On Mon, Sep 14, 2009 at 6:36 PM, Shashwat Anand
wrote:
> How do we calculate last 5-digits of 10**12 ignoring trailing zeros. The
> code i wrote works good until 10**8 and after that take ages.
> The source of problem is Project Euler :
> http://projecteuler.net/index.php?section=problems&id=160
>
This should work,
find f(6)**(10 raised to 7)
On Mon, 14 Sep 2009 18:36:57 +0530, Shashwat Anand
wrote:
How do we calculate last 5-digits of 10**12 ignoring trailing zeros. The
code i wrote works good until 10**8 and after that take ages.
The source of problem is Project Euler :
http://p
@anand:
There is a way to calculate number of leading zeros in n!
Here i had implemented it :
>>> x = 100 # number whose factorial is to be taken out
>>> s = 0 #initial count of zeros
>>> n = 1
>>> while x / 5**n:
... s += x / 5**n
... n +=1
...
>>> s
24
>>> import math
>>> len(str(math.f
@abhishek :
Can you show me the code ?
10**12 is way too much. My code gives output within a minute for 10**8 and
fails. Ofcourse I need to apply some algo here. What do you exactly mean by
taking out rightmost 7 digits ?
On Mon, Sep 14, 2009 at 8:14 PM, Abhishek Mishra wrote:
> I remember doin
I remember doing something very inaccurate a long time ago for this -
1. find everything in a loop
2.everytime you encounter some zeros, strip them
3.everytime after stripping it exceeds say 7 digits, take out
rightmost 7 digits for accuracy's sake
4. proceed till loop ends
5. print out the
Shashwat Anand wrote:
How do we calculate last 5-digits of 10**12 ignoring trailing zeros. The
code i wrote works good until 10**8 and after that take ages.
The source of problem is Project Euler :
http://projecteuler.net/index.php?section=problems&id=160
You might consider using the Stirling's
> The key point i think is we do not need factorial but modulus 10**5 without
> trailing zeros. Didn't helped though :( may be a better implementation could
> help
I think the right solution will NOT involve 10**12 computations. Think about it.
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The thought which comes to my mind is Modular exponentiation :
for a ** b % r ( if a ** b very large )
I had coded it in C++ but in python the pow() have inbuilt modular-function
made.
so pow(a ,b, r) does the trick
Here for factorial :
F(n) = n * n-1 * n-2 ...* 2 * 1
Just like in earlier case wh
> print f(1)
>
>
This is a HUGE number. I don't think you are expected to get the answer by
direct calculation. I would guess that you are expected to use mathematical
reasoning to figure out the answer (and not computation)
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BangPypers ma
> 16 fac = int(str(fac * i).strip('0')) % 10
This is ugly hack. Can you think of something more mathematical?
Anand
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How do we calculate last 5-digits of 10**12 ignoring trailing zeros. The
code i wrote works good until 10**8 and after that take ages.
The source of problem is Project Euler :
http://projecteuler.net/index.php?section=problems&id=160
The code is pasted here : http://paste.pocoo.org/show/139745/
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