14 Let $f(n)$ satisfy $f(0)=0, f(n)=n-f(f(n-1)), n=1$,
$2,3, \cdots$. Determine all real-coefficient polynomials $g(x)$ such that
$$
f(n)=[g(n)], n=0,1,2, \cdots,
$$

where $[g(n)]$ denotes the greatest integer not exceeding $g(n)$.
No answer: $g(x)=\frac{1}{2}(\sqrt{5}-1)(x+1)$.
(1) First prove
$$
f(n)=\left[\frac{1}{2}(\sqrt{5}-1)(n+1)\right] .
$$

Let $\alpha=\frac{1}{2}(\sqrt{5}-1)$. It is sufficient to prove that for $n=1,2, \cdots$,
$$
[(n+1) \alpha]=n-[([n \alpha]+1) \alpha] .
$$

Let $n \alpha=[n \alpha]+\delta$, then (note that $\alpha^{2}+\alpha=1$)
$$
[(n+1) \alpha]=[[n \alpha]+\delta+\alpha]=[n \alpha]+[\delta+\alpha],
$$
$[([n \alpha]+1) \alpha]=[(n \alpha-\delta+1) \alpha]=\left[n \alpha^{2}-\delta \alpha+\alpha\right]$
$$
\begin{array}{l}
=[n-n \alpha-\delta \alpha+\alpha]=[n-[n \alpha]-\delta-\delta \alpha+\alpha] \\
=n-[n \alpha]+[-\delta-\delta \alpha+\alpha] \\
=n-[(n+1) \alpha]+[\delta+\alpha]+[-\delta-\delta \alpha+\alpha],
\end{array}
$$

Therefore, it is sufficient to prove $[\delta+\alpha]+[-\delta-\delta \alpha+\alpha]=0$.
Case 1. $\delta+\alpha(\alpha-1)(\alpha+1)+\alpha$ $=\alpha^{2}+\alpha-1=0-\delta-\delta \alpha+\alpha \leqslant \alpha-(1+\alpha)+\alpha=-1 .
$$

Therefore, $[-\delta-\delta \alpha+\alpha]=-1=-[\delta+\alpha]$.
That is, $[\delta+\alpha]+[-\delta-\delta \alpha+\alpha]=0$.
So $g(x)=(n+1) \alpha$ meets the requirement.
(2) Prove $g(x)=a x+b, a, b \in \mathbf{R}, a \neq 0$.
Let $g(x)=a_{k} x^{k}+a_{k-1} x^{k-1}+\cdots+a_{1} x+a_{0}, a_{k} \neq 0$. By the condition $[g(n)]=n-[g(g(n-1))]$.

Therefore,
$$
\frac{[g(n)]}{n^{k^{2}}}=\frac{n}{n^{k^{2}}}-\frac{[g(g(n-1))]}{n^{k^{2}}} \text {. }
$$

If $k \geqslant 2$, then let $n \rightarrow \infty$, we have $0=0-d_{k}^{k+1}$, which contradicts $a_{k} \neq 0$. Therefore, $k \leqslant 1$. Also, by (*) we know that $g(x)$ cannot be a constant, so $k=1$, i.e., $g(x)=a_{1} x+a_{0}, a_{1} \neq 0$. (3) Prove $0 \leqslant f(n) \leqslant n$.
Use induction on $n$:
1) $n=0$ is obviously true
2) Assume it is true for $n-1$, now consider $n$. By the induction hypothesis $0 \leqslant f(n-1) \leqslant n-1$,
$$
0 \leqslant f(f(n-1)) \leqslant f(n-1) \leqslant n-1 .
$$

Therefore, $1 \leqslant f(n) \leqslant n$.
By mathematical induction, for all $n$, we have $0 \leqslant f(n) \leqslant n$.
(4) Prove $g(x)=a x+b, b \in \mathbf{R}$.
By (2) we know $g(x)=a x+b, a, b \in \mathbf{R}, a \neq 0$. In (**) let
$n \rightarrow \infty$, we have $a=1-a^{2}$. By (3) we know $a>0$, so $a=\frac{-1+\sqrt{5}}{2}=\alpha$. That is, $g(x)=\alpha x+b$.
(5) Prove $g(x)=\alpha x+\alpha$.
$$
\begin{aligned}
f(n) & =[g(n)]=[\alpha n+b]=[\alpha(n+1)+b-\alpha] \\
& =[\alpha(n+1)]+[\{\alpha(n+1)\}+b-a],
\end{aligned}
$$

By (1) we know that for all $n \geqslant 0$, $[\{\alpha(n+1)\}+b-\alpha]=0$.
Let $n=0$ we know $0 \leqslant b \leqslant 1$.
If $b-\alpha>0$, we will prove that there exists $n$ such that
$$
\{\alpha(n+1)\}>1-\frac{1}{2}(b-\alpha) .
$$

At this time, $[\{\alpha(n+1)\}+b-\alpha] \geqslant 1$.
To prove this, we need to prove the following lemma.
Lemma: Let $\alpha=\frac{1}{2}(\sqrt{5}-1), \varepsilon>0$, then there exist positive integers $n_{1} 、 n_{2}$ such that $\left\{n_{1} \alpha\right\}1-\varepsilon$.
Proof: Let $N$ be an integer, $\frac{1}{N}<\varepsilon$. Consider the set $\{k \alpha \mid k=1,2, \cdots, N\}$. By the pigeonhole principle, there exist integers $k, l$ such that $1 \leqslant k, l \leqslant N$ and $|k \alpha-l \alpha|<\frac{1}{N}$. Without loss of generality, assume $k>l$, then
$$
01-\{(k-l) \alpha\}>1-\varepsilon$.
If $k<l$, then
$$
-\frac{1}{N}<(l-k) \alpha+[k a]-[l a]<0 .
$$

That is, $-\frac{1}{N}<\{(l-k) \alpha\}-1<0$.
So $1-\varepsilon<1-\frac{1}{N}<\{(l-k) \alpha\}<1$.
Take an integer $T$ such that
$T(1-\{(l-k) \alpha\})<1<(T+1)(1-\{(l-k) \alpha\})$.

Then
$$
\begin{aligned}
T(l-k) \alpha & =T[(l-k) \alpha]+T\{(l-k) \alpha\} \\
& =T[(l-k) \alpha]+T+T(\{(l-k) \alpha\}-1)
\end{aligned}
$$

Therefore,
$$
=T[(l-k) \alpha]+T-1+1-T(1-\{(l-k) a\}),
$$
$\{T(l-k) \alpha\}=1-T(1-\{(l-k) \alpha\})$
$$
<1-\{(l-k) \alpha\}<\frac{1}{N}<\epsilon \text {. }
$$

The lemma is proved.
So $b-\alpha=0$, i.e., $b=\alpha$.
In summary,
$g(x)=\frac{1}{2}(\sqrt{5}-1)(x+1)$.