Product of divisors
Given a positive integer, return the product of its positive divisors.
- Task
Show the result for the first 50 positive integers.
Python
Finding divisors efficiently
<lang Python>def product_of_divisors(n):
assert(isinstance(n, int) and 0 < n) ans = i = j = 1 while i*i <= n: if 0 == n%i: ans *= i j = n//i if j != i: ans *= j i += 1 return ans
if __name__ == "__main__":
print([product_of_divisors(n) for n in range(1,51)])</lang>
- Output:
[1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19, 8000, 441, 484, 23, 331776, 125, 676, 729, 21952, 29, 810000, 31, 32768, 1089, 1156, 1225, 10077696, 37, 1444, 1521, 2560000, 41, 3111696, 43, 85184, 91125, 2116, 47, 254803968, 343, 125000]
Choosing the right abstraction
This is really an exercise in defining a divisors function, and the only difference between the suggested Sum and Product draft tasks boils down to two trivial morphemes:
reduce(add, divisors(n), 0) vs reduce(mul, divisors(n), 1)
The goal of Rosetta code (see the landing page) is to provide contrastive insight (rather than comprehensive coverage of homework questions :-). Perhaps the scope for contrastive insight in the matter of divisors is already exhausted by the trivially different Proper divisors task.
<lang python>Sums and products of divisors
from math import floor, sqrt from functools import reduce from operator import add, mul
- divisors :: Int -> [Int]
def divisors(n):
List of all divisors of n including n itself. root = floor(sqrt(n)) lows = [x for x in range(1, 1 + root) if 0 == n % x] return lows + [n // x for x in reversed(lows)][ (1 if n == (root * root) else 0): ]
- ------------------------- TEST -------------------------
- main :: IO ()
def main():
Product of divisors for each integer in range [1..50] print('Products of divisors:') for n in range(1, 1 + 50): print(n, '->', reduce(mul, divisors(n), 1))
print('Sums of divisors:') for n in range(1, 1 + 100): print(n, '->', reduce(add, divisors(n), 0))
- MAIN ---
if __name__ == '__main__':
main()</lang>