11. (2002 National High School Competition Question) Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$, and for any $x \in \mathbf{R}$, $f(x+5) \geqslant f(x)+5$, $f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$ .
11. Fill 1. Reason: From $g(x)=f(x)+1-x$ we get $f(x)=g(x)+x-1$, then $g(x+5)+(x+5)-1 \geqslant$ $g(x)+(x-1)+5, g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1$. Therefore, $g(x+5) \geqslant g(x), g(x+1) \leqslant$ $g(x)$, thus $g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x)$, hence $g(x+1)=g(x)$, i.e., $g(x)$ is a periodic function, and 1 is one of its periods. Also, $g(1)=1$, so $g(2002)=1$.