构造一个长度为 N 的数组，其中恰好包含可被其位置整除的 K 个元素

给定两个整数N和K，任务是构造一个长度为N的数组，其中正好包含可被其位置整除的K个元素。

例子：

Input: N = 6, K = 2
Output: {5, 1, 2, 3, 4, 6}
Explanation: Considering the above array:
At Position 1, element 5 is divisible by 1
At Position 2, element 1 is not divisible by 2
At Position 3, element 2 is not divisible by 3
At Position 4, element 3 is not divisible by 4
At Position 5, element 4 is not divisible by 5
At Position 6, element 6 is divisible by 6
Therefore, there are exactly K elements in array divisible by their positions, meeting the required criteria.
Hence the resultant array will be {5 1 2 3 4 6}.

Input: N = 5, K = 5
Output: {1, 2, 3, 4, 5}

方法：可以使用基于以下观察的贪婪方法轻松解决问题：

For any integer X, we know that:

• X will be divisible by 1 and X always.
• No integer greater than X will ever be able to divide X.

So using these observations, we can construct the array containing exactly K elements divisible by their positions, as follows:

• For position 1, place any element greater than 1, because 1 will divide all integers
• For positions greater than 1, choose K-1 positions, and place them in the array at corresponding indices.
• The remaining N-K positions can be placed at any other position to match the required criteria.

插图：

Consider an example: N = 6, K = 5

The empty array of size 6 will be: 
arr[]:         _  _ _  _ _  _
positions: 1 2 3 4 5 6

Step 1: Fill position 1 with any integer greater than 1

• For 1st value equal to its position, we have 2 options – to insert 1 at 1, and to insert some integer greater than 1 at 1. If we insert 1 at 1, there will be a case when we cannot have K=5 values same as their positions. So we will insert some other value greater than 1 at position 1 (say 5):
  • arr[]:         5 _ _  _ _  _
    positions: 1 2 3 4 5 6

Step 2: Fill K-1 (=4) positions at corresponding indices

• For 2nd value equal to its position:
  • arr[]:         5 2 _  _ _ _
    positions: 1 2 3 4 5 6
• For 3rd value equal to its position:
  • arr[]:         5 2 3  _ _ _
    positions: 1 2 3 4 5 6
• For 4th value equal to its position:
  • arr[]:         5 2 3 4 _ _
    positions: 1 2 3 4 5 6
• For 5th value equal to its position, we cannot insert 5 at position 5, as it is already used at position 1. So we will insert 1 at position 5, and 6 at position 6:
  • arr[]:         5 2 3 4 1 6
    positions: 1 2 3 4 5 6

Therefore the final array will be: 5 2 3 4 1 6

请按照以下步骤实施上述方法：

• 创建一个包含从1 到 N的N个连续正整数的数组。
• 在索引NK之后，会剩下K-1 个元素，我们不会干扰这些元素。所以，我们有K-1 个元素，它们可以被它们的位置整除。
• 我们将使数组的第一个元素等于索引NK处的元素。这也可以被它的位置整除。
• 我们将使剩余的元素（即从索引1 到 NK ）等于紧邻它们的元素。这些所有NK元素将不能被它们的位置整除，剩余的K 个元素将被它们的位置整除。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to construct an array of size
// N such that it contains all numbers from
// 1 to N and only K elements are divisible by
// their position (i.e. index+1)
vector constructArray(int N, int K)
{
    // Declaring array of size N
    vector A(N, 0);
 
    // Initializing array as {1, 2, 3....N}
    for (int i = 0; i < N; i++) {
        A[i] = i + 1;
    }
 
    // N-K index stored in a variable "target"
    // After target there will be k-1 elements
    // which are divisible by their position
    int target = N - K;
 
    // Initializing "prev" variable that helps in
    // shifting elements to their right
    int prev = A[0];
 
    // Assigning first element the value at target
    // index
    A[0] = A[target];
 
    // Making all elements from index 1 to target
    // equal to their immediate left element
    // as any number would not be divisible
    // by its next number
    for (int i = 1; i <= target; i++) {
        int temp = A[i];
        A[i] = prev;
        prev = temp;
    }
 
    return A;
}
 
// Driver Code
int main()
{
    int N = 6, K = 2;
 
    // Calling function
    // to construct the array
    vector A = constructArray(N, K);
 
    // Printing resultant array
    for (int i = 0; i < N; i++)
        cout << A[i] << " ";
    cout << endl;
}

Java

// JAVA program for above approach
import java.util.*;
class GFG
{
   
    // Function to construct an array of size
    // N such that it contains all numbers from
    // 1 to N and only K elements are divisible by
    // their position (i.e. index+1)
    public static int[] constructArray(int N, int K)
    {
       
        // Declaring array of size N
        int A[] = new int[N];
        for (int i = 0; i < A.length; ++i) {
            A[i] = 0;
        }
 
