题解 | #汽水瓶#

#本题主要考查递归算法
def bottle_num(n):
    res = n//3
    remain = n%3
    bottles = res+remain    #the bottle number of remain
    if bottles == 2: #permit borrow one bottle
        res+=1
    elif bottles < 2:   #the bottle number is not ample
        pass
    else:
        res = res + bottle_num(bottles) #bottle>2,start digui algrithm
    return res

while 1:#use while accomplishment a lot of input
    n = int(input())
    if n == 0:
        break
    else:
        print(bottle_num(n))