找出将数字分为四部分的方式数量，使得a = c和b = d

给定数字N。找出将数字分为四个部分(a，b，c，d)的方式数，使得a = c和b = d且a不等于b。

例子：

Input : N = 6
Output : 1
Expalantion : four parts are {1, 2, 1, 2}

Input : N = 20
Output : 4
Expalantion : possible ways are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}.

方法 ：
如果给定的N为奇数，则答案为0，因为四部分之和将不是偶数。

如果n被4整除，则答案将为n / 4 -1(此处a等于b是可能的，因此减去该可能性)，否则为n / 4。

下面是上述方法的实现：

C++

// C++ implementation for above approach
#include 
using namespace std;
  
// Function to find the number of ways to divide
// N into four parts such that a = c and b = d
int possibleways(int n)
{
    if (n % 2 == 1)
        return 0;
    else if (n % 4 == 0)
        return n / 4 - 1;
    else
        return n / 4;
}
  
// Driver code
int main()
{
    int n = 20;
    cout << possibleways(n);
    return 0;
}

Java

// Java implementation for above approach
class GFG
{
      
// Function to find the number of ways to divide
// N into four parts such that a = c and b = d
static int possibleways(int n)
{
    if (n % 2 == 1)
        return 0;
    else if (n % 4 == 0)
        return n / 4 - 1;
    else
        return n / 4;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 20;
    System.out.println(possibleways(n));
}
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 implementation for above approach
  
# Function to find the number of ways 
# to divide N into four parts 
# such that a = c and b = d
def possibleways(n):
  
    if (n % 2 == 1):
        return 0;
    elif (n % 4 == 0):
        return n // 4 - 1;
    else:
        return n // 4;
  
# Driver code
n = 20;
print(possibleways(n));
  
# This code is contributed by mits

C#

// C# implementation for above approach
class GFG
{
      
// Function to find the number of ways to divide
// N into four parts such that a = c and b = d
static int possibleways(int n)
{
    if (n % 2 == 1)
        return 0;
    else if (n % 4 == 0)
        return n / 4 - 1;
    else
        return n / 4;
}
  
// Driver code
static void Main()
{
    int n = 20;
    System.Console.WriteLine(possibleways(n));
}
}
  
// This code is contributed by mits

输出：

4