Let $n$ be a positive integer, the players A and B play the following game: we have $n$ balls with the numbers of $1, 2, 3, 4,...., n$ this balls will be in two boxes with the symbols $\prod$ and $\sum$.
In your turn, the player can choose one ball and the player will put this ball in some box, in the final all the balls of the box $\prod$ are multiplied and we will get a number $P$, after this all the balls of the box $\sum$ are added up and we will get a number $Q$(if the box $\prod$ is empty $P = 1$, if the box $\sum$ is empty $Q = 0$).
The player(s) play alternately, player A starts, if $P + Q$ is even player A wins, otherwise player B wins.

a)If $n= 6$, which player has the winning strategy???
b)If $n = 2012$, which player has the winning strategy???
To determine the winning strategy for players A and B, we need to analyze the game based on the parity (odd or even nature) of the numbers involved and the final sum \( P + Q \).

### Part (a): \( n = 6 \)

1. **Initial Setup**:
   - The numbers are \( 1, 2, 3, 4, 5, 6 \).
   - Player A starts the game.

2. **Parity Analysis**:
   - The numbers can be represented by their parity: \( 1, 0, 1, 0, 1, 0 \) (where \( 1 \) represents odd and \( 0 \) represents even).

3. **Strategy**:
   - The key observation is that the sum \( P + Q \) will be even if and only if the number of odd numbers in the product box \(\prod\) is even.
   - Player A can always ensure that the number of odd numbers in the product box \(\prod\) is even by placing the first odd number in the sum box \(\sum\).

4. **Game Play**:
   - Player A places the first odd number (1) in the sum box \(\sum\).
   - Player B then has to place a number, but no matter what Player B does, Player A can always respond in a way to maintain the parity of the product box \(\prod\) such that the number of odd numbers in \(\prod\) remains even.

5. **Conclusion**:
   - Since Player A can always ensure that the number of odd numbers in the product box \(\prod\) is even, \( P + Q \) will be even, and Player A will win.

### Part (b): \( n = 2012 \)

1. **Initial Setup**:
   - The numbers are \( 1, 2, 3, \ldots, 2012 \).
   - Player A starts the game.

2. **Parity Analysis**:
   - The numbers can be represented by their parity: \( 1, 0, 1, 0, \ldots, 1, 0 \) (alternating between odd and even).

3. **Strategy**:
   - Similar to the case when \( n = 6 \), the key observation is that the sum \( P + Q \) will be even if and only if the number of odd numbers in the product box \(\prod\) is even.
   - Since \( n = 2012 \) is even, there are 1006 odd numbers and 1006 even numbers.

4. **Game Play**:
   - Player A places the first odd number (1) in the sum box \(\sum\).
   - Player B then has to place a number, but no matter what Player B does, Player A can always respond in a way to maintain the parity of the product box \(\prod\) such that the number of odd numbers in \(\prod\) remains even.

5. **Conclusion**:
   - Since Player A can always ensure that the number of odd numbers in the product box \(\prod\) is even, \( P + Q \) will be even, and Player A will win.

The final answer is Player A wins for both \( \boxed{ n = 6 } \) and \( n = 2012 \).