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📜  将每个元素作为其索引的倍数或因子的数组的排列计数

📅  最后修改于: 2021-04-24 20:08:07             🧑  作者: Mango

给定一个整数N ,任务是计算生成数组的方法的数目, arr []N个整数组成,这样对于每个索引i (基于1的索引), arr [i]要么是一个因子,要么是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:

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 == 0i%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
  4. [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



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