@azhar good solution Thanks and Regards Priyaranjan code-forum.blogspot.com
On Dec 22 2010, 1:55 pm, Azhar Hussain <[email protected]> wrote: > Hi, > > Here is the simple solution > > unsigned int r = v; // r will be reversed bits of v; first get LSB of v > int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end > > for (v >>= 1; v; v >>= 1) > { > r <<= 1; > r |= v & 1; > s--; > } > r <<= s; // shift when v's highest bits are zero > if (v == r) > printf("palindrome\n"); > else > printf("not a palindrome\n"); > > source and more optimized > versionshttp://www-graphics.stanford.edu/~seander/bithacks.html#BitReverseObv... > > - > Azhar. > > On Wed, Dec 22, 2010 at 2:09 PM, pacific pacific <[email protected]>wrote: > > > > > > > > > > > On Wed, Dec 22, 2010 at 12:11 PM, mo...@ismu <[email protected]> wrote: > > >> if x is a 32 bit number > >> if((x&0x0000FFFF)==((x>>16)&0x0000FFFF))) > >> x's bit pattern is a polyndrome > > >> @snehal :Do you want to consider binary representation of 5 as 101 or > > 0000..0101 ? > > -- > > >> You received this message because you are subscribed to the Google Groups > >> "Algorithm Geeks" group. > >> To post to this group, send email to [email protected]. > >> To unsubscribe from this group, send email to > >> [email protected]<algogeeks%2bunsubscr...@googlegroups > >> .com> > >> . > >> For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to [email protected]. > > To unsubscribe from this group, send email to > > [email protected]<algogeeks%2bunsubscr...@googlegroups > > .com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" 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/algogeeks?hl=en.
