II. (50 points)
$n$ is a positive integer, let the integer closest to $\sqrt{n}$ be $a_{n}$, let $b_{n}=n+a_{n}$, remove $b_{n}$ from the set of all positive integers, and arrange the remaining positive integers in ascending order to form the sequence $\left\{c_{n}\right\}$, try to express $c_{n}$ in terms of $n$.
1. Prove $b_{n} \neq p^{2}\left(p \in \mathbb{Z}^{+}\right)$

Assume $b_{n}=p^{2}$, i.e., $p^{2}=n+\left[\frac{1}{2}+\sqrt{n}\right]$.
Let $\left[\frac{1}{2}+\sqrt{n}\right]=m$, then
$$
\begin{array}{l}
m \leqslant \frac{1}{2}+\sqrt{n}<m+1 \\
m-\frac{1}{2} \leqslant \sqrt{n}<m+\frac{1}{2} \\
m^{2}-m+\frac{1}{4} \leqslant n<m^{2}+m+\frac{1}{4}
\end{array}
$$

So $m^{2}-m+1 \leqslant n<m^{2}+m$.
So $m^{2}+1 \leqslant p^{2}=n+m<m^{2}+2 m<(m+1)^{2}$, contradiction.
2. On the other hand, it can be proven that $b_{n}=p^{2}+k, 1 \leqslant k \leqslant 2 p, p \in \mathbb{Z}^{+}$.

When $n=p^{2}$, $b_{n}=p^{2}+\left[\frac{1}{2}+\sqrt{p^{2}}\right]=p^{2}+p$.
When $n=p^{2}+q(-p+1 \leqslant q \leqslant p)$,
$$
\begin{aligned}
b_{n} & =p^{2}+q+\left[\frac{1}{2}+\sqrt{p^{2}+q}\right] \\
& =p^{2}+p+q=p^{2}+k(1 \leqslant k \leqslant 2 p) .
\end{aligned}
$$

From 1 and 2, it can be concluded that $b_{n}$ contains all non-square positive integers,
so $c_{n}=n^{2}$.