5. Given the function $f(x)$ as shown in Table 1. $a_{k}(0 \leqslant k \leqslant 4)$ equals the number of times $k$ appears in $a_{0}, a_{1}, \cdots, a_{4}$. Then $a_{0}+a_{1}+a_{2}+a_{3}=$ $\qquad$ .
5.5.

Since $a_{k}$ equals the number of times $k$ appears among the total of five numbers, and since the count must be an integer and at most 5, we have $a_{k} \in\{0,1, \cdots, 5\}$.
The following conclusions can be drawn:
(1) There does not exist a case where $a_{k}=5$, i.e.,
$$
a_{k} \in\{0,1, \cdots, 4\}.
$$

Otherwise, if $a_{k}=5$, it would mean that $k$ appears 5 times, so the five numbers would be $k, k, k, k, k$; and since $a_{k}=5$, these five numbers would have to be $5,5,5,5,5$, thus $a_{0}=5$. But 0 appears 0 times, which is a contradiction.
(2) $\left\{\begin{array}{l}a_{0}+a_{1}+\cdots+a_{4}=5, \\ 0 a_{0}+1 a_{1}+\cdots+4 a_{4}=5\end{array}\right.$.
This is because, on one hand, $a_{0}, a_{1}, \cdots, a_{4}$ consist of five numbers, and their sum is $a_{0}+a_{1}+\cdots+a_{4}$.

On the other hand, $a_{0}, a_{1}, \cdots, a_{4}$ contain $a_{0}$ zeros, $a_{1}$ ones, $a_{2}$ twos, $a_{3}$ threes, and $a_{4}$ fours. From (1), we know there are no other numbers. Therefore, there are a total of $a_{0}+a_{1}+\cdots+a_{4}$ numbers, and their sum is $0 a_{0}+1 a_{1}+\cdots+4 a_{4}$. Thus, (2) holds.
(3) $a_{4}=0, a_{0}+a_{1}+a_{2}+a_{3}=5$.
From (2), we know $a_{4} \leqslant 1$.
If $a_{4}=1$, then
$$
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}+a_{3}=4, \\
0 a_{0}+1 a_{1}+2 a_{2}+3 a_{3}=1
\end{array}\right.
$$

The only solution is $\left(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}\right)=(3,1,0,0,1)$.
Thus, $a_{0}=3$, but 0 appears 2 times, which is a contradiction.
Therefore, $a_{4}=0, a_{0}+a_{1}+a_{2}+a_{3}=5$.
In fact, it is not difficult to verify that the solution to (2) that meets the requirements is only $(2,1,2,0,0)$. In $(2,1,2,0,0)$, there are exactly 2 zeros, 1 one, 2 twos, 0 threes, and 0 fours, which satisfies the conditions.