6
\$\begingroup\$

This question already has an answer here:

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.

\$\endgroup\$

marked as duplicate by Toby Speight, Cows quack, Blue code-golf Feb 3 '17 at 12:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ I'm not sure if we should have this here, as it's basically the same as spoj.pl/SHORTEN/problems/FACTZERO \$\endgroup\$ – Nabb Feb 8 '11 at 15:29
  • \$\begingroup\$ @Nabb, I had no idea that this question was on SPOJ. I've put a significantly larger limit on the input here though. \$\endgroup\$ – Dogbert Feb 8 '11 at 15:35
  • \$\begingroup\$ @Dogbert: The limit is larger but the algorithms here are going to be exactly the same as over at SPOJ. \$\endgroup\$ – Nabb Feb 8 '11 at 16:03
  • \$\begingroup\$ for shortest code, you should use the code-golf tag \$\endgroup\$ – gnibbler Feb 8 '11 at 19:54
  • \$\begingroup\$ It's on Euler as well :) (i think) \$\endgroup\$ – st0le Feb 9 '11 at 5:50
1
\$\begingroup\$

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
\$\endgroup\$
  • \$\begingroup\$ But you still need the input and output right? \$\endgroup\$ – Quixotic Mar 8 '11 at 23:59
  • \$\begingroup\$ @Debanjan, the question states, write a function.... \$\endgroup\$ – st0le Mar 9 '11 at 5:35
  • \$\begingroup\$ @sOle:My bad!I missed that. \$\endgroup\$ – Quixotic Mar 9 '11 at 6:05
2
\$\begingroup\$

Python 50 47 Characters

n=input()
x=5
s=0
while n/x:s+=n/x;x*=5
print s
\$\endgroup\$
1
\$\begingroup\$

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

\$\endgroup\$
1
\$\begingroup\$

Haskell, 63 bytes

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

Ruby, 52 43

n=gets.to_i;a=0;i=1
a+=n/i*=5while i<n
p a
\$\endgroup\$
  • \$\begingroup\$ You could use (a+=n/i;i*=5)while i<n to save a char. \$\endgroup\$ – Dogbert Feb 8 '11 at 15:51
  • \$\begingroup\$ And ofcourse, p a as a is a number. \$\endgroup\$ – Dogbert Feb 8 '11 at 15:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.