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Copy path1518.换酒问题.py
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1518.换酒问题.py
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#
# @lc app=leetcode.cn id=1518 lang=python3
#
# [1518] 换酒问题
#
# https://leetcode.cn/problems/water-bottles/description/
#
# algorithms
# Easy (70.23%)
# Likes: 138
# Dislikes: 0
# Total Accepted: 56.2K
# Total Submissions: 80.1K
# Testcase Example: '9\n3'
#
# 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
#
#
#
# @lc code=start
class Solution:
def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
count = numBottles
left_bottle = numBottles
while left_bottle >= numExchange:
can = left_bottle//numExchange
count+=can
left_bottle = left_bottle%numExchange+can
return count
# @lc code=end