        // Initializing array as {1, 2, 3....N}
        for (int i = 0; i < N; i++) {
            A[i] = i + 1;
        }
 
        // N-K index stored in a variable "target"
        // After target there will be k-1 elements
        // which are divisible by their position
        int target = N - K;
 
        // Initializing "prev" variable that helps in
        // shifting elements to their right
        int prev = A[0];
 
        // Assigning first element the value at target
        // index
        A[0] = A[target];
 
        // Making all elements from index 1 to target
        // equal to their immediate left element
        // as any number would not be divisible
        // by its next number
        for (int i = 1; i <= target; i++) {
            int temp = A[i];
            A[i] = prev;
            prev = temp;
        }
 
        return A;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 6, K = 2;
 
        // Calling function
        // to construct the array
        int A[] = constructArray(N, K);
 
        // Printing resultant array
        for (int i = 0; i < N; i++)
            System.out.print(A[i] + " ");
        System.out.println();
    }
}
 
// This code is contributed by Taranpreet

Python3

# Python program for above approach
 
# Function to construct an array of size
# N such that it contains all numbers from
# 1 to N and only K elements are divisible by
# their position (i.e. index+1)
def constructArray(N, K):
   
    # Declaring array of size N
    A = [0]*N
 
    # Initializing array as {1, 2, 3....N}
    for i in range(N):
      A[i] = i + 1
 
    # N-K index stored in a variable "target"
    # After target there will be k-1 elements
    # which are divisible by their position
    target = N - K
 
    # Initializing "prev" variable that helps in
    # shifting elements to their right
    prev = A[0]
 
    # Assigning first element the value at target
    # index
    A[0] = A[target]
 
    # Making all elements from index 1 to target
    # equal to their immediate left element
    # as any number would not be divisible
    # by its next number
    for i in range(1,target+1):
        temp = A[i]
        A[i] = prev
        prev = temp
 
    return A
 
# Driver Code
N = 6
K = 2
 
# Calling function
# to construct the array
A = constructArray(N, K)
 
# Printing resultant array
for i in range(N):
   print(A[i],end=" ")
 
# This code is contributed by shinjanpatra

C#

// C# program for above approach
using System;
class GFG {
 
  // Function to construct an array of size
  // N such that it contains all numbers from
  // 1 to N and only K elements are divisible by
  // their position (i.e. index+1)
  static int[] constructArray(int N, int K)
  {
 
    // Declaring array of size N
    int[] A = new int[N];
    for (int i = 0; i < A.Length; ++i) {
      A[i] = 0;
    }
 
    // Initializing array as {1, 2, 3....N}
    for (int i = 0; i < N; i++) {
      A[i] = i + 1;
    }
 
    // N-K index stored in a variable "target"
    // After target there will be k-1 elements
    // which are divisible by their position
    int target = N - K;
 
    // Initializing "prev" variable that helps in
    // shifting elements to their right
    int prev = A[0];
 
    // Assigning first element the value at target
    // index
    A[0] = A[target];
 
    // Making all elements from index 1 to target
    // equal to their immediate left element
    // as any number would not be divisible
    // by its next number
    for (int i = 1; i <= target; i++) {
      int temp = A[i];
      A[i] = prev;
      prev = temp;
    }
 
    return A;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 6, K = 2;
 
    // Calling function
    // to construct the array
    int[] A = constructArray(N, K);
 
    // Printing resultant array
    for (int i = 0; i < N; i++)
      Console.Write(A[i] + " ");
    Console.WriteLine();
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach
 
    // Function to construct an array of size
    // N such that it contains all numbers from
    // 1 to N and only K elements are divisible by
    // their position (i.e. index+1)
    const constructArray = (N, K) => {
        // Declaring array of size N
        let A = new Array(N).fill(0);
 
        // Initializing array as {1, 2, 3....N}
        for (let i = 0; i < N; i++) {
            A[i] = i + 1;
        }
 
        // N-K index stored in a variable "target"
        // After target there will be k-1 elements
        // which are divisible by their position
        let target = N - K;
 
        // Initializing "prev" variable that helps in
        // shifting elements to their right
        let prev = A[0];
 
        // Assigning first element the value at target
        // index
        A[0] = A[target];
 
        // Making all elements from index 1 to target
        // equal to their immediate left element
        // as any number would not be divisible
        // by its next number
        for (let i = 1; i <= target; i++) {
            let temp = A[i];
            A[i] = prev;
            prev = temp;
        }
 
        return A;
    }
 
    // Driver Code
 
    let N = 6, K = 2;
 
    // Calling function
    // to construct the array
    let A = constructArray(N, K);
 
    // Printing resultant array
    for (let i = 0; i < N; i++)
        document.write(`${A[i]} `);
 
// This code is contributed by rakeshsahni
 

输出

5 1 2 3 4 6 

时间复杂度： O(N)
辅助空间： O(N)