## 1352. Product of the Last K Numbers

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
[[],[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.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.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`

## Rust Solution

``````#[derive(Default)]
struct ProductOfNumbers {
prefix: Vec<i32>,
}

impl ProductOfNumbers {
fn new() -> Self {
let prefix = vec![1];
ProductOfNumbers { prefix }
}

fn add(&mut self, num: i32) {
if num > 0 {
let prev = self.prefix[self.prefix.len() - 1];
self.prefix.push(prev * num);
} else {
self.prefix = vec![1];
}
}

fn get_product(&self, k: i32) -> i32 {
let k = k as usize;
let n = self.prefix.len();
if k >= n {
0
} else {
self.prefix[n - 1] / self.prefix[n - 1 - k]
}
}
}

#[test]
fn test() {
let mut obj = ProductOfNumbers::new();