重构号码

给定整数n 。检查数字是否可重构。可重构数字是一个整数n，该整数可以通过其所有除数的计数来整除。
例子 ：

Input:  n = 8
Output: yes

Explanation:
8 has 4 divisors: 1, 2, 4, 8
Since 8 is divisible by 4 therefore 8 is 
refactorable number.

Input : n = 4 
Output: no

该解决方案非常简单。这个想法是从1迭代到sqrt(n)并计算一个数字的所有除数。之后，我们只需要检查数字n是否可被总数除以即可。

C++

// C++ program to check whether number is
// refactorable or not
#include 
 
// Function to count all divisors
bool isRefactorableNumber(int n)
{
    // Initialize result
    int divCount = 0;
     
    for (int i = 1; i <= sqrt(n); ++i)
    {
        if (n % i==0)
        {
            // If divisors are equal, count
            // only one.
            if (n / i == i)
                ++divCount;
 
            // Otherwise count both
            else
                divCount += 2;
        }
    }
 
    return n % divCount == 0;
}
 
//Driver Code
int main()
{
    int n = 8;
    if (isRefactorableNumber(n))
        puts("yes");
    else
        puts("no");
 
    n = 14;
    if (isRefactorableNumber(n))
        puts("yes");
    else
        puts("no");
 
    return 0;
}

Java

// Java program to check whether number is
// refactorable or not
 
class GFG
{
    // Function to count all divisors
    static boolean isRefactorableNumber(int n)
    {
        // Initialize result
        int divCount = 0;
         
        for (int i = 1; i <= Math.sqrt(n); ++i)
        {
            if (n % i==0)
            {
                // If divisors are equal, count
                // only one.
                if (n / i == i)
                    ++divCount;
                     
                // Otherwise count both
                else
                    divCount += 2;
            }
        }
        return n % divCount == 0;
    }
     
    public static void main (String[] args)
    {
        int n = 8;
        if (isRefactorableNumber(n))
            System.out.println("yes");
        else
            System.out.println("no");
             
        n = 14;
        if (isRefactorableNumber(n))
            System.out.println("yes");
        else
            System.out.println("no");
    }
}
 
// This code is contributed by Saket Kumar

Python

# Python program to check whether number is
# refactorable or not
import math
 
def isRefactorableNumber(n):
 
    # Initialize result
    divCount = 0
 
    for i in range(1,int(math.sqrt(n))+1):
 
        if n % i == 0:
 
            # If divisors are equal, count only one
            if n/i == i:
                divCount += 1
 
            else:  # Otherwise count both
                divCount += 2
 
    return n % divCount == 0
 
 
# Driver Code
n = 8
if isRefactorableNumber(n):
    print "yes"
else:
    print "no"
 
n = 14
if (isRefactorableNumber(n)):
    print "yes"
else:
    print "no"

C#

// C# program to check whether number is
// refactorable or not
using System;
 
class GFG
{
    // Function to count all divisors
    static bool isRefactorableNumber(int n)
    {
         
        // Initialize result
        int divCount = 0;
         
        for (int i = 1; i <= Math.Sqrt(n); ++i)
        {
            if (n % i==0)
            {
                // If divisors are equal, count
                // only one.
                if (n / i == i)
                    ++divCount;
                 
                // Otherwise count both
                else
                    divCount += 2;
            }
        }
        return n % divCount == 0;
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 8;
        if (isRefactorableNumber(n))
            Console.WriteLine("yes");
        else
            Console.Write("no");
             
        n = 14;
        if (isRefactorableNumber(n))
            Console.Write("yes");
        else
            Console.Write("no");
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// JavaScript program for the above approach
  
    // Function to count all divisors
    function isRefactorableNumber(n)
    {
        // Initialize result
        let divCount = 0; 
           
        for (let i = 1; i <= Math.sqrt(n); ++i)
        {
            if (n % i==0)
            {
                // If divisors are equal, count 
                // only one.
                if (n / i == i)
                    ++divCount;
                       
                // Otherwise count both
                else
                    divCount += 2;
            }
        }
        return n % divCount == 0;
    }
 
// Driver Code
    let n = 8;
    if (isRefactorableNumber(n))
        document.write("yes" + "<br />");
    else
        document.write("no" + "<br />");
   
    n = 14;
    if (isRefactorableNumber(n))
        document.write("yes");
    else
        document.write("no");
         
        // This code is contributed by splevel62.

Output:
yes
no

时间复杂度： O(sqrt(n))
辅助空间： O(1)
关于可重构数的事实

1. 没有完美的可重构数字。
2. 没有三个连续的整数都可以重构。
3. 重构数的自然密度为零。