Let $n$ be a positive integer. A \emph{pseudo-Gangnam Style} is a dance competition between players $A$ and $B$. At time $0$, both players face to the north. For every $k\ge 1$, at time $2k-1$, player $A$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $B$ is forced to follow him; at time $2k$, player $B$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $A$ is forced to follow him.

After time $n$, the music stops and the competition is over. If the final position of both players is north or east, $A$ wins. If the final position of both players is south or west, $B$ wins. Determine who has a winning strategy when:

(a) $n=2013^{2012}$

(b) $n=2013^{2013}$
To determine who has a winning strategy, we need to analyze the final position of both players after \( n \) moves. The key observation is that the net rotation of both players depends on the parity of \( n \).

1. **Understanding the Moves:**
   - At odd times \( 2k-1 \), player \( A \) can either stay stationary or turn \( 90^\circ \) clockwise, and player \( B \) follows.
   - At even times \( 2k \), player \( B \) can either stay stationary or turn \( 90^\circ \) clockwise, and player \( A \) follows.

2. **Net Rotation Analysis:**
   - Each pair of moves (one by \( A \) and one by \( B \)) results in a net rotation of \( 0^\circ \) or \( 90^\circ \) clockwise.
   - If \( A \) and \( B \) both choose to turn \( 90^\circ \) clockwise, the net rotation is \( 180^\circ \).
   - If one chooses to turn and the other stays, the net rotation is \( 90^\circ \).
   - If both stay, the net rotation is \( 0^\circ \).

3. **Modulo 4 Analysis:**
   - The direction faced by the players after \( n \) moves depends on \( n \mod 4 \):
     - \( n \equiv 0 \pmod{4} \): Final direction is North.
     - \( n \equiv 1 \pmod{4} \): Final direction is East.
     - \( n \equiv 2 \pmod{4} \): Final direction is South.
     - \( n \equiv 3 \pmod{4} \): Final direction is West.

4. **Case (a): \( n = 2013^{2012} \)**
   - Calculate \( 2013^{2012} \mod 4 \):
     \[
     2013 \equiv 1 \pmod{4} \implies 2013^{2012} \equiv 1^{2012} \equiv 1 \pmod{4}
     \]
   - Since \( 2013^{2012} \equiv 1 \pmod{4} \), the final direction is East.
   - Therefore, \( A \) wins.

5. **Case (b): \( n = 2013^{2013} \)**
   - Calculate \( 2013^{2013} \mod 4 \):
     \[
     2013 \equiv 1 \pmod{4} \implies 2013^{2013} \equiv 1^{2013} \equiv 1 \pmod{4}
     \]
   - Since \( 2013^{2013} \equiv 1 \pmod{4} \), the final direction is East.
   - Therefore, \( A \) wins.

The final answer is \( \boxed{ A } \) wins in both cases.