酒瓶换酒

"""
一开始有5瓶啤酒
两个空酒瓶换一瓶啤酒
四个瓶盖换一瓶啤酒
问你最后能够喝到多少瓶啤酒
"""


def get(n):
    bottle, head = n, n
    drink = 0
    while 1:
        drink += bottle
        use = bottle // 2 + head // 4
        if use == 0:
            break
        bottle %= 2
        head %= 4
        bottle += use
        head += use
    return drink


def recursive(bottle, head):
    """
    use=b//2+head//4
    f(b,h)=f(use+bottle%2,use+head%4)+bottle
    
    利用这个递推式可能能找到O(1)的算法
    
    :param bottle: 
    :param head: 
    :return: 
    """
    use = bottle // 2 + head // 4
    if use == 0:
        return bottle
    return recursive(use + bottle % 2, use + head % 4) + bottle

for i in range(1, 100):
    print(get(i), recursive(i, i))

"""
如果每瓶啤酒中奖的概率是p
那么期望得到多少瓶啤酒
"""