Python/project_euler/problem_21/sol1.py
Bruno Simas Hadlich 267b5eff40 Added doctest and more explanation about Dijkstra execution. (#1014)
* Added doctest and more explanation about Dijkstra execution.

* tests were not passing with python2 due to missing __init__.py file at number_theory folder

* Removed the dot at the beginning of the imported modules names because 'python3 -m doctest -v data_structures/hashing/*.py' and 'python3 -m doctest -v data_structures/stacks/*.py' were failing not finding hash_table.py and stack.py modules.

* Moved global code to main scope and added doctest for project euler problems 1 to 14.

* Added test case for negative input.

* Changed N variable to do not use end of line scape because in case there is a space after it the script will break making it much more error prone.

* Added problems description and doctests to the ones that were missing. Limited line length to 79 and executed python black over all scripts.

* Changed the way files are loaded to support pytest call.

* Added __init__.py to problems to make them modules and allow pytest execution.

* Added project_euler folder to test units execution

* Changed 'os.path.split(os.path.realpath(__file__))' to 'os.path.dirname()'
2019-07-17 01:09:53 +02:00

62 lines
1.3 KiB
Python

# -.- coding: latin-1 -.-
from math import sqrt
"""
Amicable Numbers
Problem 21
Let d(n) be defined as the sum of proper divisors of n (numbers less than n
which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and
each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55
and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and
142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
"""
try:
xrange # Python 2
except NameError:
xrange = range # Python 3
def sum_of_divisors(n):
total = 0
for i in xrange(1, int(sqrt(n) + 1)):
if n % i == 0 and i != sqrt(n):
total += i + n // i
elif i == sqrt(n):
total += i
return total - n
def solution(n):
"""Returns the sum of all the amicable numbers under n.
>>> solution(10000)
31626
>>> solution(5000)
8442
>>> solution(1000)
504
>>> solution(100)
0
>>> solution(50)
0
"""
total = sum(
[
i
for i in range(1, n)
if sum_of_divisors(sum_of_divisors(i)) == i
and sum_of_divisors(i) != i
]
)
return total
if __name__ == "__main__":
print(solution(int(str(input()).strip())))