no i didn't mean that in first test u checking if count of X should be either equal of one more than that of O
and in last u r checking if both are winning or only only one but what i meant is if O has already won but no of moves of X are greater than O the answer should be No but your solution will print yes Correct me if i m wrong Test case: XXO XXO ..O Ans: No tell me what is your output for this case On Fri, Jun 17, 2011 at 1:56 PM, KK <kunalkapadi...@gmail.com> wrote: > @sunny: > This test: > if(! ( (countx == counto + 1) || (countx == counto) ) ) > cout << "no" << endl; > prints no if countx > counto > > and this one > if(o && x) > cout << "no" << endl; > else > cout << "yes" << endl; > > prints no if both have won or else yes.... > correct me if m wrong... > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- Sunny Aggrawal B-Tech IV year,CSI Indian Institute Of Technology,Roorkee -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
