Five. (20 points) Given that $a$, $b$, and $c$ are real numbers, and 
$$
\sqrt{2 x^{2} + a x + b} > x - c
$$
holds if and only if $x \leqslant 0$ or $x > 1$.

Find the range of values for $c$.
From the problem, we get the inequality
$$
\sqrt{2 x^{2}+a x+b}>x-c
$$

The solution set is $x \leqslant 0$ or $x>1$, and equation (1) is equivalent to
$$
\left\{\begin{array}{l}
2 x^{2}+a x+b \geqslant 0, \\
2 x^{2}+a x+b>(x-c)^{2}
\end{array}\right.
$$

or $\left\{\begin{array}{l}2 x^{2}+a x+b \geqslant 0, \\ x-c(x-c)^{2}\right\} \\
=\left\{x \mid xy_{2}\right\},
\end{array}
$$

where $y_{1} 、 y_{2}$ are the two roots of the equation $2 x^{2}+a x+b=(x-c)^{2}$ $\left(y_{1} \leqslant y_{2}\right) ;$
$$
C=\{x \mid x-c1\} .
\end{array}
$$

Thus, substituting $x_{1}=0, y_{2}=1$ into the equation, we get
$$
\begin{array}{l}
b=0, x_{2}=-\frac{a}{2}>0, \\
a=(c-1)^{2}-2, y_{1}=-c^{2} .
\end{array}
$$

Then $A=\left\{x \mid x \leqslant 0\right.$ or $\left.x \geqslant-\frac{a}{2}\right\}$,
$$
B \cup C=\left\{x \mid x1\right\} \text {. }
$$

Therefore, $0<\max \left\{-c^{2}, c\right\} \leqslant-\frac{a}{2} \leqslant 1$.
Thus, $0<c \leqslant 1$.