小區便利店正在促銷，用 numExchange 個空酒瓶可以兌換一瓶新酒。你購入了 numBottles 瓶酒。

如果喝掉了酒瓶中的酒，那麼酒瓶就會變成空的。

請你計算 最多 能喝到多少瓶酒。

示例 1：

輸入：numBottles = 9, numExchange = 3
輸出：13
解釋：你可以用 3 個空酒瓶兌換 1 瓶酒。
所以最多能喝到 9 + 3 + 1 = 13 瓶酒。
示例 2：

輸入：numBottles = 15, numExchange = 4
輸出：19
解釋：你可以用 4 個空酒瓶兌換 1 瓶酒。
所以最多能喝到 15 + 3 + 1 = 19 瓶酒。
示例 3：

輸入：numBottles = 5, numExchange = 5
輸出：6
示例 4：

輸入：numBottles = 2, numExchange = 3
輸出：2

提示：

1 <=numBottles <= 100
2 <=numExchange <= 100

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/water-bottles

class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        if numBottles<numExchange:return numBottles
        
 
if numBottles==numExchange:return numBottles+1
        gain=numBottles//numExchange
        left=numBottles%numExchange
        res=gain*numExchange+gain+left
        while True:
            if gain+left>=numExchange:
                numBottles=gain+left
                gain=numBottles//numExchange
                left
 
=numBottles%numExchange
                res+=gain
            else:
                break
    
        return res

class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        if numBottles<numExchange:return numBottles
        left=numBottles%numExchange
        return numBottles-left+self.numWaterBottles(left+(numBottles-left)//numExchange,numExchange)

class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        gain=numBottles
        while gain//numExchange>0:
            numBottles+=gain//numExchange
            gain=gain//numExchange+gain%numExchange
        return numBottles