12. Given the function $f(x)=a+x-b^{x}$ has a zero $x_{0} \in(n, n+1)(n \in \mathbf{Z})$, where the constants $a, b$ satisfy the conditions $2019^{a}=2020,2020^{b}=2019$, then the value of $n$ is
Solve
$2019^{a b}=2020^{b}=2019 \Rightarrow a b=1, a>1,00, \\
f(-1)=a-1-b^{-1}=-1<0
\end{array} \Rightarrow x_{0} \in(-1,0) \Rightarrow n=-1\right.
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