找出自然数N的第k个最小除数

系统会为您提供数字N和数字K。我们的任务是找到N的^第k^个最小除数。
例子：

Input : N = 12, K = 5
Output : 6
The divisors of 12 after sorting are 1, 2, 3, 4, 6 and 12. 
Where the value of 5th divisor is equal to 6.

Input : N = 16, K 2
Output : 2

简单方法：一个简单的方法是从1到√N循环，找到N的所有因子并将其推入向量。最后，对向量排序并从向量打印第K个值。
注意：向量中的元素最初不会被排序，因为我们同时推动了(i)和(n / i)两个因子。这就是为什么需要在打印第K个因子之前对向量进行排序的原因。
下面是上述方法的实现：

C++

// C++ program to find K-th smallest factor
 
#include 
using namespace std;
 
// function to find the k'th divisor
void findkth(int n, int k)
{
    // initialize a vector v
    vector v;
 
    // store all the divisors
    // so the loop will needs to run till sqrt ( n )
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            v.push_back(i);
            if (i != sqrt(n))
                v.push_back(n / i);
        }
    }
 
    // sort the vector in an increasing order
    sort(v.begin(), v.end());
 
    // if k is greater than the size of vector
    // then no divisor can be possible
    if (k > v.size())
        cout << "Doesn't Exist";
    // else print the ( k - 1 )th value of vector
    else
        cout << v[k - 1];
}
 
// Driver code
int main()
{
    int n = 15, k = 2;
 
    findkth(n, k);
 
    return 0;
}

Java

// Java program to find K-th smallest factor
import java.util.*;
 
class GFG{
 
// function to find the k'th divisor
static void findkth(int n, int k)
{
    // initialize a vector v
    Vector v = new Vector();
 
    // store all the divisors
    // so the loop will needs to run till sqrt ( n )
    for (int i = 1; i <= Math.sqrt(n); i++) {
        if (n % i == 0) {
            v.add(i);
            if (i != Math.sqrt(n))
                v.add(n / i);
        }
    }
 
    // sort the vector in an increasing order
    Collections.sort(v);
 
    // if k is greater than the size of vector
    // then no divisor can be possible
    if (k > v.size())
        System.out.print("Doesn't Exist");
     
    // else print the ( k - 1 )th value of vector
    else
        System.out.print(v.get(k - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15, k = 2;
 
    findkth(n, k);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find K-th smallest factor
from math import sqrt
 
# function to find the k'th divisor
def findkth(n, k):
     
    # initialize a vector v
    v = []
 
    # store all the divisors so the loop
    # will needs to run till sqrt ( n )
    p = int(sqrt(n)) + 1
    for i in range(1, p, 1):
        if (n % i == 0):
            v.append(i)
            if (i != sqrt(n)):
                v.append(n / i);
         
    # sort the vector in an increasing order
    v.sort(reverse = False)
 
    # if k is greater than the size of vector
    # then no divisor can be possible
    if (k > len(v)):
        print("Doesn't Exist")
         
    # else print the (k - 1)th
    # value of vector
    else:
        print(v[k - 1])
 
# Driver code
if __name__ == '__main__':
    n = 15
    k = 2
 
    findkth(n, k)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find K-th smallest factor
using System;
using System.Collections.Generic;
 
class GFG{
  
// function to find the k'th divisor
static void findkth(int n, int k)
{
    // initialize a vector v
    List v = new List();
  
    // store all the divisors
    // so the loop will needs to run till sqrt ( n )
    for (int i = 1; i <= Math.Sqrt(n); i++) {
        if (n % i == 0) {
            v.Add(i);
            if (i != Math.Sqrt(n))
                v.Add(n / i);
        }
    }
  
    // sort the vector in an increasing order
    v.Sort();
  
    // if k is greater than the size of vector
    // then no divisor can be possible
    if (k > v.Count)
        Console.Write("Doesn't Exist");
      
    // else print the ( k - 1 )th value of vector
    else
        Console.Write(v[k - 1]);
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 15, k = 2;
  
    findkth(n, k);
}
}
 
// This code is contributed by PrinciRaj1992

---

输出：

3

时间复杂度：√Nlog(√N)
高效的方法：一种有效的方法是将因子存储在两个单独的向量中。也就是说，对于从1到√N的所有i，因子i将存储在单独的向量中，而N / i将存储在单独的向量中。
现在，如果仔细观察，可以看出第一矢量已经按升序排序，第二矢量已经按降序排序。因此，反转第二个矢量并从其所在的两个矢量中打印第K个元素。
下面是上述方法的实现：

