1. The quadratic trinomial
$$
x^{2}+a x+b(a 、 b \in \mathbf{N})
$$

has real roots, and $a b=2^{2011}$. Then the number of such quadratic trinomials is $\qquad$.
$-1.1341$.
It is known that the numbers $a$ and $b$ are powers of 2 with non-negative integer exponents, i.e., $a=2^{k}, b=2^{2011-k}$. Therefore, we have
$$
\begin{array}{l}
\Delta=a^{2}-4 b \geqslant 0 \\
\Rightarrow 2 k \geqslant 2013-k \Rightarrow k \geqslant \frac{2013}{3}=671 .
\end{array}
$$

But $k \leqslant 2011$, so $k$ can take
$$
2011-671+1=1341
$$

different integer values.
Each $k$ corresponds exactly to one of the required quadratic trinomials, hence there are 1341 such quadratic trinomials.