THEORIE UND PRAXIS

Eingabe und Ausgabe

Aufgaben

Bedingungen

Aufgaben

For-Schleife

Aufgaben

Zeichenketten

Aufgaben

While-Schleife

Aufgaben

Listen

Aufgaben

2D-Arrays

Aufgaben

Wörterbücher

Aufgaben

Mengen

Aufgaben

Funktionen und Rekursion

Aufgaben

Lektion 5. While cycle

Schwierigkeit:

Aufgabe«Denis Ansammlungen»

💻 Python

Denis ist ein Anfängerprogrammierer und träumt von einem neuen leistungsstarken Smartphone für Arbeit und Unterhaltung. Er hat beschlossen, dass er dafür 20% seines monatlichen Gehalts sparen wird.

Schreiben Sie ein Programm, das Denis dabei hilft zu berechnen, wie viel Zeit er benötigt, um den erforderlichen Betrag anzusparen. Das Programm soll zwei Werte vom Benutzer abfragen: Denis' monatliches Gehalt und den vollen Preis des Smartphones.

& nbsp;

rn rn

Wenn Denis' Gehalt null oder negativ ist, soll das Programm melden, dass es unmöglich ist, für ein Smartphone zu sparen. Wenn das Smartphone günstiger ist als das, was Denis durch das Zurücklegen von 20% eines Gehalts sparen kann, soll das Programm die entsprechende Meldung ausgeben.

In anderen Fällen soll das Programm die genaue Anzahl der vollen Jahre und Monate berechnen, die für das Ansparen erforderlich sind.

Die Ausgabe soll grammatikalisch korrekt sein. Zum Beispiel "1 Jahr", "2 Jahre", "5 Jahre"; "1 Monat", "3 Monate", "10 Monate".

Wenn das Ansparen genau n Jahre (ohne Monate) oder genau n Monate (weniger als ein Jahr) erfordert, geben Sie nur die Jahre bzw. nur die Monate aus.

Eingabeformat

In der ersten Zeile wird eine Zahl eingegeben — die Höhe des monatlichen Gehalts.
In der zweiten Zeile wird eine Zahl eingegeben — die Gesamtkosten eines Smartphones.

Ausgabeformat

Linie & mdash; Die Anzahl der Monate oder Jahre, die erforderlich sind, um den richtigen Betrag anzusammeln.

Beispiel

Eingabe

50000
25000

Ausgabe

Der Zugriff dauert: 3 Monate

Hinweis

While in python

In the Python programming language, the while loop construction can have an optional else block. This block is executed only when the loop completes in its natural way, that is, when its condition becomes false, and not as a result of a forced interruption using the break operator.

Syntax:

while condition:
    # the body of the loop
    # ...the code that runs on each iteration...
else:
    # else block
    # ...the code that is executed if the loop ended without a break...

You can think of the else block as a "plan B": it is triggered if the main goal within the loop (which is usually completed through a break) has not been achieved.

Usage example: Searching for an element

Consider a classic problem where we search for a number in a list. If a number is found, we immediately break the loop using break. If we have checked all the elements and still haven't found what we are looking for, the else block is executed.

numbers = [1, 2, 3, 4, 5]
target = 8 # Let's try to find a number that is not in the list.

index = 0
while index < len(numbers):
    if numbers[index] == target:
        print(f"Success! {target} found at {index}.")
break # Goal reached, exit the loop
    index += 1
else:
    # This block will be executed because break was not called
    print(f"The search is completed. {target} not found in the list.")

How it works:

The whileloop is executed as long as the index < len(numbers) condition is true.
Inside the loop, we check whether the current element is equal to the desired one (target).
If we find a value, a success message is displayed and the loop is immediately interrupted by the break statement. The else block is ignored in this case.
If the loop reaches the end (checked all the elements), it means that break has never worked. The index < len(numbers) condition becomes false, and then the else block is executed, reporting a failed search.

