$5 \cdot 151 n$ is a positive integer, find the number of integer coefficient polynomials $f(x)$, such that its coefficients belong to the set $\{0,1,2,3\}$, and satisfy $f(2)=n$.
Let
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{k} x^{k},
$$

where $k$ is a non-negative integer, $a_{j}(0 \leqslant j \leqslant k) \in\{0,1,2,3\}$, and $a_{k} \neq 0$.
From $f(2)=n$ we get
$$
n=a_{0}+a_{1} \cdot 2+a_{2} \cdot 2^{2}+\cdots+a_{k} \cdot 2^{k} .
$$

Let
$$
b_{j}=\left[\frac{a_{j}}{2}\right], \quad(j=0,1,2, \cdots, k)
$$

where $\left[\frac{a_{j}}{2}\right]$ denotes the greatest integer not exceeding $\frac{a_{j}}{2}$, then
$$
b_{j} \in\{0,1\}
$$

Let
$$
m=b_{0}+b_{1} \cdot 2+b_{2} \cdot 2^{2}+\cdots+b_{k} \cdot 2^{k},
$$

obviously
$$
0 \leqslant m \leqslant \frac{n}{2} .
$$

Thus, for a polynomial $f(x)$ that satisfies the conditions of the problem, there corresponds a non-negative integer $m$ in the closed interval $\left[0, \frac{n}{2}\right]$, denoted as:
$$
g(f(x))=m,
$$

where the domain of the mapping $g$ is all polynomials that satisfy the conditions of the problem, and the range is $\left[0, \frac{n}{2}\right]$.
Next, we prove: if $m$ is a non-negative integer in $\left[0, \frac{n}{2}\right]$, then there exists a unique integer-coefficient polynomial $f(x)$ that satisfies the conditions of the problem, such that
$$
g(f(x))=m .
$$

In fact, first express $m$ in binary as
$$
m=c_{0}+c_{1} \cdot 2+c_{2} \cdot 2^{2}+\cdots+c_{k} \cdot 2^{k},
$$

where $k$ is a non-negative integer, $c_{j} \in\{0,1\}, j=1,2, \cdots, k$, and when $m \neq 0$, $c_{k}=1$.
If there exists a polynomial that satisfies the conditions of the problem
$$
f(x)=a_{0}+a_{1} \cdot x+a_{2} \cdot x^{2}+\cdots+a_{t} \cdot x^{t}
$$

such that
$$
g(f(x))=m .
$$
then
$$
\begin{array}{l}
n=a_{0}+a_{1} \cdot 2+a_{2} \cdot 2^{2}+\cdots+a_{t} \cdot 2^{t}, \\
c_{j}=\left[\frac{a_{j}}{2}\right], \quad(j=0,1,2, \cdots, k) ; \\
{\left[\frac{a_{j}}{2}\right]=0, \quad(j=k+1, \cdots, t) t \geqslant k .}
\end{array}
$$

If $c_{0}=0$, then $a_{0}=0$ or 1. Since $n-a_{0}$ is a multiple of 2, $a_{0}$ has the same parity as $n$. When $n$ is even, $a_{0}=0$; when $n$ is odd, $a_{0}=1$.

If $c_{0}=1$, then $a_{0}=2$ or 3. Similarly, when $n$ is even, $a_{0}=2$; when $n$ is odd, $a_{0}=3$.
Therefore, $a_{0}$ can be uniquely determined by $c_{0}$ and the parity of $n$. And $n-a_{0}$ is a multiple of 2.
Assume $a_{0}, a_{1}, \cdots, a_{j-1}$ can be uniquely determined by $c_{0}, c_{1}, \cdots, c_{j-1}$ and $n$, and $n-a_{0}-a_{1} \cdot 2-\cdots-a_{j-1} \cdot 2^{j-1}$ is a multiple of $2^{j}$.
When $c_{j}=0$, then $a_{j}=0$ or 1. Since
$$
n-a_{0}-a_{1} \cdot 2-\cdots-a_{j-1} \cdot 2^{j-1}-a_{j} \cdot 2^{j}
$$

is a multiple of $2^{j+1}$, so, if
$n-a_{0}-a_{1} \cdot 2-\cdots-a_{j-1} \cdot 2^{j-1}$ is a multiple of $2^{j+1}$, then $a_{j}=0$; otherwise $a_{j}=1$.
Similarly, when $c_{j}=1$, then $a_{j}=2$ or 3. If
$$
n-a_{0}-a_{1} \cdot 2-\cdots-a_{j-1} \cdot 2^{j-1} \text { is a multiple of } 2^{j+1} \text {, then } a_{j}=2 \text {; }
$$

otherwise $a_{j}=3$.
Therefore, $a_{j}$ is also uniquely determined, and
$n-a_{0}-a_{1} \cdot 2-\cdots-a_{j-1} \cdot 2^{j-1}-a_{j} \cdot 2^{j}$ is a multiple of $2^{j+1}$.
Thus $a_{0}, a_{1}, a_{2} \cdots, a_{k}$ can all be uniquely determined in sequence.
At this point,
$$
A=n-\left(a_{0}+a_{1} \cdot 2+a_{2} \cdot 2^{2}+\cdots+a_{k} \cdot 2^{k}\right) \text {, is a multiple of } 2^{k+1}
$$

Since
$$
\left[\frac{a_{j}}{2}\right]=0(j=k+1, k+2, \cdots, t)
$$

Therefore
$$
a_{j} \in\{0,1\}(j=k+1, k+2, \cdots, t) .
$$

Thus
$$
a_{k+1} \cdot 2^{k+1}+a_{k+2} \cdot 2^{k+2}+\cdots+a_{t} \cdot 2^{t} \text { is the binary representation of } A \text {, }
$$

so $c_{k+1}, a_{k+2}, \cdots, a_{t}$ are all uniquely determined.
Thus the mapping $g$ is a one-to-one correspondence between all integer-coefficient polynomials that satisfy the conditions of the problem and all non-negative integers in $\left[0, \frac{n}{2}\right]$, hence the number of all integer-coefficient polynomials that satisfy the conditions of the problem is equal to the number of all non-negative integers in $\left[0, \frac{n}{2}\right]$, which is $\left[\frac{n}{2}\right]+1$, where $\left[\frac{n}{2}\right]$ denotes the greatest integer not exceeding $\frac{n}{2}$.