构造一个不包含三元组(i，j，k)的前N个自然数的数组，使得a [i] + a [j] = 2 * a [k]，其中i <j <k

给定正整数N ，任务是使用前N个自然数构造一个数组a [] ，该自然数不包含满足a [k] * 2 = a [i] + a [j]的三元组(i，j，k)并且我 。

例子：

Input: N = 3 
Output: {2, 3, 1 } 
Explanation: 
Since no such triplet exists in the array satisfying the condition, the required output is { 2, 3, 1 }.

Input: N = 10 
Output: { 8, 4, 6, 10, 2, 7, 3, 5, 9, 1 }

方法：可以使用贪婪技术解决问题。请按照以下步骤解决问题：

• 递归地找到结果数组的前(N / 2)个元素和结果数组的最后(N / 2)个元素。
• 合并数组的两半，以使数组的前半部分包含偶数，而数组的后半部分包含奇数。
• 最后，打印结果数组。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
vector constructArray(int N)
{
 
    // Base case
    if (N == 1) {
 
        return { 1 };
    }
 
    // Stores the first half
    // of the array
    vector first
        = constructArray(N / 2);
 
    // Stores the last half
    // of the array
    vector last
        = constructArray(N - (N / 2));
 
    // Stores the merged array
    vector ans;
 
    // Insert even numbers
    for (auto e : first) {
 
        // Insert 2 * e
        ans.push_back(2 * e);
    }
 
    // Insert odd numbers
    for (auto o : last) {
 
        // Insert (2 * o - 1)
        ans.push_back((2 * o) - 1);
    }
 
    return ans;
}
 
// Function to print the resultant array
void printArray(vector ans, int N)
{
 
    // Print resultant array
    cout << "{ ";
    for (int i = 0; i < N; i++) {
 
        // Print current element
        cout << ans[i];
 
        // If i is not the last index
        // of the resultant array
        if (i != N - 1) {
            cout << ", ";
        }
    }
 
    cout << " }";
}
 
// Driver Code
int main()
{
    int N = 10;
 
    // Store the resultant array
    vector ans
        = constructArray(N);
 
    printArray(ans, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
static ArrayList constructArray(int N)
{
     
    // Base case
    if (N == 1)
    {
        ArrayList a = new ArrayList(1);
        a.add(1);
        return a;
    }
     
    // Stores the first half
    // of the array
    ArrayList first = new ArrayList(N);
    first = constructArray(N / 2);
     
    // Stores the last half
    // of the array
    ArrayList last = new ArrayList(N);
    last = constructArray(N - N / 2);
     
    ArrayList ans = new ArrayList(N);
     
    // Insert even numbers
    for(int i = 0; i < first.size(); i++)
    {
         
        // Insert 2 * first[i]
        ans.add(2 * first.get(i));
    }
     
    // Insert odd numbers
    for(int i = 0; i < last.size(); i++)
    {
         
        // Insert (2 * last[i] - 1)
        ans.add(2 * last.get(i) - 1);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
     
    ArrayList answer = new ArrayList(N);
    answer = constructArray(N);
     
    System.out.print("{");
    for(int i = 0; i < answer.size(); i++)
    {
        System.out.print(answer.get(i));
        System.out.print(", ");
    }
    System.out.print("}");
}   
}
 
// This code is contributed by koulick_sadhu

Python3

# Python3 program to implement
# the above approach
 
# Function to construct the array of size N
# that contains no such triplet satisfying
# the given conditions
def constructArray(N) :
 
    # Base case
    if (N == 1) :
        a = []
        a.append(1)
        return a;
 
    # Stores the first half
    # of the array
    first = constructArray(N // 2);
 
    # Stores the last half
    # of the array
    last = constructArray(N - (N // 2));
 
    # Stores the merged array
    ans = [];
 
    # Insert even numbers
    for e in first :
 
        # Insert 2 * e
        ans.append(2 * e);
 
    # Insert odd numbers
    for o in last:
 
        # Insert (2 * o - 1)
        ans.append((2 * o) - 1);
 
    return ans;
 
# Function to print the resultant array
def printArray(ans, N) :
 
    # Print resultant array
    print("{ ", end = "");
    for i in range(N) :
 
        # Print current element
        print(ans[i], end = "");
 
        # If i is not the last index
        # of the resultant array
        if (i != N - 1) :
            print(", ",end = "");
 
    print(" }", end = "");
 
# Driver Code
if __name__ == "__main__" :
    N = 10;
 
    # Store the resultant array
    ans = constructArray(N);
 
    printArray(ans, N);
 
       # This code is contributed by AnkThon

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
static List constructArray(int N)
{
     
    // Base case
    if (N == 1)
    {
        List a = new List(1);
        a.Add(1);
        return a;
    }
      
    // Stores the first half
    // of the array
    List first = new List();
    first = constructArray(N / 2);
      
    // Stores the last half
    // of the array
    List last = new List();
 
    last = constructArray(N - N / 2);
      
    List ans = new List();
      
    // Insert even numbers
    for(int i = 0; i < first.Count; i++)
    {
         
        // Insert 2 * first[i]
        ans.Add(2 * first[i]);
    }
      
    // Insert odd numbers
    for(int i = 0; i < last.Count; i++)
    {
         
        // Insert (2 * last[i] - 1)
        ans.Add(2 * last[i] - 1);
    }
    return ans;
}
  
// Driver code
public static void Main()
{
    int N = 10;
     
    List answer = new List(N);
    answer = constructArray(N);
      
    Console.Write("{");
    for(int i = 0; i < answer.Count; i++)
    {
        Console.Write(answer[i]);
        Console.Write(", ");
    }
    Console.Write("}");
}   
}
 
// This code is contributed by sanjoy_62

输出：

{ 8, 4, 6, 10, 2, 7, 3, 5, 9, 1 }

时间复杂度： O(N * log(N))
辅助空间： O(N)