Read the question properly. There are total n moves and A uses n/2 of these moves and B n/2. If n is odd then A gets one more move than B as he starts first
a=a*pow(2,(n/2)); b= b*(pow(2,(n/2)));
One word. Overflow. Please instead of calculating exactly whats asked, try and see if this fraction can be reduced into a prettier/easy/simpler form.
Got your point. But still not getting AC for subtask 2. Should i change my approach to the question?
New Solution -
Sorry i didn’t see the constraints earlier. Try what vijju said.