doubts of logic in Python

Question

I am a beginner in the world of programming and also in python.

Solving some exercises I came across this

Make a Program for a paint shop. The program should ask for the size in square meters of the area to be painted. Consider that the paint cover is 1 liter for every 6 square meters and that the paint is sold in 18 liter cans, which cost R $ 80,00 or 3,6 liter gallons, which cost R $ 25,00. Inform the user the quantities of ink to be purchased and the respective prices in 3 situations: buy only 18-litre cans; buy only 3.6 litre gallons; mix cans and gallons so that the waste of paint is less. Add 10% off and always round up the values, that is, consider full cans.

My resolution works, I arrive at the expected result, but the logic I used is very extensive. how could it resolve this issue more efficiently? follows below the code. Thanks in advance.

import math

m2 = float(input('informe a quantidade de METRO QUADRADO (m²) a ser pintado: '))
consumo_litro = m2 / 6
print('\no consumo de tinta é: {:.2f} LITROS\n'.format(consumo_litro))
qtd_latas18 = consumo_litro / 18
qtd_latas36 = consumo_litro / 3.6
print('a quantidade de GALOES de 18 LITROS a ser usado é: {}'.format(math.ceil(qtd_latas18)))
print('a quantidade de LATAS de 3,6 LITROS a ser usado é: {}\n'.format(math.ceil(qtd_latas36)))
valor_total18 = math.ceil(qtd_latas18) * 80
valor_total36 = math.ceil(qtd_latas36) * 25
print('o valor total em GALOES de 18 LITROS é: R${}'.format(valor_total18))
print('o valor total em LATAS de 3,6 LITROS é: R$ {}\n'.format(valor_total36))
print('considerando o menor desperdíciode tinta, temos: \n')

qtd_latas_mistas18 = ((consumo_litro * 0.10) + consumo_litro) / 18
qtd_litros18 = math.trunc(qtd_latas_mistas18) * 18
resto18 = ((consumo_litro * 0.10) + consumo_litro) - qtd_litros18
qtd_latas_mistas36 = resto18 / 3.6
qtd_latas_mistas_total = math.trunc(qtd_latas_mistas18) + math.ceil(qtd_latas_mistas36)
valor_misto18 = math.trunc(qtd_latas_mistas18) * 80
valor_misto36 = math.ceil(qtd_latas_mistas36) * 25
total_misto = valor_misto18 + valor_misto36

print('quantidade de latas de 18 litros: {}'.format(math.trunc(qtd_latas_mistas18)))
print('qtd latas de 3,6: {}'.format(math.ceil(qtd_latas_mistas36)))
print('qtd latas mistas: {}'.format(qtd_latas_mistas_total))
print('\no valor total considerando GALOES e LATAS é (acresentando 10% de quebra): R$ {}'.format(total_misto))

Answer 1

First thing that pops out the eyes is the amount of magic numbers. I’ve turned everything into constants (which are actually not constants, they’re just variables with high-box names). This increases code readability and facilitates future changes.

Note that the safety margin is 1.1, nay 0.1, 'cause it’ll make it a little easier.

M2_POR_LITRO = 6
PRECO_GALAO = 80
LITROS_GALAO = 18
PRECO_LATA = 25
LITROS_LATA = 3.6
MARGEM_SEGURANCA = 1.1

After that, I take the user input exactly as you did, but I already add the safety margin right there, because in my interpretation of the problem statement this margin should be applied to all situations, not just the third.

m2 = float(input('informe a quantidade de METRO QUADRADO (m²) a ser pintado: '))

consumo_litro = m2 / M2_POR_LITRO * MARGEM_SEGURANCA

Moving on, it’s time to process the data.

I decided to do the math before printing the results. This does not change the output of the pro user program at all, but leaves the code separated into "logical blocks": first the input, then the processing, then the output. I have the opinion that this makes the flow of code clearer.

Note that I simplified the calculation of mixed inks, I took advantage of the operator //, sometimes called integer Division.

qtd_galao_apenas = math.ceil(consumo_litro / LITROS_GALAO)
valor_galao_apenas = qtd_galao_apenas * PRECO_GALAO

qtd_lata_apenas = math.ceil(consumo_litro / LITROS_LATA)
valor_lata_apenas = qtd_lata_apenas * PRECO_LATA

qtd_galao_misto = consumo_litro // LITROS_GALAO
qtd_lata_misto = math.ceil((consumo_litro - qtd_galao_misto * LITROS_GALAO) / LITROS_LATA)
valor_galao_misto = qtd_galao_misto * PRECO_GALAO
valor_lata_misto = qtd_lata_misto * PRECO_LATA

To finish, I printed all the results on the screen.

The biggest difference to your code is that I used the f-strings instead of the method format.

Another difference is that I used several commands print() to generate blank line instead of \n inside the strings. I feel this makes the code easier to understand, as it looks a lot like output in practice. This is purely a matter of taste and you can keep putting the \n to generate blank lines without any problem.

print()
print(f'o consumo de tinta é: {consumo_litro:.2f} LITROS')
print()
print(f'a quantidade de GALOES de 18 LITROS a ser usado é: {qtd_galao_apenas:.0f}')
print(f'o valor total em GALOES de 18 LITROS é: R$ {valor_galao_apenas:.2f}')
print()
print(f'a quantidade de LATAS de 3,6 LITROS a ser usado é: {qtd_lata_apenas:.0f}')
print(f'o valor total em LATAS de 3,6 LITROS é: R$ {valor_lata_apenas:.2f}')
print()
print('considerando o menor desperdíciode tinta, temos:')
print(f'quantidade galões: {qtd_galao_misto:.0f}')
print(f'quantidade latas: {qtd_lata_misto:.0f}')
print(f'quantidade total mistas: {qtd_galao_misto + qtd_lata_misto:.0f}')
print(f'valor total considerando GALOES e LATAS é: R$ {valor_galao_misto + valor_lata_misto:.2f}')

The next step would be to break the program into several functions, and I encourage you to do so from now on, especially in longer programs.