1518. 换酒问题

小区便利店正在促销，用 numExchange 个空酒瓶可以兑换一瓶新酒。你购入了 numBottles 瓶酒。

如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。

请你计算 最多 能喝到多少瓶酒。

示例 1：

输入：numBottles = 9, numExchange = 3
输出：13
解释：你可以用 3 个空酒瓶兑换 1 瓶酒。
所以最多能喝到 9 + 3 + 1 = 13 瓶酒。
示例 2：

输入：numBottles = 15, numExchange = 4
输出：19
解释：你可以用 4 个空酒瓶兑换 1 瓶酒。
所以最多能喝到 15 + 3 + 1 = 19 瓶酒。
示例 3：

输入：numBottles = 5, numExchange = 5
输出：6
示例 4：

输入：numBottles = 2, numExchange = 3
输出：2
 

提示：

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

来源：力扣（LeetCode）
链接：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