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

Javascript

    // Javascript implementation of the above code
     
    // Function to count the number of ways
    function 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);
        }
    }
     
    let n = 5;
       
    document.write("Number of ways = " + ways(n));
 
// This code is contributed by suresh07.

输出：

Number of ways = 8

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