未验证 提交 d8f573c0 编写于 作者: G GGn0 提交者: GitHub

Add solution to problem 74 (#3110)

* Add solution to problem 74

* Fix typo

* Edit unnecessary comment

* Rename folder, add default params in solution()

* Rename file to solve conflicts

* Fix doctests
上级 d3ead538
"""
Project Euler Problem 074: https://projecteuler.net/problem=74
Starting from any positive integer number
it is possible to attain another one summing the factorial of its digits.
Repeating this step, we can build chains of numbers.
It is not difficult to prove that EVERY starting number
will eventually get stuck in a loop.
The request is to find how many numbers less than one million
produce a chain with exactly 60 non repeating items.
Solution approach:
This solution simply consists in a loop that generates
the chains of non repeating items.
The generation of the chain stops before a repeating item
or if the size of the chain is greater then the desired one.
After generating each chain, the length is checked and the counter increases.
"""
def factorial(a: int) -> int:
"""Returns the factorial of the input a
>>> factorial(5)
120
>>> factorial(6)
720
>>> factorial(0)
1
"""
# The factorial function is not defined for negative numbers
if a < 0:
raise ValueError("Invalid negative input!", a)
# The case of 0! is handled separately
if a == 0:
return 1
else:
# use a temporary support variable to store the computation
temporary_computation = 1
while a > 0:
temporary_computation *= a
a -= 1
return temporary_computation
def factorial_sum(a: int) -> int:
"""Function to perform the sum of the factorial
of all the digits in a
>>> factorial_sum(69)
363600
"""
# Prepare a variable to hold the computation
fact_sum = 0
""" Convert a in string to iterate on its digits
convert the digit back into an int
and add its factorial to fact_sum.
"""
for i in str(a):
fact_sum += factorial(int(i))
return fact_sum
def solution(chain_length: int = 60, number_limit: int = 1000000) -> int:
"""Returns the number of numbers that produce
chains with exactly 60 non repeating elements.
>>> solution(60,1000000)
402
>>> solution(15,1000000)
17800
"""
# the counter for the chains with the exact desired length
chain_counter = 0
for i in range(1, number_limit + 1):
# The temporary list will contain the elements of the chain
chain_list = [i]
# The new element of the chain
new_chain_element = factorial_sum(chain_list[-1])
""" Stop computing the chain when you find a repeating item
or the length it greater then the desired one.
"""
while not (new_chain_element in chain_list) and (
len(chain_list) <= chain_length
):
chain_list += [new_chain_element]
new_chain_element = factorial_sum(chain_list[-1])
""" If the while exited because the chain list contains the exact amount of elements
increase the counter
"""
chain_counter += len(chain_list) == chain_length
return chain_counter
if __name__ == "__main__":
import doctest
doctest.testmod()
print(f"{solution()}")
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