Drinking soda scenario 2: recursion

Buy soda alone , One yuan, one bottle of soda , Three bottle caps can replace a bottle of soda , Two empty bottles can be exchanged for a soda , ask 20 How much soda can I buy for yuan ？

#python -version: python3

def soda(n, bottle, cap):
    print([' The current number of bottles you can drink ', ' Current number of empty bottles remaining ', ' Current number of remaining caps '], [n, bottle, cap])

    #  Exchange the remaining empty bottles , Add the empty bottle left after drinking the drink 
    bottle = bottle + n

    #  Exchange the remaining bottle caps , Add the bottle cap left after drinking the drink 
    cap = cap + n

    #  After a drink , When the number of empty bottles is less than 2, At the same time, the cap is smaller than 3 when , Then return directly   The number of bottles currently drunk 
    if bottle < 2 and cap < 3:
        return n
    else:  #  Add up all the drinks you drink , And the number of bottles exchanged for drinks , Remaining empty bottles , The number of remaining caps enters recursion 
        return n + soda(bottle//2+cap//3, bottle % 2, cap % 3) #  recursive 


def main():
    for i in [1, 3, 5, 20]:
        # Money = int(input('Please input money: '))
        Money = i
        print('\n')
        print(str(Money) + '  Yuan ')
        soda_total = soda(Money, 0, 0)
        print(str(Money)+'  Yuan '+' The total number of soda bottles you can drink is : '+str(soda_total))


if __name__ == '__main__':
    main()

Running results ：

Um. ~~~, Empty bottles are not borrowed , The bottle cap is not borrowed

Game Over~~~