Let $f$ be a function from $R \rightarrow R$, and
(1) for any $x, y \in R$,
$$
f(x)+f(y)+1 \geqslant f(x+y) \geqslant f(x)+f(y) ;
$$
(2) for any $x \in [0,1)$, $f(0) \geqslant f(x)$;
$$
\text{(3) } -f(-1)=f(1)=1 \text{.}
$$

Find all functions that satisfy the conditions.
First, prove that $f(0)=0$. From (1) and (2), we can deduce
$$
2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0),
$$

thus $f(0) \geqslant 0 \geqslant f(0)$, which means $f(0)=0$.
Next, prove that for any $x \in(0,1), f(x)=0$. From (2), for such an $x, f(x) \leqslant 0$. If $f(x)f(x)+f(1-x)+1 \geqslant f(1)=1$,
$$

a contradiction. Therefore, $f(x)=0$.
Finally, prove that for any $x \in R, f(x+1)=f(x)+1$. This leads to: for $x \in R, f(x)=[x]$. From (1), we can deduce
$$
f(x+1) \geqslant f(x)+f(1)=f(x)+1 \text {. }
$$

On the other hand, from (1) and (3), we get
$$
f(x) \geqslant f(x+1)+f(-1)=f(x+1)-1 \text {. }
$$

Therefore, $f(x+1)=f(x)+1$, for any $x \in R$.