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English Version

题目描述

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

解法

直接模拟空瓶兑新酒即可。

Python3

class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        ans = numBottles
        while numBottles >= numExchange:
            numBottles -= (numExchange - 1)
            ans += 1
        return ans

Java

class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int ans = numBottles;
        while (numBottles >= numExchange) {
            numBottles -= (numExchange - 1);
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numWaterBottles(int numBottles, int numExchange) {
        int ans = numBottles;
        while (numBottles >= numExchange)
        {
            numBottles -= (numExchange - 1);
            ++ans;
        }
        return ans;
    }
};

Go

func numWaterBottles(numBottles int, numExchange int) int {
	ans := numBottles
	for numBottles >= numExchange {
		numBottles -= (numExchange - 1)
		ans++
	}
	return ans
}

...