题解 | #汽水瓶#递归函数/循环条件嵌套语句

'''
1/3=0---1        0

2/3=0---2        0
(0+2+1)/3=1---0  0+1=1

3/3=1---0        1

4/3=1---1        1
(1+1+1)/3=1---0  1+1=2

5/3=1---2        1
(1+2)/3=1---0    1+1=2

6/3=2---0        2
(2+0+1)/3=1---0  2+1=3

10/3=3---1       3
(3+1)/3=1---1    1
(1+1+1)/3=1---0  3+1+1=5

11/3=3---2       3
(3+2)/3=1---2    1
(1+2)/3=1---0    3+1+1=5

12/3=4---0       4
(4+0)/3=1---1    1
(1+1+1)/3=1---0  4+1+1=6
'''
'''1、循环语句：计算过程较为复杂
while True:
    drink=0
    n=int(input())
    if n>0:
        if n==1:
            drink += 0
        elif n>1:
            drink += 1
        while n//3+n%3>1:
            drink += n//3
            if n//3+n%3>=3:
                borrow=0
            else:
                borrow=3-n//3-n%3
            n = n//3+n%3+borrow
    elif n==0:
        print()
        break
    print(drink)
'''
#2、简化思路：观察n与瓶数的关系，发现n//2=瓶数
'''
while True:
    n=int(input())
    if n>0:
        print(n//2)
    if n==0:
        #print()
        break

while True:
    n=int(input())
    if n==0:         # break条件放在前面，更简洁。
        break
    print(n//2)
'''
#3、递归思路：
'''
import sys

def f(n):
    if n == 0: return 0
    if n == 1: return 0
    if n >=2: return f(n-2) + 1

if __name__ == '__main__':
    data = sys.stdin
    for x in data:
        x = int(x.strip())
        if x != 0:
            print(f(x))
'''

def max_bottles(n):
    result = n // 3
    bottles = n // 3 + n % 3
    if bottles == 2:  # 空瓶数量为2，可以借一瓶，此后不能再兑换
        result += 1
    elif bottles < 2:  # 空瓶数小于2，无法兑换
        result += 0
    else:
        result += max_bottles(bottles)
    return result
while True:
    try:
        n = int(input())
        if n == 0:
            break
        else:
            print(max_bottles(n))
    except:
        break