C++

// C++ program to find the K-th smallest factor
 
#include 
using namespace std;
 
// Function to find the k'th divisor
void findkth ( int n, int k)
{
    // initialize vectors v1 and v2
    vector  v1;
    vector  v2;
     
    // store all the divisors in the two vectors
    // accordingly
    for( int i = 1 ; i <= sqrt( n ); i++ )
    {
        if ( n % i == 0 )
        {
            v1.push_back ( i );
             
            if ( i != sqrt ( n ) )
                v2.push_back ( n / i );
        }
    }
     
    // reverse the vector v2 to sort it
    // in increasing order
    reverse(v2.begin(), v2.end());
     
    // if k is greater than the size of vectors
    // then no divisor can be possible
    if ( k > (v1.size() + v2.size()))
        cout << "Doesn't Exist" ;
    // else print the ( k - 1 )th value of vector
    else
    {
        // If K is lying in first vector
        if(k <= v1.size())
            cout<

Java

// Java program to find the K-th smallest factor
import java.util.*;
 
class GFG
{
 
// Function to find the k'th divisor
static void findkth ( int n, int k)
{
    // initialize vectors v1 and v2
    Vector v1 = new Vector();
    Vector  v2 = new Vector();
     
    // store all the divisors in the two vectors
    // accordingly
    for( int i = 1 ; i <= Math.sqrt( n ); i++ )
    {
        if ( n % i == 0 )
        {
            v1.add ( i );
             
            if ( i != Math.sqrt ( n ) )
                v2.add ( n / i );
        }
    }
     
    // reverse the vector v2 to sort it
    // in increasing order
    Collections.reverse(v2);
     
    // if k is greater than the size of vectors
    // then no divisor can be possible
    if ( k > (v1.size() + v2.size()))
        System.out.print("Doesn't Exist");
     
    // else print the ( k - 1 )th value of vector
    else
    {
        // If K is lying in first vector
        if(k <= v1.size())
            System.out.print(v1.get(k - 1));
         
        // If K is lying in second vector
        else
            System.out.print(v2.get(k-v1.size() - 1));
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15, k = 2;
     
    findkth ( n, k) ;
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find the K-th
# smallest factor
import math as mt
 
# Function to find the k'th divisor
def findkth (n, k):
 
    # initialize vectors v1 and v2
    v1 = list()
    v2 = list()
     
    # store all the divisors in the
    # two vectors accordingly
    for i in range(1, mt.ceil(n**(.5))):
     
        if (n % i == 0):
         
            v1.append(i)
             
            if (i != mt.ceil(mt.sqrt(n))):
                v2.append(n // i)
         
    # reverse the vector v2 to sort it
    # in increasing order
    v2[::-1]
     
    # if k is greater than the size of vectors
    # then no divisor can be possible
    if ( k > (len(v1) + len(v2))):
        print("Doesn't Exist", end = "")
         
    # else print the ( k - 1 )th value of vector
    else:
     
        # If K is lying in first vector
        if(k <= len(v1)):
            print(v1[k - 1])
             
        # If K is lying in second vector
        else:
            print(v2[k - len(v1) - 1])
     
# Driver code
n = 15
k = 2
findkth (n, k)
 
# This code is contributed by Mohit kumar

C＃

// C# program to find
// the K-th smallest factor
using System;
using System.Collections.Generic;
class GFG{
 
// Function to find the k'th divisor
static void findkth (int n, int k)
{
  // initialize vectors v1 and v2
  List v1 = new List();
  List  v2 = new List();
 
  // store all the divisors in the
  // two vectors accordingly
  for(int i = 1; i <= Math.Sqrt(n); i++)
  {
    if (n % i == 0)
    {
      v1.Add (i);
 
      if (i != Math.Sqrt (n))
        v2.Add (n / i);
    }
  }
 
  // reverse the vector v2 to sort it
  // in increasing order
  v2.Reverse();
 
  // if k is greater than the
  // size of vectors then no
  // divisor can be possible
  if (k > (v1.Count + v2.Count))
    Console.Write("Doesn't Exist");
 
  // else print the (k - 1)th
  // value of vector
  else
  {
    // If K is lying in first vector
    if(k <= v1.Count)
      Console.Write(v1[k - 1]);
 
    // If K is lying in second vector
    else
      Console.Write(v2[k - v1.Count - 1]);
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 15, k = 2;
  findkth (n, k);
}
}
 
// This code is contributed by gauravrajput1

输出：

3

时间复杂度：√N