将给定的数字分成两个偶数部分

给定一个数字N ，任务是检查这个数字是否可以分成 2 个偶数部分。

例子：

Input: N = 8
Output: YES
Explanation: 8 can be divided into two even parts in two ways, 2, 6 or 4, 4 as both are even.

Input: N = 5
Output: NO

天真的方法：检查直到N的所有数字对，使得它们都是偶数并且它们的总和都为N 。如果可能，打印 Yes，否则打印 No。

时间复杂度： O(N ² )
辅助空间： O(1)

有效方法：观察可以看出，任何偶数都可以表示为除2之外的两个偶数之和，并且没有两个偶数相加可以形成奇数。

下面是上述方法的实现。

C++

// C++ code to implement above approach
#include 
using namespace std;
  
// Divide number into 2 even parts
bool divNum(int n)
{
    return (n <= 2
                ? false
                : (n % 2 == 0
                       ? true
                       : false));
}
  
// Driven Program
int main()
{
    int n = 8;
    cout << (divNum(n) ? "YES" : "NO")
         << endl;
    return 0;
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {
  
  // Divide number into 2 even parts
  static boolean divNum(int n)
  {
    return (n <= 2 ? false
            : (n % 2 == 0 ? true : false));
  }
  
  // Driven Program
  public static void main(String args[])
  {
    int n = 8;
    System.out.println(divNum(n) ? "YES" : "NO");
  }
}
  
// This code is contributed by Samim Hossain Mondal.

Python

# Pyhton program for above approach
  
# Divide number into 2 even parts
def divNum(n):
    ans = False if n <= 2 else True if n % 2 == 0 else False                       
    return ans
  
# Driver Code
n = 8
  
if(divNum(n) == True):
    print("YES")
else:
    print("NO")
  
# This code is contributed by Samim Hossain Mondal.

C#

// C# code to implement above approach
using System;
class GFG {
  
    // Divide number into 2 even parts
    static bool divNum(int n)
    {
        return (n <= 2 ? false
                       : (n % 2 == 0 ? true : false));
    }
  
    // Driven Program
    public static void Main()
    {
        int n = 8;
        Console.WriteLine(divNum(n) ? "YES" : "NO");
    }
}
  
// This code is contributed by ukasp.

Javascript

// Javascript code to implement above approach
  
// Divide number into 2 even parts
function divNum(n)
{
    return (n <= 2
                ? false
                : (n % 2 == 0
                       ? true
                       : false));
}
  
// Driven Program
let n = 8;
document.write((divNum(n) ? "YES" : "NO") + "\n");
  
// This code is contributed by Samim Hossin Mondal.

输出

YES

时间复杂度： O(1)
辅助空间： O(1)