On Feb 20, 5:44 pm, Mark Dickinson <dicki...@gmail.com> wrote: > On Feb 20, 11:17 am, mukesh tiwari <mukeshtiwari.ii...@gmail.com> > wrote: > > > Hello everyone. I think it is related to the precision with double > > arithmetic so i posted here.I am trying with this problem > > (https://www.spoj.pl/problems/CALCULAT) and the problem say that "Note : for > > all test cases whose N>=100, its K<=15." I know precision of doubles > > in c is 16 digits. Could some one please help me with this precision > > issue.I used stirling (http://en.wikipedia.org/wiki/ > > Stirling's_approximation) to calculate the first k digits of N. > > Thank you. > > If I understand you correctly, you're trying to compute the first k > digits in the decimal expansion of N!, with bounds of k <= 15 and 100 > <= N < 10**8. Is that right? > > Python's floats simply don't give you enough precision for this: > you'd need to find the fractional part of log10(N!) with >= 15 digits > of precision. But the integral part needs ~ log10(log10(N!)) digits, > so you end up needing to compute log10(N!) with at least 15 + > log10(log10(N!)) ~ 15 + log10(N) + log10(log10(N)) digits (plus a few > extra digits for safety); so that's at least 25 digits needed for N > close to 10**8. > > The decimal module would get you the results you need (are you allowed > imports?). Or you could roll your own log implementation based on > integer arithmetic. > > -- > Mark
Yes i am trying to first k digits of N!.I will try your method. Thank you -- http://mail.python.org/mailman/listinfo/python-list
