Problem 1266

【Analysis】Answer: Player A has a winning strategy.
A's strategy is to subtract the smaller real root of the original equation from the constant term and add 1 to the coefficient of the linear term in each operation.

The following proof shows that under A's strategy, B will inevitably choose the first option after a finite number of rounds.

Let the two roots of the equation before A's operation be $x_{1}$ and $x_{2}$ $\left(x_{1} \leqslant x_{2}\right)$, at this point, the coefficient of the linear term $p=-\left(x_{1}+x_{2}\right)$, then the equation can be expressed as
$$\begin{array}{l} \left(x-x_{1}\right)\left(x-x_{2}\right) \\ =x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}=0 . \end{array}$$

The new equation after A's operation is
$$x^{2}-\left(x_{1}+x_{2}-1\right) x+x_{1} x_{2}-x_{1}=0 \text {. }$$

At this point, the roots of the new equation are $x_{1}$ and $x_{2}-1$, and the corresponding coefficient of the linear term is $p+1$, meaning that the two roots of the new equation obtained after A's operation are simply the larger root of the original equation reduced by 1 while the smaller root remains unchanged, and the coefficient of the linear term is increased by 1.

If B chooses the second option next, let the two roots of the equation before B's operation be $x_{1}^{\prime}$ and $x_{2}^{\prime}\left(x_{1}^{\prime} \leqslant x_{2}^{\prime}\right)$, at this point, the coefficient of the linear term $p^{\prime}=-\left(x_{1}^{\prime}+x_{2}^{\prime}\right)$, then the equation can be expressed as
$$\begin{array}{l} \left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right) \\ =x^{2}-\left(x_{1}^{\prime}+x_{2}^{\prime}\right) x+x_{1}^{\prime} x_{2}^{\prime}=0, \end{array}$$

The new equation after B's operation is
$$x^{2}-\left(x_{1}^{\prime}+x_{2}^{\prime}+1\right) x+x_{1}^{\prime} x_{2}^{\prime}+x_{2}^{\prime}=0 \text {. }$$

At this point, the roots of the new equation are $x_{1}^{\prime}+1$ and $x_{2}^{\prime}$, and the corresponding coefficient of the linear term is $p^{\prime}-1$, meaning that after B's operation, the smaller root of the equation increases by 1 while the larger root remains unchanged, and the coefficient of the linear term returns to the coefficient of the linear term $p$ before A's operation in this round.

Thus, if B always chooses the second option, then the difference between the two roots of the equation will inevitably be no more than 1 after a finite number of rounds (even if the equation remains so after another operation by A). Therefore, in the next round, B can only choose the first option. At this point, the coefficient of the linear term is $p+1$.

Thereafter, if A adopts the same strategy, B will still not be able to always choose the second option, so after a finite number of rounds, B will eventually choose the first option. At this point, the coefficient of the linear term is $p+2$.
If the game continues without A winning, the coefficient of the linear term will inevitably become a non-negative real number at some point, at which point A will still win.
Therefore, under the aforementioned strategy, A will inevitably win.