you & mdash; A special agent under the cover that is trying to penetrate the secret object. The only way to get inside & mdash; Enter the correct code phrase into the access terminal. The security system of the object is very cunning: it does not check the phrase itself, but analyzes its structure. The door will open only if the number of vowels in the introduced phrase is equal to the secret number & mdash; 7.
rnYour task: & nbsp; write a program that imitates the work of this terminal. The program must request the user input of the code phrase. Then it should calculate the number of vowels in this phrase and, depending on the result, withdraw one of two messages: “access is allowed” or “access is prohibited”.
rnRules:
rnPassol phrase (type: line)
The result of the check (type: line)
a Quick Brown Fox Jumps High
Access is allowed
EN
RU
EN
RU
# Создаем переменную-счетчик для гласных и устанавливаем ее в 0
vowel_count = 0
# Определяем строку, содержащую все гласные в нижнем регистре для удобства проверки
vowels = "aeiou"
# Запрашиваем у пользователя ввод кодовой фразы
secret_phrase = input()
# Запускаем цикл, который перебирает каждый символ во введенной строке
for char in secret_phrase:
# Проверяем, является ли текущий символ (приведенный к нижнему регистру) одной из гласных
if char.lower() in vowels:
# Если это гласная, увеличиваем наш счетчик на 1
vowel_count += 1
# После завершения цикла проверяем, равно ли итоговое количество гласных секретному числу 7
if vowel_count == 7:
# Если равно, выводим сообщение об успехе
print("Доступ разрешен")
else:
# Если не равно, выводим сообщение об отказе
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