Why do results differ from the Java version of this loop?

Asked Feb 06 '17 at 00:18

Active Feb 06 '17 at 00:26

Viewed 48 times

I'm relatively new to Python and am trying to write a piece of code in which should return the number of trailing zeroes in n!

The following is the code that I've written:

def countzeroes(n):
    count = 0
    i = 5
    while n/i >= 1:
        count += n/i
        i *=5
    return count

print(countzeroes(100))

Now, this code returns a value of 20 to me while it should actually return 24.

I implemented the exact same logic in Java and it returns me a value of 24. Is there a difference in the way the while loop is implemented in Python which is causing my loop to exit after the first iteration.

edited Feb 06 '17 at 00:26

martineau

119,623
25
170
301

asked Feb 06 '17 at 00:18

Kaushik Rajan

It would help if we could see the Java code that you used (although I have a hunch). – David Z Feb 06 '17 at 00:21
Are you sure about the 20 ? i get a 24 – karthikr Feb 06 '17 at 00:21
use `print()` to see values in variables - if it is Python 2 then dividing may gives wrong results - it rounts to int. – furas Feb 06 '17 at 00:22
I tested in Python 2.7 and 3.4 and got 24 from both. – David Z Feb 06 '17 at 00:24
Oh, there are mixed tabs and spaces in your source. Use Python 3 (strongly recommended) or run with `python2 -tt`. – Ry- Feb 06 '17 at 00:24
Keep it simple: "100".count("0") – gregory Feb 06 '17 at 00:35
@gregory: It’s to count the zeroes in the factorial of 100, not just 100. – Ry- Feb 06 '17 at 00:39
@Ryan, so why not: str(math.factorial(100)).count('0')? – gregory Feb 06 '17 at 00:48
Out of curiosity, what is the logic here? – juanpa.arrivillaga Feb 06 '17 at 00:54
@gregory: Because that’s too slow and uses too much memory for factorial(1000000). – Ry- Feb 06 '17 at 01:03
@Ryan, Sure; but a simple solution for small scales seems reasonable when the OP had a trail of only 24 zeros which is most likely just factorial(76) not factorial(1000000). – gregory Feb 06 '17 at 01:28
@gregory: Yeah, we can also simplify it to `print 24` for scales between 100 and 100 – Ry- Feb 06 '17 at 01:42
Thank you! http://stackoverflow.com/questions/28021409/indentation-in-function-does-not-work-as-expected - This link had the exact solution for my problem. I had used tab to indent it instead of space which caused the problem. I'm getting the expected results now. – Kaushik Rajan Feb 06 '17 at 03:12
@juanpa.arrivillaga - The logic is to count the number of 5s in the prime factors of n! – Kaushik Rajan Feb 06 '17 at 03:17
Also, in case you missed it: I really recommend upgrading from Python 2 to Python 3. It catches problems with indentation and much more. – Ry- Feb 06 '17 at 03:56

Why do results differ from the Java version of this loop?

0 Answers0