human
I found an interesting problem on https://leetcode.com/problems/clumsy-factorial/:
Normally, the factorial of a positive integer n is the product of all positive integers less than or equal to n.  For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.
We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
For example, clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1.  However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.
Additionally, the division that we use is floor division such that 10 * 9 / 8 equals 11.  This guarantees the result is an integer.
Implement the clumsy function as defined above: given an integer N, it returns the clumsy factorial of N.
 
Example 1:
Input: 4
Output: 7
Explanation: 7 = 4 * 3 / 2 + 1

Example 2:
Input: 10
Output: 12
Explanation: 12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1

 
Note:

1 <= N <= 10000
-2^31 <= answer <= 2^31 - 1  (The answer is guaranteed to fit within a 32-bit integer.)

I tried it in Python, but could not do it. Can you solve it?

 Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def clumsy(self, N: int) -> int:
        
        res = 0
        for i in range(N, 0, -4):
            if i >= 4:
                if i == N:
                    res += i * (i-1) // (i-2) + (i-3)
                else:
                    res -= i * (i-1) // (i-2) - (i-3)
            #print(res)
        
        neg = [1, -1][int(N >= 4)]
        
        if N % 4 == 3:
            res += 6 * neg
        elif N % 4 == 2:
            res += 2 * neg
        elif N % 4 == 1:
            res += 1 * neg
        return int(res)