Three. (25 points) Given that $\sqrt{1-|x|}+\sqrt{2-2|y|}$ is meaningful, $M=\sqrt{x^{2}+2 x y+y^{2}}+\sqrt{x^{2}+2 x+1}+|2 x-y-4|$ has maximum and minimum values of $a$ and $b$ respectively. If the five-digit integer $1 a b c d$ is divisible by 99, find the values of $c$ and $d$.

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Given that $\sqrt{1-|x|}+\sqrt{2-2|y|}$ is meaningful, $M=\sqrt{x^{2}+2 x y+y^{2}}+\sqrt{x^{2}+2 x+1}+|2 x-y-4|$ has maximum and minimum values of $a$ and $b$ respectively. If the five-digit integer $1 a b c d$ is divisible by 99, find the values of $c$ and $d$.
Three, it is known that $1-|x| \geqslant 0, 2-2|y| \geqslant 0$.
Therefore, $-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1$.
Then $-2 \leqslant x+y \leqslant 2, 0 \leqslant x+1 \leqslant 2$,
$$
-7 \leqslant 2 x-y-4 \leqslant -1 \text{. }
$$

When $x+y \geqslant 0$,
$$
\begin{array}{l}
M=|x+y|+|x+1|+|2 x-y-4| \\
=x+y+x+1-2 x+y+4=2 y+5 .
\end{array}
$$

By $3 \leqslant 2 y+5 \leqslant 7$, we get
$$
3 \leqslant M \leqslant 7 \text{. }
$$

When $x+y \leqslant 0$,
$$
\begin{array}{l}
M=|x+y|+|x+1|+|2 x-y-4| \\
=-x-y+x+1-2 x+y+4=-2 x+5 .
\end{array}
$$

By $3 \leqslant -2 x+5 \leqslant 7$, we get $3 \leqslant M \leqslant 7$.
When $x=1, y=1$, $M=7$.
When $x=1, y=-1$, $M=3$.
Therefore, the maximum value of $M$ is 7, and the minimum value is 3, i.e., $a=7, b=3$.
Since the five-digit integer $\overline{1 a b c d}$ can be divided by 99, $99=9 \times 11$, and 9 and 11 are coprime, so $\overline{173 c d}$ can be divided by 9 and 11.
Thus, $1+7+3+c+d$ can be divided by 9.
Therefore, $2+c+d$ can be divided by 9.
Since $2 \leqslant 2+c+d \leqslant 20$, then
$2+c+d=9$ or 18.
Hence, $c+d=7$ or 16.
Also, $1-7+3-c+d$ can be divided by 11, then $-3-c+d$ can be divided by 11.
Since $-12 \leqslant -3-c+d \leqslant 6$, then
$-3-c+d=-11$ or 0.
Hence, $-c+d=-8$ or 3.
Noting that the parity of $c+d$ and $-c+d$ is the same, from equations (1) and (2) we get
$$
\left\{\begin{array}{l}
c+d=7, \\
-c+d=3
\end{array} \text{ or } \left\{\begin{array}{l}
c+d=16, \\
-c+d=-8
\end{array}\right.\right.
$$

Solving, we get $\left\{\begin{array}{l}d=5, \\ c=2\end{array}\right.$ or $\left\{\begin{array}{l}d=4, \\ c=12\end{array}\right.$ (not valid, discard).
Therefore, $c=2, d=5$.