1. $S(x)$ denotes the sum of the digits of the natural number $x$. Then the set of $x$ that satisfies $x+S(x)+S(S(x))=2014$
is $\qquad$ .
$-1 . \varnothing$.
For any natural number $x$, $x+S(x)+S(S(x))$ is divisible by 3, and 2014 leaves a remainder of 1 when divided by 3. Therefore, the solution set is $\varnothing$.