生成一个 N 大小的唯一元素数组，其中相邻对的 GCD 为 X

给定两个整数N （总是偶数）和X ，任务是找到一个大小为N且具有不同数字的数组，使得相邻对的GCD之和（其中每个元素仅是一对的一部分）为X 。如果没有这样的数组是可能的，则返回 -1。

注意：如果有超过 1 个可能的序列，则返回其中任何一个

例子 ：

Input: N = 6, X = 6
Output: 4 8 9 10 11 12
Explanation: Starting with first two numbers, gcd of (4, 8) = 4, then gcd of (9, 10) = 1, gcd of (11, 12) = 1.
Thus, gcd sums up to 4 + 1 + 1  = 6 which is equal to X.

Input: N = 4, X = 1
Output: -1

方法：这个想法是

If the value of X is less than N/2, it’s obvious that it’s not possible to make a minimum GCD of N/2. 
So, if X < N/2, answer = -1. 

Otherwise, lets denote extra  = X – N/2 + 1, which signifies how much is the difference between X and sum (sum = the sum of GCD when all the adjacent pairs have GCD = 1 except for the first one) i.e. this must be GCD of the first pair
Thus, based upon the above fact, extra + sum of GCD of other N/2-1 pairs would be equal to X. 

extra 的公式可以推导如下：

For N numbers there are a total of N/2 pairs.

If the first pair is left and all the other pairs are made to have a GCD of 1. 
So the total sum of GCDs of these pairs are (N/2 -1)*1 = (N/2 -1)

So extra = X – (N/2 – 1) = X – N/2 + 1

请按照以下步骤解决问题：

• 将第一对设为extra和(extra*2) 。
• 使其他对由连续整数组成，使得它们的 GCD = 1。

下面是上述方法的实现：

C++

// C++ Program of the above approach.
 
#include 
using namespace std;
 
// Function to get the
// Resultant array
void solve(int n, int x)
{
    // Formula generated to calculate
    // Extra gcd required
    int extra = x - n / 2 + 1;
 
    // Print extra required in the
    // Starting of the array sequence
    cout << extra << " ";
 
    // Multiplied starting element by 2
    // Such that after taking pair of both
    // Starting and next - starting
    // Element it will be equal to extra
    extra = extra * 2;
 
    // Printing the leftover pairs
    // Or elements with gcd = 1
    for (int i = 1; i < n; i++) {
        cout << (extra++) << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 4, X = 3;
    solve(N, X);
    return 0;
}

Java

// Java Program of the above approach.
import java.util.*;
public class GFG {
 
  // Function to get the
  // Resultant array
  static void solve(int n, int x)
  {
 
    // Formula generated to calculate
    // Extra gcd required
    int extra = x - n / 2 + 1;
 
    // Print extra required in the
    // Starting of the array sequence
    System.out.print(extra + " ");
 
    // Multiplied starting element by 2
    // Such that after taking pair of both
    // Starting and next - starting
    // Element it will be equal to extra
    extra = extra * 2;
 
    // Printing the leftover pairs
    // Or elements with gcd = 1
    for (int i = 1; i <= n; i++) {
      System.out.print((extra++) + " ");
    }
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 4, X = 3;
    solve(N, X);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to get the
# Resultant array
def solve( n,  x):
 
    # Formula generated to calculate
    # Extra gcd required
    extra = x - (int)(n / 2) + 1;
 
    # Print extra required in the
    # Starting of the array sequence
    print(extra, end= " ");
 
    # Multiplied starting element by 2
    # Such that after taking pair of both
    # Starting and next - starting
    # Element it will be equal to extra
    extra = extra * 2;
 
    # Printing the leftover pairs
    # Or elements with gcd = 1
    for i in range(1,n+1):
        print(extra, end = " ");
        extra = extra + 1;
 
# Driver Code
N = 4
X = 3
solve(N, X);
    
# This code is contributed by Potta Lokesh

C#

// C# Program of the above approach.
using System;
class GFG {
 
  // Function to get the
  // Resultant array
  static void solve(int n, int x)
  {
 
    // Formula generated to calculate
    // Extra gcd required
    int extra = x - n / 2 + 1;
 
    // Print extra required in the
    // Starting of the array sequence
    Console.Write(extra + " ");
 
    // Multiplied starting element by 2
    // Such that after taking pair of both
    // Starting and next - starting
    // Element it will be equal to extra
    extra = extra * 2;
 
    // Printing the leftover pairs
    // Or elements with gcd = 1
    for (int i = 1; i <= n; i++) {
      Console.Write((extra++) + " ");
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 4, X = 3;
    solve(N, X);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript Program of the above approach.
 
    // Function to get the
    // Resultant array
    const solve = (n, x) => {
     
        // Formula generated to calculate
        // Extra gcd required
        let extra = x - parseInt(n / 2) + 1;
 
        // Print extra required in the
        // Starting of the array sequence
        document.write(`${extra} `);
 
        // Multiplied starting element by 2
        // Such that after taking pair of both
        // Starting and next - starting
        // Element it will be equal to extra
        extra = extra * 2;
 
        // Printing the leftover pairs
        // Or elements with gcd = 1
        for (let i = 1; i < n; i++) {
            document.write(`${(extra++)} `);
        }
    }
 
    // Driver Code
    let N = 4, X = 3;
    solve(N, X);
 
// This code is contributed by rakeshsahni
 

输出

2 4 5 6 7 

时间复杂度： O(N)
空间复杂度： O(1)