A useful tip: Using while...else helps to avoid "flagged" variables. Without this block, the code would look like this:

# Alternative version with "flag"
numbers = [1, 2, 3, 4, 5]
target = 8
found = False # Flag variable

index = 0
while index < len(numbers):
    if numbers[index] == target:
        found = True
        break
    index += 1

if found:
    print(f"Success! {target} found.")
else:
print(f"Search completed. {target} not found in the list.")

As you can see, the while...else construction makes the code cleaner, more concise, and more "Pythonic", eliminating the need to manually control the flag.

If `break` is missing

If there is no break operator in the body of the while loop, then the else block will always be executed after the loop ends.

count = 0
while count < 3:
print(f"Iteration number {count}")
count += 1
else:
print("The loop ended naturally, so the else block is executed.")

Here, the while loop runs until count is less than 3. As soon as count becomes 3, the loop condition becomes false, and control is transferred to the else block.

Important nuances

What about continue? The continue operator does not interrupt the loop, but only ends the current iteration ahead of schedule. It does not affect the execution of the else block in any way.
If the loop was never executed? If the while condition is initially false, the loop body will not be executed even once. However, the else block will still run, since the loop ended "normally" (it was not interrupted via break).

value = 10
while value < 5: # The condition is immediately false
    print("This text will never be printed")
else:
print("The cycle didn't even start, but it ended normally, so else worked.")

This also works for for loops! The same logic applies to for...else loops. The else block after for is executed if the cycle has passed through all the elements and has not been interrupted using break.

Output

The while...else construction is a powerful and elegant Python tool. It is ideal for situations where you need to perform an action only if the loop has completed its work completely without being interrupted. This is especially useful in search algorithms, condition checking, or when working with data streams, where break signals a special event, and else indicates a regular termination without any.

main.py

Test 1

Test 2

Test 3

Test 4

Test 5

Test 6

Test 7

Test 8

Test 9

Test 10

Entwicklerlösung

# Запрашиваем у пользователя исходные данные
salary = float(input())
smartphone_cost = float(input())

# Проверяем, может ли Денис в принципе копить
if salary <= 0:
    print("При такой зарплате Денис никогда не сможет накопить на смартфон.")
else:
    # Рассчитываем, сколько Денис откладывает каждый месяц
    monthly_saving = salary * 0.20

    # Проверяем, может ли он купить смартфон с одной зарплаты
    if monthly_saving >= smartphone_cost:
        print("Денис может купить смартфон сразу, с одной зарплаты!")
    else:
        # Используем цикл while для симуляции процесса накопления
        accumulated_sum = 0  # Переменная для хранения накопленной суммы
        total_months = 0     # Счетчик месяцев

        while accumulated_sum < smartphone_cost:
            accumulated_sum += monthly_saving  # Каждый месяц добавляем к накоплениям
            total_months += 1                  # Увеличиваем счетчик месяцев

        # Рассчитываем полные годы и оставшиеся месяцы
        years = total_months // 12
        months = total_months % 12

        # Готовим строки для вывода с правильными окончаниями
        year_str = ""
        month_str = ""

        # Формируем строку для лет (используем условия для правильной грамматики)
        if years > 0:
            if years % 10 == 1 and years % 100 != 11:
                year_str = f"{years} год"
            elif 2 <= years % 10 <= 4 and (years % 100 < 10 or years % 100 >= 20):
                year_str = f"{years} года"
            else:
                year_str = f"{years} лет"

        # Формируем строку для месяцев (используем условия для правильной грамматики)
        if months > 0:
            if months == 1:
                month_str = "1 месяц"
            elif 2 <= months <= 4:
                month_str = f"{months} месяца"
            else:
                month_str = f"{months} месяцев"

        # Собираем и выводим итоговый результат
        result_parts = []
        if year_str:
            result_parts.append(year_str)
        if month_str:
            result_parts.append(month_str)
        
        print("Накопление займет: " + " и ".join(result_parts))