1. If $\{x\}=x-[x]([x]$ represents the greatest integer not exceeding $x$), then the number of real solutions to the equation $2005 x+\{x\}=\frac{1}{2006}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
- 1.C.

Since $x=[x]+\{x\}$, we have
$$
2005[x]+2006\{x\}=\frac{1}{2006} \text {. }
$$

Also, $0 \leqslant 2006\{x\}<2006$, so, $[x]=0$ or -1.
If $[x]=0$, then $\{x\}=\frac{1}{2006^{2}}$, i.e., $x=\frac{1}{2006^{2}}$;
If $[x]=-1$, then $2006\{x\}=2005+\frac{1}{2006}$, i.e.,
$$
\begin{array}{l}
\{x\}=\frac{2005}{2006}+\frac{1}{2006^{2}}=\frac{2005 \times 2006+1}{2006^{2}} \\
=\frac{2006^{2}-2005+1}{2006^{2}}=1-\frac{2005}{2006^{2}}<1 .
\end{array}
$$

Thus, $x=-1+1-\frac{2005}{2006^{2}}=-\frac{2005}{2006^{2}}$.