Python/project_euler/problem_014/sol1.py
Joyce bcfca67faa
[mypy] fix type annotations for all Project Euler problems (#4747)
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3

* [mypy] fix type annotations for project euler problem007/sol2

* [mypy] fix type annotations for project euler problem008/sol2

* [mypy] fix type annotations for project euler problem009/sol1

* [mypy] fix type annotations for project euler problem014/sol1

* [mypy] fix type annotations for project euler problem 025/sol2

* [mypy] fix type annotations for project euler problem026/sol1.py

* [mypy] fix type annotations for project euler problem037/sol1

* [mypy] fix type annotations for project euler problem044/sol1

* [mypy] fix type annotations for project euler problem046/sol1

* [mypy] fix type annotations for project euler problem051/sol1

* [mypy] fix type annotations for project euler problem074/sol2

* [mypy] fix type annotations for project euler problem080/sol1

* [mypy] fix type annotations for project euler problem099/sol1

* [mypy] fix type annotations for project euler problem101/sol1

* [mypy] fix type annotations for project euler problem188/sol1

* [mypy] fix type annotations for project euler problem191/sol1

* [mypy] fix type annotations for project euler problem207/sol1

* [mypy] fix type annotations for project euler problem551/sol1
2021-10-12 00:33:44 +08:00

61 lines
1.5 KiB
Python

"""
Problem 14: https://projecteuler.net/problem=14
Problem Statement:
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains
10 terms. Although it has not been proved yet (Collatz Problem), it is thought
that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
"""
def solution(n: int = 1000000) -> int:
"""Returns the number under n that generates the longest sequence using the
formula:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
# The code below has been commented due to slow execution affecting Travis.
# >>> solution(1000000)
# 837799
>>> solution(200)
171
>>> solution(5000)
3711
>>> solution(15000)
13255
"""
largest_number = 0
pre_counter = 0
for input1 in range(n):
counter = 1
number = input1
while number > 1:
if number % 2 == 0:
number //= 2
counter += 1
else:
number = (3 * number) + 1
counter += 1
if counter > pre_counter:
largest_number = input1
pre_counter = counter
return largest_number
if __name__ == "__main__":
print(solution(int(input().strip())))