
  
    // Driver code     
    public static void main (String[] args) 
    {
      
        int n = 16, k = 4; 
        System.out.println( countDivisors(n, k)); 
          
    }
}
//This Code is Contributed by akt_mit

的Python 3

# Python 3 program to count number of 
# divisors of N which are divisible by K
import math
  
# Function to count number of divisors
# of N which are divisible by K
def countDivisors(n, k):
      
    # integer to count the divisors
    count = 0
  
    # Traverse from 1 to sqrt(N)
    for i in range(1, int(math.sqrt(n)) + 1):
  
        # Check if i is a factor
        if (n % i == 0) :
              
            # increase the count if i
            # is divisible by k
            if (i % k == 0) :
                count += 1
  
            # (n/i) is also a factor check
            # whether it is divisible by k
            if ((n // i) % k == 0) :
                count += 1
  
      
      
    # If the number is a perfect square
    # and it is divisible by k
    # if i is sqrt reduce by 1
    if ((i * i == n) and (i % k == 0)) :
        count -= 1  
  
    return count
  
# Driver code
if __name__ == "__main__":
    n = 16
    k = 4
  
    print(countDivisors(n, k))
  
# This code is contributed 
# by ChitraNayal

C＃

// C# program to count number of divisors 
// of N which are divisible by K
using System;
  
class GFG
{
      
// Function to count number of divisors 
// of N which are divisible by K 
static int countDivisors(int n, int k) 
{ 
    // integer to count the divisors 
    int count = 0, i; 
  
    // Traverse from 1 to sqrt(N) 
    for (i = 1; i <= Math.Sqrt(n); i++)
    { 
  
        // Check if i is a factor 
        if (n % i == 0) 
        { 
            // increase the count if i 
            // is divisible by k 
            if (i % k == 0) 
            { 
                count++; 
            } 
  
            // (n/i) is also a factor check
            // whether it is divisible by k 
            if ((n / i) % k == 0) 
            { 
                count++; 
            } 
        } 
    } 
    
    i--;
  
    // If the number is a perfect square 
    // and it is divisible by k 
    if ((i * i == n) && (i % k == 0))
    { 
        count--; 
    } 
  
    return count; 
} 
  
// Driver code 
static public void Main ()
{
    int n = 16, k = 4; 
    Console.WriteLine( countDivisors(n, k)); 
}
}
  
// This code is contributed by ajit

输出

3

时间复杂度：O(√(n))