所有顶点的度数之和为L的树数

给定一个整数L ，它是某棵树的所有顶点的度数之和。任务是找到所有这些不同的树(标记为树)的计数。如果两棵树至少具有一个不同的边缘，则它们是不同的。

例子：

Input: L = 2
Output: 1

Input: L = 6
Output: 16

一个简单的解决方案：一个简单的解决方案是找到所有顶点的度数之和为L的树的节点数。如本文所述，这种树中的节点数为n =(L / 2 + 1) 。
现在的解决方案是形成可以使用n个节点形成的所有标记树。这种方法非常复杂，并且对于较大的n值，无法使用此过程找出树木的数量。

高效的解决方案：高效的解决方案是使用Cayley公式来计算节点数，该公式指出存在n ^{(n – 2)}棵树，其中n个标注了顶点。因此，现在代码的时间复杂度降低为O(n) ，可以使用模块化幂运算将其进一步降低为O(logn) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
  
// Iterative Function to calculate (x^y) in O(log y)
ll power(int x, ll y)
{
  
    // Initialize result
    ll res = 1;
  
    while (y > 0) {
  
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x);
  
        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}
  
// Function to return the count 
// of required trees
ll solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;
  
    ll ans = power(n, n - 2);
  
    // Return the result
    return ans;
}
  
// Driver code
int main()
{
    int L = 6;
  
    cout << solve(L);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG
{
      
// Iterative Function to calculate (x^y) in O(log y)
static long power(int x, long y)
{
  
    // Initialize result
    long res = 1;
  
    while (y > 0)
    {
  
        // If y is odd, multiply x with result
        if (y==1)
            res = (res * x);
  
        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}
  
// Function to return the count 
// of required trees
static long solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;
  
    long ans = power(n, n - 2);
  
    // Return the result
    return ans;
}
  
// Driver code
public static void main (String[] args)
{
  
    int L = 6;
    System.out.println (solve(L));
}
}
  
// This code is contributed by ajit.

Python3

# Python implementation of the approach
  
# Iterative Function to calculate (x^y) in O(log y)
def power(x, y):
  
    # Initialize result
    res = 1;
  
    while (y > 0):
  
        # If y is odd, multiply x with result
        if (y %2== 1):
            res = (res * x);
  
        # y must be even now
        #y = y / 2
        y = int(y) >> 1;
        x = (x * x);
    return res;
  
  
# Function to return the count 
# of required trees
def solve(L):
      
    # number of nodes
    n = L / 2 + 1;
  
    ans = power(n, n - 2);
  
    # Return the result
    return int(ans);
  
L = 6;
print(solve(L));
  
# This code has been contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
// Iterative Function to calculate (x^y) in O(log y)
static long power(int x, long y)
{
  
    // Initialize result
    long res = 1;
  
    while (y > 0)
    {
  
        // If y is odd, multiply x with result
        if (y == 1)
            res = (res * x);
  
        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}
  
// Function to return the count 
// of required trees
static long solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;
  
    long ans = power(n, n - 2);
  
    // Return the result
    return ans;
}
  
// Driver code
static public void Main ()
{
    int L = 6;
    Console.WriteLine(solve(L));
}
}
  
// This code is contributed by Tushil.

输出：

16