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 where we needed to do was, a * a * a ..(b times) % r, modulation at each step help reduce computation since what we need is F(n) % 10**5 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 On Mon, Sep 14, 2009 at 6:51 PM, Navin Kabra <navin.ka...@gmail.com> wrote: > > print f(1000000000000) >> >> > 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) > > > _______________________________________________ > BangPypers mailing list > BangPypers@python.org > http://mail.python.org/mailman/listinfo/bangpypers > >
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