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74 lines
2.2 KiB
Markdown
74 lines
2.2 KiB
Markdown
# [1518. Water Bottles](https://leetcode.com/problems/water-bottles/)
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## 题目
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Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.
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The operation of drinking a full water bottle turns it into an empty bottle.
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Return the maximum number of water bottles you can drink.
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**Example 1**:
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Input: numBottles = 9, numExchange = 3
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Output: 13
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Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
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Number of water bottles you can drink: 9 + 3 + 1 = 13.
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**Example 2**:
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Input: numBottles = 15, numExchange = 4
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Output: 19
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Explanation: You can exchange 4 empty bottles to get 1 full water bottle.
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Number of water bottles you can drink: 15 + 3 + 1 = 19.
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**Example 3**:
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Input: numBottles = 5, numExchange = 5
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Output: 6
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**Example 4**:
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Input: numBottles = 2, numExchange = 3
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Output: 2
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**Constraints:**
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- 1 <= numBottles <= 100
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- 2 <= numExchange <= 100
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## 题目大意
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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。
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如果喝掉了酒瓶中的酒,那么酒瓶就会变成空的。
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请你计算 最多 能喝到多少瓶酒。
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## 解题思路
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- 模拟。首先我们一定可以喝到 numBottles 瓶酒,剩下 numBottles 个空瓶。接下来我们可以拿空瓶子换酒,每次拿出 numExchange 个瓶子换一瓶酒,然后再喝完这瓶酒,得到一个空瓶。这样模拟下去,直到所有的空瓶子小于numExchange结束。
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## 代码
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```go
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package leetcode
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func numWaterBottles(numBottles int, numExchange int) int {
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if numBottles < numExchange {
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return numBottles
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}
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quotient := numBottles / numExchange
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reminder := numBottles % numExchange
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ans := numBottles + quotient
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for quotient+reminder >= numExchange {
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quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
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ans += quotient
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}
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return ans
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}
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``` |