human
Solve in Python:
Implement the class ProductOfNumbers that supports two methods:
1. add(int num)

Adds the number num to the back of the current list of numbers.

2. getProduct(int k)

Returns the product of the last k numbers in the current list.
You can assume that always the current list has at least k numbers.

At any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.
 
Example:
Input
["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]

Output
[null,null,null,null,null,null,20,40,0,null,32]

Explanation
ProductOfNumbers productOfNumbers = new ProductOfNumbers();
productOfNumbers.add(3).
       // [3]
productOfNumbers.add(0).
       // [3,0]
productOfNumbers.add(2).
       // [3,0,2]
productOfNumbers.add(5).
       // [3,0,2,5]
productOfNumbers.add(4).
       // [3,0,2,5,4]
productOfNumbers.getProduct(2).
// return 20. The product of the last 2 numbers is 5 * 4 = 20
productOfNumbers.getProduct(3).
// return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
productOfNumbers.getProduct(4).
// return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
productOfNumbers.add(8).
       // [3,0,2,5,4,8]
productOfNumbers.getProduct(2).
// return 32. The product of the last 2 numbers is 4 * 8 = 32 

 
Constraints:

There will be at most 40000 operations considering both add and getProduct.
0 <= num <= 100
1 <= k <= 40000
assistant
#Here is the Python program you requested.

class ProductOfNumbers:

    def __init__(self):
        self.products = []
        

    def add(self, num: int) -> None:
        
        if num == 0:
            self.products = []    
        else:
            self.products.append(num if not len(self.products) else self.products[-1] * num)
           
    def getProduct(self, k: int) -> int:
        if k > len(self.products):
            return 0
        elif k == len(self.products):
            return self.products[-1]
        else:
            return self.products[-1] // self.products[-k-1]

# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)