u can do a lot of stuff with the bitwise operators , u just gotta experiment
with it

On Tue, May 10, 2011 at 2:21 AM, Morgan Bauer <[email protected]>wrote:

> XOR is the bread and butter of Symmetric Cryptography.
> ~mhb
>
> On Mon, May 9, 2011 at 4:44 PM, Marcelo Ramires
> <[email protected]> wrote:
> > I understood how to solve this, but how does one come to this solution ?
> >
> > If the xor of all numbers is zero, you can pick any candy, and the xor to
> > this number is going to be equal to the xor from the rest of them.
> >
> > I get this, if I have 9 numbers with XOR 3, XORing it with 3 will get me
> > zero.
> >
> > How has everybody thought of this at the same time ? have I skipped a
> logics
> > class ? is this concept so disseminated among coders ?
> >
> > I had never XORed nubmers before this code jam, only booleans, and I
> didn't
> > know you could.
> >
> > As a side questions, can anybody tell me any alternative uses for XORing
> > integers other than 1 and 0 ?
> >
> > Thanks!
> >
> > Marcelo Ramires
> >
> > On Sun, May 8, 2011 at 1:50 AM, vivek dhiman <[email protected]>
> wrote:
> >>
> >> Lucky!
> >>
> >> You are right.
> >>
> >> if xor of two lists is same. (say xor1 = xo2)
> >>
> >> So the exor of these two wil be 0  (xor (xor1,xor2) = 0)
> >> Or in other words lists can be divided if the xor of all the elements is
> >> zero.
> >>
> >> :)
> >>
> >>
> >>
> >> On Sun, May 8, 2011 at 8:17 AM, keshav agarwal <[email protected]>
> >> wrote:
> >>>
> >>> please tell me of my logic was correct or i just got lucky to get it
> >>> correct
> >>>
> >>> if xor to a list of nos. is zero only then the division is possible
> >>> in this case patrick can be given the one candy with lowest value while
> >>> sean keeps the rest
> >>>
> >>> if      xor(n nos.)=0
> >>> then       (nth no.) xor (xor of n-1 nos.)=0
> >>>
> >>> so patrick gets the nth candy and sean keeps the rest
> >>>
> >>> --
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> >>
> >>
> >>
> >> --
> >> Regards
> >> Vivek Dhiman
> >>
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