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Katrin_Kolarska avatar Katrin_Kolarska 0 Точки

Programming Basics; Phyton - Fuel Tank - Part 2

Здравейте, 
решавам задачата Fuel Tank - Part 2 - последната от допълнителните за Conditional Statements. В джъдж ми дава 90/100, но не мога да си открия грешката. Ще може ли някой да ми подскаже или да ми даде неговото решение :)
Благодаря. 

fuel_type = str(input())
fuel_quantity = float(input())
discount_card = str(input())
price= float()

if discount_card == "Yes" and fuel_type == "Gas":
    if fuel_quantity > 25:
        price = 0.9*(0.85 * fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92*(0.85 * fuel_quantity)
    else:
        price = 0.85 * fuel_quantity
if discount_card == "No" and fuel_type == "Gas":
    if fuel_quantity > 25:
        price = 0.9 * (0.93*fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92 * (0.93*fuel_quantity)
    else:
        price = 0.93*fuel_quantity
if discount_card == "Yes" and fuel_type == "Diesel":
    if fuel_quantity > 25:
        price = 0.9*(2.21 * fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92*(2.21 * fuel_quantity)
    else:
        price = 2.21 * fuel_quantity
if discount_card == "No" and fuel_type == "Diesel":
    if fuel_quantity > 25:
        price = 0.9 * (2.33 * fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92 * (2.33 * fuel_quantity)
    else:
        price = 2.33 * fuel_quantity
if discount_card == "Yes" and fuel_type == "Gasoline":
    if fuel_quantity > 25:
        price = 0.9*(2.10 * fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92 * (2.10 * fuel_quantity)
    else:
        price = 2.10 * fuel_quantity
if discount_card == "No" and fuel_type == "Gasoline":
    if fuel_quantity > 25:
        price = 0.9 * (2.22 * fuel_quantity)
    elif 20 < fuel_quantity <= 25:
        price = 0.92 * (2.22 * fuel_quantity)
    else:
        price = 2.22 * fuel_quantity

print(f'{price:.2f} lv.')
Тагове:
0
Programming Basics 12/07/2020 20:16:31
roshavi4ak avatar roshavi4ak 4 Точки

Здравей :)
От това което видях, грешката ти е в цената на Gasoline с карта. Намалението трябва да е с 18 ст. на литър, а ти го смяташ с 12. Т.е. вместо 2.10 трябва да е 2.04. Така ми го дава 100/100.

И си мисля че между 20 и 25 включително трябва да обхваща и 20. т.е. трябва да е навсякъде 20 <= х <= 25

Обаче си мисля че това решение е доста трудоемко за писане и както си личи, лесно за грешки. Сега ще реша задачата и ще покажа моето решение.

Приятна седмица и приятно програмиране! 

EDIT: Ето го моето решение: https://repl.it/@KrastyoKrastev/Fuel-Tank-2
Така всички цени и дискаунти са на едно място. С dictionary щеше да стане дори малко по-лесно.

0
13/07/2020 16:16:39
Katrin_Kolarska avatar Katrin_Kolarska 0 Точки

Много Ви благодаря, Вие сте супер smiley - не знам как не го видях.
Да, много се чудих как да я разпиша - другата седмица ни предстои да вземем вложените ифове и с времето се надявам да се науча да ги пиша по умно.
Благодаря Ви и приятен ден!
 

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