Five, let $f(x)$ be a function that satisfies the following conditions:
(1) If $x>y$ and $f(x)+x \geqslant w \geqslant f(y)+y$, then there exists a real number $z \in[y, x]$, such that $f(z)=w-z$;
(2) The equation $f(x)=0$ has at least one solution, and among these solutions, there is one that is not greater than all the others;
(3) $f(0)=1$;
(4) $f(-1999) \leqslant 2000$;
(5)
$$
\begin{array}{l}
f(x) f(y) \\
=f(x f(y)+y f(x)+x y) .
\end{array}
$$

Find the value of $f(-1999)$.
Let $F(x)=f(x)+x$, then $F(0)=1$.
Let $u$ be the smallest root of $f(x)=0$, then $F(u)=u$.
If $u0$
For any real number $x$, by (5) we have
$$
\begin{aligned}
0 & =f(x) f(u)=f(x f(u)+u f(x)+x u) \\
& =f(u f(x)+u u) .
\end{aligned}
$$
$\therefore u f(x)+x u$ is a root of $f(x)=0$.
$\because u$ is the smallest root of $f(x)=0$,
$$
\begin{array}{l}
\therefore u f(x)+x u \geqslant u . \\
\text { Thus, } f(x)+x \geqslant 1,
\end{array}
$$

i.e., $f(x) \geqslant 1-x$.
$$
\begin{array}{l}
\therefore f(-1999) \geqslant 2000 . \\
\text { Also, } \because f(-1999) \leqslant 2000, \\
\therefore f(-1999)=2000 .
\end{array}
$$