Thanks Renato for the information. Can you please provide some input on "how to improve on big size data processing "?
Any pointer for it is most welcome Thanks -Amit Jain On Mon, Feb 27, 2012 at 5:50 PM, Renato Parente <[email protected]> wrote: > a/b % mod = a* (b^-1) % mod > Since mod = 10^9 + 7 is a prime, so it's 'cheap' to find (b^-1) % mod > instead of checking every possible value. > One of the ways it's noticing that, for any b,n such that n is a prime and > b is not divisble by n, b^(n-1) = 1 (mod n). If you notice, b^(n-1) = > b*b^(n-2), therefore, b^(n-2) is the modular inverse of b, that's why the > formula is like that > > I hope you've understood :) > > > On 27 February 2012 09:13, Mitesh Joshi <[email protected]> wrote: > >> problem find a/b modulo 10^9+7 .... >> mod=10^9+7 >> bcoz a and b some value which more than mod... >> >> so the solution uses a/b= ( (a%mod) * ( (b%mod) ^ mod-2 ) ) % mod >> why..... >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Google Code Jam" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/google-code?hl=en. >> >> > -- > You received this message because you are subscribed to the Google Groups > "Google Code Jam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-code?hl=en. > -- You received this message because you are subscribed to the Google Groups "Google Code Jam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
