Three. (25 points) As shown in Figure 5, a particle moves within the region $\{(x, y) \mid x \geqslant 0, y \geqslant 0\}$. In the first second, it moves from the origin to point $B_{1}(0,1)$, then from point $B_{1} \rightarrow C_{1}$ $\rightarrow A_{1}$, and subsequently moves along the $x$-axis, $y$-axis, and their parallel lines in the direction indicated by the arrows in the figure, moving 1 unit length per second. Find the time required for the particle to move from the origin to point $P(16,44)$.
Three, suppose the time taken for a particle to reach points $A_{n}, B_{n}, C_{n}$ from the origin are $a_{n}, b_{n}, c_{n}$, respectively. Then it is clear that
$$
\begin{array}{l}
a_{1}=3, a_{2}=a_{1}+1, \\
a_{3}=a_{1}+12=a_{1}+3 \times 4, a_{4}=a_{3}+1, \\
a_{5}=a_{3}+20=a_{3}+5 \times 4, a_{6}=a_{5}+1, \\
\cdots \cdots \\
a_{2 n-1}=a_{2 n-3}+(2 n-1) \times 4, a_{2 n}=a_{2 n-1}+1 . \\
\text { Hence } a_{2 n-1}=a_{1}+4[3+5+\cdots+(2 n-1)]=4 n^{2}-1, \\
a_{2 n}=a_{2 n-1}+1=4 n^{2}, \\
b_{2 n-1}=a_{2 n-1}-2(2 n-1)=4 n^{2}-4 n+1, \\
b_{2 n}=a_{2 n}+2 \times 2 n=4 n^{2}+4 n, \\
c_{2 n-1}=b_{2 n-1}+(2 n-1)=4 n^{2}-2 n \\
=(2 n-1)^{2}+(2 n-1), \\
c_{2 n}=a_{2 n}+2 n=4 n^{2}+2 n=(2 n)^{2}+2 n .
\end{array}
$$

Therefore, $c_{n}=n^{2}+n$.
From Figure 5, we know that the time taken for the particle to move from the origin to point $P(16,44)$ is the time taken to reach point $C_{4}$ plus $44-16$ $=28$ (s), so,
$$
t=44^{2}+44+28=2008(\mathrm{~s}) \text {. }
$$