在给定的操作之后打印 N 个不同的数字

给定一个始终是偶数的整数 N ，我们的任务是按照给定的条件打印N 个不同的数字：

• 前半数为偶数，后半数为奇数
• 前半数元素之和与后半数元素之和应相等

如果上述条件满足，则打印数组，否则输出“-1”。
例子：

Input: N = 4 
Output: 2 4 1 5 
Explanation: 
Given number 4 we are required to print 4 numbers. First Half = 2, 4 and their sum is 6, other half = 1, 5 and their sum is also 6.
Input: N = 22 
Output: -1 
Explanation: 
It is not possible to print the required array. 
 

方法：
为了解决上面提到的问题，我们必须观察到整数 N 必须是 4 的倍数。

• 我们知道前 N/2 个偶数的和是偶数，所以如果其他 N/2 个整数的和也是偶数，那么 N/2 一定是偶数，因为奇数个奇数的和是总是奇怪。
• 如果 N/2 是偶数，那么 N 是 4 的倍数，所以如果 n 不能被 4 整除，那么答案是“-1”，否则，就会有一个可能的数组。
• 为了打印数组，我们将考虑两部分，即 N/2 个元素的前半部分将是 2 的倍数，另一半将是 2 – 1 的倍数。对于数组中的最后一个元素，我们将计算通过应用直接公式 N + N / 2 – 1 来计算整数，因为我们应该使两半的和相等。

下面是上述方法的实现：

C++

// C++ implementation to Print N distinct numbers
#include 
using namespace std;
 
// Function to print the required array
bool printArr(int n)
{
 
    // Check if number is a multiple of 4
    if (n % 4 == 0) {
        // Printing Left Half of the array
        for (int i = 1; i <= n / 2; i++)
            cout << i * 2 << ' ';
 
        // Printing Right Half of the array
        for (int i = 1; i < n / 2; i++)
            cout << i * 2 - 1 << ' ';
        cout << n + n / 2 - 1 << '\n';
    }
 
    else
        cout << "-1";
}
 
// Driver code
int main()
{
    int n = 22;
 
    printArr(n);
 
    return 0;
}

Java

// Java implementation to print N distinct numbers
import java.util.*;
 
class GFG{
 
// Function to print the required array
static void printArr(int n)
{
 
    // Check if number is a multiple of 4
    if(n % 4 == 0)
    {
       // Printing left half of the array
       for(int i = 1; i <= n / 2; i++)
           System.out.print(i * 2 + " ");
 
       // Printing Right Half of the array
       for(int i = 1; i < n / 2; i++)
           System.out.print(i * 2 - 1 + " ");
 
       System.out.println(n + n / 2 - 1);
    }
    else
        System.out.print("-1");
}
 
// Driver code
public static void main(String[] args)
{
    int n = 22;
    printArr(n);
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to print
# N distinct numbers
 
# Function to print the required array
def printArr(n):
 
    # Check if number is a multiple of 4
    if (n % 4 == 0):
         
        # Printing Left Half of the array
        for i in range(1, (n / 2) + 1):
            print (i * 2, end = " ")
 
        # Printing Right Half of the array
        for i in range(1, n / 2):
            print (i * 2 - 1, end = " ")
             
        print (n + n / 2 - 1, end = "\n")
 
    else:
        print ("-1")
 
# Driver code
n = 22
printArr(n)
 
# This code is contributed by PratikBasu

C#

// C# implementation to print N distinct numbers
using System;
 
public class GFG{
 
// Function to print the required array
static void printArr(int n)
{
 
    // Check if number is a multiple of 4
    if(n % 4 == 0)
    {
         
    // Printing left half of the array
    for(int i = 1; i <= n / 2; i++)
        Console.Write(i * 2 + " ");
 
    // Printing Right Half of the array
    for(int i = 1; i < n / 2; i++)
        Console.Write(i * 2 - 1 + " ");
 
    Console.WriteLine(n + n / 2 - 1);
    }
     
    else
        Console.Write("-1");
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 22;
    printArr(n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation to Print N distinct numbers
 
// Function to print the required array
function printArr(n)
{
 
    // Check if number is a multiple of 4
    if (n % 4 == 0) {
        // Printing Left Half of the array
        for (var i = 1; i <= n / 2; i++)
            document.write( i * 2 + ' ');
 
        // Printing Right Half of the array
        for (var i = 1; i < n / 2; i++)
            document.write( i * 2 - 1 + ' ');
        document.write( n + n / 2 - 1 + '<br>');
    }
 
    else
        document.write( "-1");
}
 
// Driver code
var n = 22;
printArr(n);
 

输出：

-1

时间复杂度： O(N)

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