Calculate the number of trailing zeros in n!

Question

Write a function which takes n as a parameter, and returns the number of trailing zeros in n!.

Input Constraints

0 <= n <= 10^100

Output Constraints

Should be return the result in less than 10 seconds.

Test Input

1
213
45678
1234567
78943533
4567894123
121233112233112231233112323123

Test Output

0
51
11416
308638
19735878
1141973522
30308278058278057808278080759

Shortest code by character count wins.

Answer 1

Python

36 Chars Ripped from my earlier answer

f=lambda n:n//5+(n//5>0and f(n//5)or 0) #Py3k #39 Chars
f=lambda n:n/5+(n/5>0and f(n/5)or 0) #Before 3.0 #36 Chars

Answer 2

Python 50 47 Characters

n=input()
x=5
s=0
while n/x:s+=n/x;x*=5
print s

Answer 3

Ruby: 52 43

n=gets.to_i;a=0;i=1
a+=n/i*=5while i<n
p a

Answer 4

Haskell (63)

f i=length$filter(\x->x`mod`5/=0)[1..i];main=interact$show.f.read

Answer 5

J - 26

f=:3 :'+/<.y%5^(1+i.144x)'

eg

   f 121233112233112231233112323123x
     30308278058278057808278080759
   f 4567894123
     1141973522
   f 0
     0
   f 10^100x
     2499999999999999...99999999999999999982

in less than a second for all input examples