When $n$ is not a perfect power of $2$ then $A$ wins. If $n$ is a perfect power of $2,$ then $B$ wins.

Solved with PC Claim: If there are an odd number of stones then $A$ wins. Because then $A$ takes $1$ stone and then $B$ is forced to take only $1$ stone. And then $A$ takes $1$ stone and so on. Since we have odd number of stones. $A$ will be the last person to take the stone, hence $A$ wins. Claim: If $v_2(n)=1,~~n\ne 2,$ then $A$ wins. $A$ starts with taking $2$ stones. Now, $B$ won't take $1$ stone, because then $A$ will get odd stones, and we know having odd stones means we can win. Hence $B$ will take $2$ stones. Since $v_2(n)=1,$ we know that $A$ will be the last person to take the stone. Claim: If $v_2(n)=2,n\ne 4$ then $A$ wins $A$ starts with taking $4$ stones. Now, $B$ knows that if $A$ gets a stones with $v_2(n)=0,1;$ $A$ wins. Hence $B$ will take $4$ stones and will try to maintain $v_2(n)=2.$ But then we know $n=4k,~k\ne 1 $ odd. Hence, $A$ will win. So this gives induction vibes. Claim: If $n$ not a perfect power of $2,$ $A$ wins. We prove it using induction. Consider $v_2(n).$ For $v_2(n)=0,1,2$ $A$ wins. Suppose $A$ wins for $v_2(n)=0,1,\dots k.$ Now we will show that $A$ wins for $v_2(n)=k+1\implies n=2^{k+1}\cdot m,~~m>1$ is odd. $A$ will subtract $2^{k+1}.$ $B$ knows that $A$ will win if $A$ has $v_2(n)<k+1.$ Hence $B$ will try to maintain $v_2(n)=k+1.$ So $B$ subtarcts $2^{k+1}.$ And then again $A$ subtracts $2^{k+1}$ and so on. But $A$ will win since $m>1$ is odd. And we are done. Claim: When $n$ is a perfect power of $2.$ Then $B$ wins. If $n=2^k,$ $A$ has to take less than $2^{k-1}$ stones because if not in the next move $B$ can take everything. But then whatever $A$ takes, $B$ will receive a "non-perfect power of $2$" number of stones. Which we know is a winnable situation. So $B$ wins.