从前N个数字计算对与相邻元素的对的可能组合

给定数字N，任务是计算使用相邻元素形成的对的所有可能组合。

注意：如果一个元素对已经存在，则不能在下一个元素对中进行选择。例如：对于{1,2,3}：{1,2}和{2,3}将不被视为正确的组合。

例子：

Input : N = 4
Output : 5
Explanation : If N = 4, the possible combinations are:
{1}, {2}, {3}, {4}
{1, 2}, {3, 4}
{1}, {2, 3}, {4}
{1}, {2}, {3, 4}
{1, 2}, {3}, {4}

Input : N = 5
Output : 8

方法：将问题分解为较小的子问题。如果有N个数字，并且有两种情况，一个数字是单独的，或者是成对的；如果一个数字是单独的，请找出将剩余的(n-1)个数字配对的方法，或者是成对的，找到剩下的配对(n-2)个数字的方式。如果只剩下2个数字，则它们可以单独或成对生成2个组合，如果剩下一个数字，则将是单例，因此只有1个组合。

下面是上述方法的实现：

C++

#include 
using namespace std;
// Function to count the number of ways
int ways(int n)
{
    // If there is a single number left 
    // it will form singleton
    if (n == 1) {
        return 1;
    }
    // if there are just 2 numbers left, 
    // they will form a pair
    if (n == 2) {
        return 2;
    }
    else {
        return ways(n - 1) + ways(n - 2);
    }
}
  
// Driver Code
int main()
{
    int n = 5;
  
    cout << "Number of ways = " << ways(n);
  
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
  
class GFG
{
      
// Function to count the number of ways
static int ways(int n)
{
    // If there is a single number left 
    // it will form singleton
    if (n == 1) 
    {
        return 1;
    }
      
    // if there are just 2 numbers left, 
    // they will form a pair
    if (n == 2) 
    {
        return 2;
    }
    else
    {
        return ways(n - 1) + ways(n - 2);
    }
}
  
// Driver Code
public static void main (String[] args) 
{
    int n = 5;
      
    System.out.println("Number of ways = " + ways(n));
}
}

Python3

# Python3 code implementation of the above program
  
# Function to count the number of ways 
def ways(n) :
      
    # If there is a single number left 
    # it will form singleton 
    if (n == 1) :
        return 1; 
      
    # if there are just 2 numbers left, 
    # they will form a pair 
    if (n == 2) :
        return 2; 
      
    else :
        return ways(n - 1) + ways(n - 2); 
  
# Driver Code 
if __name__ == "__main__" : 
  
    n = 5; 
  
    print("Number of ways = ", ways(n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above code
using System;
  
class GFG 
{ 
      
// Function to count the number of ways 
static int ways(int n) 
{ 
    // If there is a single number left 
    // it will form singleton 
    if (n == 1) 
    { 
        return 1; 
    } 
      
    // if there are just 2 numbers left, 
    // they will form a pair 
    if (n == 2) 
    { 
        return 2; 
    } 
    else
    { 
        return ways(n - 1) + ways(n - 2); 
    } 
} 
  
// Driver Code 
public static void Main() 
{ 
    int n = 5; 
      
    Console.WriteLine("Number of ways = " + ways(n)); 
} 
} 
  
// This code is contributed by AnkitRai01

输出：

Number of ways = 8