将一个数分成两个不相等的偶数部分

给定一个正整数N 。任务是决定整数是否可以分成两个不相等的正偶数部分。

例子：

Input: N = 8
Output: YES
Explanation: 8 can be divided into two different even parts i.e. 2 and 6.

Input: N = 5
Output: NO
Explanation: 5 can not be divided into two even parts in any way.

Input: N = 4
Output: NO
Explanation: 4 can be divided into two even parts, 2 and 2. Since the numbers are equal, the output is NO.

编程需要懂一点英语

先决条件：了解 if-else 条件语句。

方法：问题的核心概念在于以下观察：

The sum of any two even numbers is always even. Conversely any even number can be expressed as sum of two even numbers.

编程需要懂一点英语

但这里有两个例外

数字 2 在这里是个例外。它只能表示为两个奇数的和（1 + 1）。
数字 4 只能表示为相等偶数 (2 + 2) 的总和。

因此，只有当N 为偶数且不等于 2 或 4时，才有可能将N表示为两个偶数之和。如果 N 是奇数，则不可能将其分成两个偶数部分。请按照以下步骤操作：

检查 N = 2 还是 N = 4。
如果是，则打印 NO。
否则检查 N 是否为偶数（即 2 的倍数）
如果是，则打印 YES。
否则，打印 NO。

下面是上述方法的实现。

C++

// C++ code to implement above approach
#include
using namespace std;
  
// Function to check if N can be divided 
// into two unequal even parts
bool evenParts(int N)
{   
    // Check if N is equal to 2 or 4  
    if(N == 2 || N == 4)
        return false;
  
    // Check if N is even
    if(N % 2 == 0)
        return true;
    else
        return false;
}
   
//Driver code
int main(){
   int N = 8;
      
   // Function call
   bool ans = evenParts(N);
  
   if(ans)
       std::cout << "YES" << '\n';
   else
       std::cout << "NO" << '\n';
   
   return 0;
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {
  
  // Function to check if N can be divided 
  // into two unequal even parts
  static boolean evenParts(int N)
  {   
  
    // Check if N is equal to 2 or 4  
    if(N == 2 || N == 4)
      return false;
  
    // Check if N is even
    if(N % 2 == 0)
      return true;
    else
      return false;
  }
  
  // Driver code
  public static void main(String args[])
  {
    int N = 8;
  
    // Function call
    boolean ans = evenParts(N);
  
    if(ans)
      System.out.println("YES");
    else
      System.out.println("NO");
  
  }
}
  
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
  
# Function to check if N can be divided
# into two unequal even parts
def evenParts(N):
  
    # Check if N is equal to 2 or 4
    if (N == 2 or N == 4):
        return False
  
    # Check if N is even
    if (N % 2 == 0):
        return True
    else:
        return False
  
# Driver code
N = 8
  
# Function call
ans = evenParts(N)
if (ans):
    print("YES")
else:
    print("NO")
  
# This code is contributed by Saurabh Jaiswal.

C#

// C# code to implement above approach
using System;
class GFG {
  
  // Function to check if N can be divided 
  // into two unequal even parts
  static bool evenParts(int N)
  {   
      
    // Check if N is equal to 2 or 4  
    if(N == 2 || N == 4)
      return false;
  
    // Check if N is even
    if(N % 2 == 0)
      return true;
    else
      return false;
  }
  
  // Driver code
  public static void Main()
  {
    int N = 8;
  
    // Function call
    bool ans = evenParts(N);
  
    if(ans)
      Console.Write("YES" + '\n');
    else
      Console.Write("NO" + '\n');
  
  }
}
  
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

YES

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