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We can help only after competition is over. Sorry.
but do pls clear my doubt after the competition
thanx for the reply
Hey. The Editorial for the problem is here. You can check it out:
EQUALCOIN - Editorial - editorial - CodeChef Discuss
There had to be five cases. Below is my solution
So many cases are not required.
if( x%2 | ( !x & y%2 ) ) cout<<"NO"<<"\n"; else cout<<"YES"<<"\n";
I think this should suffice just 3 conditions.