将每个元素作为其索引的倍数或因子的数组的排列计数

给定的整数，N，任务进行计数的方式的数量，以产生一个阵列，其由N个整数的常用3 []，使得对于每一个索引i(从1分度)，ARR [I]可以是一个因素或i的倍数，或两者皆是。 arr []必须是[1，N]范围内所有数字的排列。

例子：

Input: N=2
Output: 2
Explanation:
Two possible arrangements are {1, 2} and {2, 1}
Input: N=3
Output: 3
Explanation:
The 6 possible arrangements are {1, 2, 3}, {2, 1, 3}, {3, 2, 1}, {3, 1, 2}, {2, 3, 1} and {1, 3, 2}.
Among them, the valid arrangements are {1, 2, 3}, {2, 1, 3} and {3, 2, 1}.

方法：可以使用Backtracking技术解决问题，并使用递归打印所有排列的概念。请按照以下步骤查找重复关系：

1. 遍历范围[1，N] 。
2. 对于当前索引pos ，如果i％pos == 0且i％pos == 0 ，则将i插入到排列中，并使用“回溯”概念查找有效的排列。
3. 删除我。
4. 对[1，N]范围内的所有值重复上述步骤，最后打印有效排列的计数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the count of
// desired permutations
int findPermutation(unordered_set& arr,
                    int N)
{
    int pos = arr.size() + 1;
 
    // Base case
    if (pos > N)
        return 1;
 
    int res = 0;
 
    for (int i = 1; i <= N; i++) {
 
        // If i has not been inserted
        if (arr.find(i) == arr.end()) {
 
            // Backtrack
            if (i % pos == 0 or pos % i == 0) {
 
                // Insert i
                arr.insert(i);
 
                // Recur to find valid permutations
                res += findPermutation(arr, N);
 
                // Remove i
                arr.erase(arr.find(i));
            }
        }
    }
 
    // Return the final count
    return res;
}
 
// Driver Code
int main()
{
    int N = 5;
    unordered_set arr;
    cout << findPermutation(arr, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to find the count of
// desired permutations
static int findPermutation(Setarr,
                           int N)
{
    int pos = arr.size() + 1;
 
    // Base case
    if (pos > N)
        return 1;
 
    int res = 0;
 
    for(int i = 1; i <= N; i++)
    {
         
        // If i has not been inserted
        if (! arr.contains(i))
        {
             
            // Backtrack
            if (i % pos == 0 || pos % i == 0)
            {
                 
                // Insert i
                arr.add(i);
 
                // Recur to find valid permutations
                res += findPermutation(arr, N);
 
                // Remove i
                arr.remove(i);
            }
        }
    }
 
    // Return the final count
    return res;
}
 
// Driver Code
public static void main(String []args)
{
    int N = 5;
    Set arr = new HashSet();
     
    System.out.print(findPermutation(arr, N));
}
}
 
// This code is contributed by chitranayal

Python3

# Python3 program to implement
# the above approach
 
# Function to find the count of
# desired permutations
def findPermutation(arr, N):
 
    pos = len(arr) + 1
 
    # Base case
    if(pos > N):
        return 1
 
    res = 0
 
    for i in range(1, N + 1):
 
        # If i has not been inserted
        if(i not in arr):
 
            # Backtrack
            if(i % pos == 0 or pos % i == 0):
 
                # Insert i
                arr.add(i)
 
                # Recur to find valid permutations
                res += findPermutation(arr, N)
 
                # Remove i
                arr.remove(i)
 
    # Return the final count
    return res
 
# Driver Code
N = 5
arr = set()
 
# Function call
print(findPermutation(arr, N))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the count of
// desired permutations
static int findPermutation(HashSetarr,
                           int N)
{
    int pos = arr.Count + 1;
 
    // Base case
    if (pos > N)
        return 1;
 
    int res = 0;
 
    for(int i = 1; i <= N; i++)
    {
         
        // If i has not been inserted
        if (! arr.Contains(i))
        {
             
            // Backtrack
            if (i % pos == 0 || pos % i == 0)
            {
                 
                // Insert i
                arr.Add(i);
 
                // Recur to find valid permutations
                res += findPermutation(arr, N);
 
                // Remove i
                arr.Remove(i);
            }
        }
    }
 
    // Return the readonly count
    return res;
}
 
// Driver Code
public static void Main(String []args)
{
    int N = 5;
    HashSet arr = new HashSet();
     
    Console.Write(findPermutation(arr, N));
}
}
 
// This code is contributed by gauravrajput1

输出：

10

时间复杂度： O(N×N！)
辅助空间： O(N)