从特定节点开始的大小为k的循环数

给定两个正整数n，k 。考虑一个完整连接图中n个节点的无向完整连接图。任务是计算可以从任何节点开始并通过访问K个节点返回该节点的方式的数目。
例子：

Input : n = 3, k = 3
Output : 2

Input : n = 4, k = 2
Output : 3

令f(n，k)为一个函数，该函数返回可以从任何节点开始并通过访问K个节点返回的方式的数目。
如果我们从一个节点开始和结束，那么对于中间节点，我们将有K – 1个选择，因为我们已经在开始时选择了一个节点。对于每个中间选择，您都有n – 1个选项。因此，这将产生(n – 1) ^{k – 1，}但是随后我们必须删除所有导致较小循环的选择，因此我们减去f(n，k – 1) 。
因此，递归关系变为
f(n，k)=(n – 1) ^{k – 1} – f(n，k – 1)，基本情况为f(n，2)= n – 1。
在扩展时，
f(n，k)=(n – 1) ^{k – 1} –(n – 1) ^{k – 2} +(n – 1) ^{k – 3} …..(n – 1)(例如等式1)
将f(n，k)除以(n – 1)，
f(n，k)/(n – 1)=(n – 1) ^{k – 2} –(n – 1) ^{k – 3} +(n – 1) ^{k – 4} ….. 1(等式2)
在添加eqn 1和eqn 2时
f(n，k)+ f(n，k)/(n – 1)=(n – 1) ^{k – 1} +( ^-1 ) ^k
f(n，k)*((n -1)+1)/(n – 1)=(n – 1) ^{k – 1} +( ^-1 ) ^k
$f(n, k) = \frac{(n-1)^{k} + (-1)^{k}(n-1)}{n}$
下面是此方法的实现：

C++

// C++ Program to find number of cycles of length
// k in a graph with n nodes.
#include 
using namespace std;
 
// Return the Number of ways from a
// node to make a loop of size K in undirected
// complete connected graph of N nodes
int numOfways(int n, int k)
{
    int p = 1;
 
    if (k % 2)
        p = -1;
 
    return (pow(n - 1, k) + p * (n - 1)) / n;
}
 
// Driven Program
int main()
{
    int n = 4, k = 2;
    cout << numOfways(n, k) << endl;
    return 0;
}

Java

// Java Program to find number of
// cycles of length k in a graph
// with n nodes.
public class GFG {
     
    // Return the Number of ways
    // from a node to make a loop
    // of size K in undirected
    // complete connected graph of
    // N nodes
    static int numOfways(int n, int k)
    {
        int p = 1;
     
        if (k % 2 != 0)
            p = -1;
     
        return (int)(Math.pow(n - 1, k)
                    + p * (n - 1)) / n;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 4, k = 2;
     
        System.out.println(numOfways(n, k));
    }
}
 
// This code is contributed by Sam007.

Python3

# python Program to find number of
# cycles of length k in a graph
# with n nodes.
 
# Return the Number of ways from a
# node to make a loop of size K in
# undirected complete connected
# graph of N nodes
def numOfways(n,k):
     
    p = 1
 
    if (k % 2):
        p = -1
 
    return (pow(n - 1, k) +
                   p * (n - 1)) / n
 
# Driver code
n = 4
k = 2
print (numOfways(n, k))
 
# This code is contributed by Sam007.

C#

// C# Program to find number of cycles
// of length k in a graph with n nodes.
using System;
 
class GFG {
     
    // Return the Number of ways from
    // a node to make a loop of size
    // K in undirected complete
    // connected graph of N nodes
    static int numOfways(int n, int k)
    {
        int p = 1;
     
        if (k % 2 != 0)
            p = -1;
     
        return (int)(Math.Pow(n - 1, k)
                     + p * (n - 1)) / n;
    }
     
    // Driver code
    static void Main()
    {
        int n = 4, k = 2;
         
        Console.Write( numOfways(n, k) );
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// JavaScript Program to find number of
// cycles of length k in a graph
// with n nodes.
 
    // Return the Number of ways
    // from a node to make a loop
    // of size K in undirected
    // complete connected graph of
    // N nodes
    function numOfways(n, k)
    {
        let p = 1;
       
        if (k % 2 != 0)
            p = -1;
       
        return (Math.pow(n - 1, k)
                    + p * (n - 1)) / n;
    }
  
// Driver code
         let n = 4, k = 2;
        document.write(numOfways(n, k));
     
    // This code is contributed by code_hunt.

输出：

3