Given a positive integer $c$, let the sequence $x_{1}, x_{2}, \cdots$ satisfy $x_{1}=$ $c$, and $x_{n}=x_{n-1}+\left[\frac{2 x_{n-1}-(n+2)}{n}\right]+1, n=2,3, \cdots$, where $[x]$ denotes the greatest integer not exceeding $x$. Find the general term formula for the sequence $\left\{x_{n}\right\}$.
(Huang Yumin)
Obviously, when $n \geqslant 2$,
$$
x_{n}=x_{n-1}+\left[\frac{2\left(x_{n-1}-1\right)}{n}\right] \text {. }
$$

Let $a_{n}=x_{n}-1$, then
$$
\begin{array}{l}
a_{1}=c-1, \\
a_{n}=a_{n-1}+\left[\frac{2 a_{n-1}}{n}\right]=\left[\frac{n+2}{n} \cdot a_{n-1}\right], n=2,3, \cdots .
\end{array}
$$

Let $u_{n}=A \cdot \frac{(n+1)(n+2)}{2}, n=1,2, \cdots, A$ be a non-negative integer. Since when $n \geqslant 2$,
$$
\begin{array}{l}
{\left[\frac{n+2}{n} \cdot u_{n-1}\right]=\left[A \cdot \frac{(n+2)}{2 n} \cdot n(n+1)\right]} \\
=A \cdot \frac{(n+1)(n+2)}{2}=u_{n},
\end{array}
$$

Therefore, the sequence $\left\{u_{n}\right\}$ satisfies equation (1).
Let $y_{n}=n, n=1,2, \cdots$. Since when $n \geqslant 2$,
$$
\begin{array}{l}
{\left[\frac{n+2}{n} \cdot y_{n-1}\right]=\left[\frac{(n+2)(n-1)}{2}\right]} \\
=\left[n^{2}+1-\frac{2}{n}\right]=n^{2}=y_{n},
\end{array}
$$

Therefore, $\left\{y_{n}\right\}$ also satisfies equation (1).
Let $z_{n}=\left[\frac{(n+2)^{2}}{4}\right], n=1,2, \cdots$.
When $n=2 m$ and $m \geqslant 1$,
$$
\begin{array}{l}
{\left[\frac{n+2}{n} \cdot z_{n-1}\right]=\left[\frac{m+1}{m} \cdot\left[\frac{(2 m+1)^{2}}{4}\right]\right]} \\
=\left[\frac{m+1}{m} \cdot m(m+1)\right]=(m+1)^{2}=z_{n} .
\end{array}
$$

When $n=2 m+1$ and $m \geqslant 1$,
$$
\begin{array}{l}
{\left[\frac{n+2}{n} \cdot z_{n-1}\right]=\left[\frac{2 m+3}{2 m+1} \cdot\left[\frac{(2 m+2)^{2}}{4}\right]\right]} \\
=\left[\frac{2 m+3}{2 m+1}(m+1)^{2}\right] \\
=\left[(m+1)(m+2)+\frac{m+1}{2 m+1}\right] \\
=(m+1)(m+2)=\left[\frac{(2 m+3)^{2}}{4}\right]=z_{n} .
\end{array}
$$

Thus, $\left\{z_{n}\right\}$ also satisfies equation (1).
For any non-negative integer $A$, let
$$
\begin{array}{l}
v_{n}=u_{n}+y_{n}=A \cdot \frac{(n+1)(n+2)}{2}+n, \\
w_{n}=u_{n}+z_{n}=A \cdot \frac{(n+1)(n+2)}{2}+\left[\frac{(n+2)^{2}}{4}\right],
\end{array}
$$
$n=1,2, \cdots$, clearly $\left\{v_{n}\right\}$ and $\left\{w_{n}\right\}$ both satisfy equation (1).
Since $u_{1}=3 A, y_{1}=1, z_{1}=\left[\frac{9}{4}\right]=2$, therefore,
when $3 \mid a_{1}$, $a_{n}=\frac{a_{1}}{6}(n+1)(n+2)$;
when $a_{1} \equiv 1(\bmod 3)$,
$a_{n}=\frac{a_{1}-1}{6}(n+1)(n+2)+n ;$
when $a_{1} \equiv 2(\bmod 3)$,
$a_{n}=\frac{a_{1}-2}{6}(n+1)(n+2)+\left[\frac{(n+2)^{2}}{4}\right]$.
In summary,
when $c \equiv 1(\bmod 3)$, $x_{n}=\frac{c-1}{6}(n+1)(n+2)+1$;
when $c \equiv 2(\bmod 3)$, $x_{n}=\frac{c-2}{6}(n+1)(n+2)+n+1$;
when $c \equiv 0(\bmod 3)$, $x_{n}=\frac{c-3}{6}(n+1)(n+2)+$ $\left[\frac{(n+2)^{2}}{4}\right]+1$