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 rightmost 5 digits of what remains
Highly inaccurate but helped crossing icpc prelims in 1st year because there were humans who checked my solution with smaller inputs :) On Mon, Sep 14, 2009 at 6:36 PM, Shashwat Anand <anand.shash...@gmail.com> 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 > > The code is pasted here : http://paste.pocoo.org/show/139745/ > > 1 ''' > 2 For any N, let f(N) be the last five digits before the trailing zeroes in > N!. > 3 For example, > 4 > 5 9! = 362880 so f(9)=36288 > 6 10! = 3628800 so f(10)=36288 > 7 20! = 2432902008176640000 so f(20)=17664 > 8 > 9 Find f(1,000,000,000,000) > 10 ''' > 11 def f(n): > 12 fac = 1 > 13 i = 1 > 14 #for i in range(1, n+1): > 15 while i < n + 1: > 16 fac = int(str(fac * i).strip('0')) % 100000 > 17 i += 1 > 18 return fac > 19 > 20 print f(1000000000000) > > PS. hope posting algorithmic doubts will not be considered spamming :) > > > _______________________________________________ > BangPypers mailing list > BangPypers@python.org > http://mail.python.org/mailman/listinfo/bangpypers > _______________________________________________ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
