Simplify this python code
python
Solution
A straight-forward implementation that doesn't involve calculating the factorial (so that it works with big numbers, ie 2000000!) (edited):
fives = 0
twos = 0
for i in range(10, 201):
while i % 5 == 0:
fives = fives + 1
i /= 5
while i % 2 == 0:
twos = twos + 1
i /= 2
print(min(fives, twos))
Problem
I've written a program to check if my thought about solution on paper is right (and it is). The task: how many zeros is in the back of multiplication of all numbers from 10 to 200. It is 48 and it is a simple to calculate manually. I never write on python seriously and this is what I get: ``` mul = 1 for i in range(10, 200 + 1): mul *= i string = str(mul) string = string[::-1] count = 0; for c in str(string): if c == '0': count += 1 else: break print count print mul ``` I bet it is possible to write the same more elegant in such language like a python. ps: yes, it is a homework, but not mine - i just helped a guy ;-)