从N面多边形中找到与顶点M对角线对角的顶点

给定两个整数N和M ，任务是找到与N边多边形的^第M^个顶点对角线对角的顶点。

例子：

Input: N = 6, M = 2 
Output: 5 
Explanation:

It can be observed from the image above that the vertex opposite to vertex 5 is 2.

Input: N = 8, M = 5 
Output: 1 
Explanation:

It can be observed from the image above that the vertex opposite to vertex 8 is 1.

方法：要解决给定的问题，需要考虑以下两种情况：

1. 如果M> N / 2：顶点将始终为M —(N / 2) 。
2. 如果M≤N / 2：顶点将始终为M +(N / 2) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to return the
// required vertex
int getPosition(int N, int M)
{
 
    // Case 1:
    if (M > (N / 2)) {
        return (M - (N / 2));
    }
 
    // Case 2:
    return (M + (N / 2));
}
 
// Driver Code
int main()
{
    int N = 8, M = 5;
    cout << getPosition(N, M);
 
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
// Function to return the
// required vertex
static int getPosition(int N,
                       int M)
{
  // Case 1:
  if (M > (N / 2))
  {
    return (M - (N / 2));
  }
 
  // Case 2:
  return (M + (N / 2));
}
 
// Driver Code
public static void main(String[] args)
{
  int N = 8, M = 5;
  System.out.print(getPosition(N, M));
 
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the
# above approach
 
# Function to return the
# required vertex
def getPosition(N, M):
     
  # Case 1:
  if (M > (N // 2)):
    return (M - (N // 2))
    
  # Case 2:
  return (M + (N // 2))
   
# Driver Code
N = 8
M = 5
 
print(getPosition(N, M))
 
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to return the
// required vertex
static int getPosition(int N, int M)
{
     
    // Case 1:
    if (M > (N / 2))
    {
        return (M - (N / 2));
    }
     
    // Case 2:
    return (M + (N / 2));
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 8, M = 5;
     
    Console.Write(getPosition(N, M));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
 
// Javascript program for the above approach
 
// Function to return the
// required vertex
function getPosition(N, M)
{
 
    // Case 1:
    if (M > parseInt(N / 2)) {
        return (M - parseInt(N / 2));
    }
 
    // Case 2:
    return (M + parseInt(N / 2));
}
 
// Driver Code
var N = 8, M = 5;
document.write(getPosition(N, M));
 

输出：

1

时间复杂度： O(1)
辅助空间： O(1)