/
ssmih
/
Python
Обзор
Документация
Войти
/
ssmih
/
Python
Код
Запросы
0
Задачи
Вики
Пакеты
0
Релизы
0
Аналитика
Безопасность
master
project_euler/problem_057/sol1.py
48 строк
1 KB
Benjamin Smith
Project Euler 57 - Square root convergents (#3259)
16 окт 2020, 18:47
Не верифицирован
16 окт 2020, 18:47
5887567
Код
Авторство
О чём код?
""" Project Euler Problem 57: https://projecteuler.net/problem=57 It is possible to show that the square root of two can be expressed as an infinite continued fraction. sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + ...))) By expanding this for the first four iterations, we get: 1 + 1 / 2 = 3 / 2 = 1.5 1 + 1 / (2 + 1 / 2} = 7 / 5 = 1.4 1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17 / 12 = 1.41666... 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/ 29 = 1.41379... The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator. In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator? """ def solution(n: int = 1000) -> int: """ returns number of fractions containing a numerator with more digits than the denominator in the first n expansions. >>> solution(14) 2 >>> solution(100) 15 >>> solution(10000) 1508 """ prev_numerator, prev_denominator = 1, 1 result = [] for i in range(1, n + 1): numerator = prev_numerator + 2 * prev_denominator denominator = prev_numerator + prev_denominator if len(str(numerator)) > len(str(denominator)): result.append(i) prev_numerator = numerator prev_denominator = denominator return len(result) if __name__ == "__main__": print(f"{solution() = }")