这是一个创建于 3343 天前的主题，其中的信息可能已经有所发展或是发生改变。
2 元钱一瓶啤酒， 4 个瓶子可以换 1 瓶啤酒， 2 个盖子可以换一瓶啤酒。问 10 元钱可以喝到多少批啤酒。这个用python怎么写出公式计算？
 |
1
aluo1 2015-10-31 15:30:11 +08:00
def how_many_beer(money):
# The price of the beer beer_price = 2 # The number of beer bought by the given money beer_bought = money / beer_price beer = beer_bought # Bottle and lid number is the same as the beer bought initially bottle = beer_bought lid = beer_bought # Check whether can trade in more beer get_more_beer = tradable(bottle, lid) while get_more_beer: # Get the trade-in result on_hand = trade_in(bottle, lid) # Get the number of trade-in beer more_beer = on_hand["beer"] # Beer drunk so far beer += more_beer bottle = on_hand["bottle"] + more_beer lid = on_hand["lid"] + more_beer get_more_beer = tradable(bottle, lid) return beer def tradable(bottle,lid): # How many bottle can trade in one beer bottle_to_beer = 4 # How many lid can trade in one beer lid_to_beer = 2 return (bottle / bottle_to_beer) | (lid/lid_to_beer) def trade_in(bottle, lid): # How many bottle can trade in one beer bottle_to_beer = 4 # How many lid can trade in one beer lid_to_beer = 2 beer_from_bottle = bottle / bottle_to_beer beer_from_lid = lid / lid_to_beer beer = beer_from_bottle + beer_from_lid remained_bottle = bottle % bottle_to_beer remained_lid = bottle % lid_to_beer on_hand = {"beer": beer, "bottle": remained_bottle, "lid": remained_lid} return on_hand print how_many_beer(10) 应该没错吧。。。求大神指教 |
|
2
aluo1 2015-10-31 15:32:25 +08:00
def how_many_beer(money):
# The price of the beer beer_price = 2 # The number of beer bought by the given money beer_bought = money / beer_price beer = beer_bought # Bottle and lid number is the same as the beer bought initially bottle = beer_bought lid = beer_bought # Check whether can trade in more beer get_more_beer = tradable(bottle, lid) while get_more_beer: # Get the trade-in result on_hand = trade_in(bottle, lid) # Get the number of trade-in beer more_beer = on_hand["beer"] # Beer drunk so far beer += more_beer bottle = on_hand["bottle"] + more_beer lid = on_hand["lid"] + more_beer get_more_beer = tradable(bottle, lid) return beer def tradable(bottle,lid): # How many bottle can trade in one beer bottle_to_beer = 4 # How many lid can trade in one beer lid_to_beer = 2 return (bottle / bottle_to_beer) | (lid/lid_to_beer) def trade_in(bottle, lid): # How many bottle can trade in one beer bottle_to_beer = 4 # How many lid can trade in one beer lid_to_beer = 2 beer_from_bottle = bottle / bottle_to_beer beer_from_lid = lid / lid_to_beer beer = beer_from_bottle + beer_from_lid remained_bottle = bottle % bottle_to_beer remained_lid = bottle % lid_to_beer on_hand = {"beer": beer, "bottle": remained_bottle, "lid": remained_lid} return on_hand print how_many_beer(10) |
|
3
aluo1 2015-10-31 15:32:40 +08:00
为什么空格不对。。。
|
Select Language