python的编码题怎么写？

Question

Three empty cans can be exchanged for a new one. Suppose you have N cans of soda, try to use the program to solve how many cans of soda you can drink in the end?

Input description: Input a positive integer N. ex.5 / ex.100

Output description: The maximum number of sodas that can be drunk, and must have a newline character at the end. ex.7 / ex.149 `

n = int(input())
a = n-3 
sum = 0
while a > 2 :
  sum += 1 
  a -= 3 
print(f'{n+sum}')

if a == 2 :
  print(f'{n+sum+1}')

`

I used while to finish the code which is on above, but I input 5 and output 6,and it is actually to be 7.The other side, I input 100 and output 132. Actually, the correct answer is 149.

Answer 1

You can try this way -

def get_total_cans(n):
    s = n # we can get at least n cans
    while n > 2:
        s += 1
        n -= 3 # 3 exchanged
        n += 1 # got 1 for the exchange
    return s

n = int(input())
print(get_total_cans(n))

The logic is simple and comments are added to explain.

In your code, the first thing to notice it generates wrong output when n is less than 3. For example for n = 2, your output is 3 which is not possible. Also in the while loop you are decrementing a by 3 for the exchange but you fail to add 1 to a for the one soda you can exchanging 3 empty cans. That is the issue. My above code addresses those issues

Answer 2

Here's a recursive approach:

def dr_cans(full, empty=0):
    # if we have at least 3 empty cans, exchange them for a full one
    if empty >=3:
        return dr_cans(full+1,empty-3)
    # no full cans, and not enough empty ones
    if full == 0:
        return 0
    # at least one full can: drink it and gain an empty one
    return 1 + dr_cans(full-1, empty+1)

Answer 3

If I understand corretly the question is as follows: Let k be the index of the exchange process. Since not all N can be divided by 3 we have N[k] = floor(M/3) , where M=N[k-1]+R[k-1] new cans in each step. Plus some R[k] = M%3 leftover cans, where % is the modulo operator...

With this is should be quite easy...

def compute_num_cans(empty_cans: int, exchange: int = 3) -> tuple:
    """
    :param empty_cans: The number of cans to exchange
    :return: tuple of (full_cans, empty_cans), where the empty cans are < exchange rate
    """
    leftovers = empty_cans % exchange
    full = empty_cans // exchange
    return full, leftovers


EXCHANGE = 3
NUM_CANS = 51

print(f'Start with {NUM_CANS} and an exchange rate of {EXCHANGE}:1')
current_cans = NUM_CANS
drunk_cans = NUM_CANS
leftovers = 0
steps = 0
while current_cans >= EXCHANGE:
    full, leftovers = compute_num_cans(current_cans, exchange=EXCHANGE)
    current_cans = full + leftovers
    drunk_cans += full
    steps += 1

print(f'Cans drunk: {drunk_cans}, leftover cans: {leftovers}.')
print(f'A total of {steps} exchanges was needed.')

This yields as output

# Start with 51 and an exchange rate of 3:1
# Cans drunk: 76, leftover cans: 0.
# A total of 4 exchanges was needed.

Answer 4

General Answer:- Accepted also if we change the numExchange also.

Code:-

def numWaterBottles(numBottles: int, numExchange: int) -> int:
    ans=numBottles  #Initial bottles he/she will drink
    while numBottles>=numExchange: #If numBottles<numExchange exit the while loop
        remainder=numBottles%numExchange #remaining bottles which is not change
        numBottles//=numExchange #The bottles which are changed
        ans+=numBottles #The bottles which are changed added to the answer
        numBottles+=remainder #Remaining bottles==The bottles which is not change+The bottles which are changed
    return ans    #Return The answer
    
print(numWaterBottles(5,3))
print(numWaterBottles(100,3)) 
print(numWaterBottles(32,4)) #numexchange when different

Output:-

7
149
42