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# 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
["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`

实现两个API: add(num)在末尾添加一个数，productOfNumbers.getProduct(k)返回最近添加的k个元素的积。要求实时返回subarray性质 + 连乘 => presum套路。将乘上每次遇到的新数后把当前乘积存下来，后K个就是res\[-1] / res\[size - k + 1]。但乘法和加法不同的是遇到0后从后面再往前乘到超过0位置时乘积都会是0，也就是遇到0后要重置res。

```cpp
class ProductOfNumbers {
public:
    vector<int> nums;
    ProductOfNumbers() {
        nums = {1};
    }
    
    void add(int num) {
        if (num != 0) nums.push_back(nums.back() * num);
        else nums = {1};
    }
    
    int getProduct(int k) {
        return k >= nums.size() ? 0 : nums.back() / nums[nums.size() - k - 1];
    }
};

/**
 * Your ProductOfNumbers object will be instantiated and called as such:
 * ProductOfNumbers* obj = new ProductOfNumbers();
 * obj->add(num);
 * int param_2 = obj->getProduct(k);
 */
```