mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 13:31:07 +00:00
42 lines
1001 B
Python
42 lines
1001 B
Python
|
#-.- coding: latin-1 -.-
|
||
|
|
from __future__ import print_function
|
||
|
|
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
|
||
|
|
|
||
|
|
sums = []
|
||
|
|
total = 0
|
||
|
|
|
||
|
|
for i in xrange(1, 10000):
|
||
|
|
n = sum_of_divisors(i)
|
||
|
|
|
||
|
|
if n < len(sums):
|
||
|
|
if sums[n-1] == i:
|
||
|
|
total += n + i
|
||
|
|
|
||
|
|
sums.append(n)
|
||
|
|
|
||
|
|
print(total)
|