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…e, ruff checks were passed
Godrik0
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Dec 13, 2025
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ISBN: 5/10
Бинарный поиск: 5/5
Оформление: 10/10
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В задании требовалось также реализовать консольную утилиту.
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| def modulo_11_check_sum(isbn_number: str): | ||
| if not isbn_number: # Обработка случая пустой строки | ||
| return False | ||
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| # Извлекаем только цифры, без других символов | ||
| digits = [int(char) for char in isbn_number if char.isdigit()] | ||
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| # Проверяем что цифр достаточно | ||
| if len(digits) < 2: | ||
| return False | ||
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| check_digit = digits[-1] | ||
| total = 0 | ||
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| # Неправильно написан алгоритм | ||
| # Веса должны уменьшаться с 10 до 1, на шаг -1 | ||
| for i in range(len(digits) - 1): | ||
| weight = 10 - i # Веса уменьшаются от 10 до 1 | ||
| digit = digits[i] | ||
| total += digit * weight | ||
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| # Проверка контрольной суммы | ||
| checksum = total % 11 | ||
| return checksum == check_digit |
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У Вас пропущена обработка символа X, который может служить контрольной цифрой.
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| checksum = total % 11 | ||
| return checksum == check_digit |
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Ошибка в логике.
Нужно было либо оставить checksum = total + checkDigit и сравнивать остаток от деления на 11 с 0.
Либо высчитывать вот так (11 - (sum % 11)) % 11 и тогда уже сравнивать с контрольной цифрой.
| digits = [int(char) for char in isbn_number if char.isdigit()] | ||
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| # Проверяем что цифр достаточно | ||
| if len(digits) < 2: |
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Сдача контрольной работы. Долженко Александра Борисовна
Выполненные задания:
1. Настройка репозитория
2. Функция бинарного поиска (bin_search)
3. Функция проверки контрольной суммы checksum