11. Given that the domain of the function $f(x)$ is $\mathbf{R}$, for any real number $x$, it always holds that $f(1+x)=f(3-x)$, and $f(2+x)=-f(1-x)$. Then $f(1)+f(2)+\cdots+f(100)=$ $\qquad$ .
11. 0. From $f(1+x)=f(3-x)$ we get $f(4-x)=f(x)$, so $f(x+2)=-f(x)$, which gives a period $T=4$. From $f(x+2)+f(x)=0$ we get $f(1)+f(3)=0$, $f(2)+f(4)=0, \cdots$, hence $f(1)+f(2)+\cdots+f(100)=0$.