Question 4 Let $m, n, p$ be given positive integers, and $m \leqslant n \leqslant p \leqslant m+n$. Find the minimum value of $q=f(m, n, p)$ such that the following conclusion always holds: If $q$ students all participate in three mock exams of a mathematics competition, and the score of each student in the first exam is an integer from 0 to $m$, the score in the second exam is an integer from 0 to $n$, and the score in the third exam is an integer from 0 to $p$, then there must be two students $A, B$ such that for each exam, $A$'s score is not less than $B$'s score.
$$
\begin{array}{l}
\text { The solution to problem 4, the minimum value of } q \text { is } \\
q_{0}=(m+1)(n+1)-\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+2}{2}\right]+1 . \\
\text { First, we prove that when } q=q_{0}-1 \text {, it does not guarantee the existence of } \\
\text { two students satisfying the requirements of the problem. } \\
\text { Let } S=\{(a, b, c) \mid a, b, c \in \mathbf{N}, 0 \leqslant a \leqslant m, \\
0 \leqslant b \leqslant n, 0 \leqslant c \leqslant p\} \\
\text { where, }(a, b, c) \text { represents the scores of a student in the first, second, and third exams } \\
\text { as } a, b, c \text {, respectively, and let } \\
t_{0}=\left[\frac{m+p-n}{2}\right], t_{1}=\left[\frac{m+p-n+1}{2}\right] . \\
\text { Then } t_{0}+t_{1}=m+p-n \text { (whether } r \text { is even or odd, } \\
\text { we have } \left.\left[\frac{r}{2}\right]+\left[\frac{r+1}{2}\right]=r\right) \text {, and } \\
m-t_{0}=t_{1}-p+n=\left[\frac{m+n-p+1}{2}\right] \text {, } \\
\text { Let the set } M \text { of scores of some students in three exams } \\
\text { be the union of the following } p+1 \text { disjoint sets: } \\
S_{i}=|(a, b, c)|(a, b, c) \in S, a+b=n+ \\
\text { and } \left|S_{i}\right|=\left\{\begin{array}{ll}
m-t_{0}+i+1, & 0 \leqslant i \leqslant t_{0}-1 ; \\
m+1, & t_{0} \leqslant i \leqslant t_{0}+n-m ; \\
n+t_{0}-i+1, & t_{0}+n-m+1 \leqslant i \leqslant p,
\end{array}\right. \\
\text { then }|M|=\left|\cup_{i=0}^{B} S_{i}\right|=\sum_{i=0}^{n}\left|S_{i}\right| \\
=\left(\sum_{i=1}^{m} i-\sum_{i=1}^{m-10} i\right)+(n-m+1)(m+1)+ \\
\left(\sum_{i=1}^{m} i-\sum_{i=1}^{n+t+p} i\right) \\
=2 \sum_{i=1}^{m} i+(n-m+1)(m+1)-\sum_{i=1}^{m-2} i-\sum_{i=1}^{m-1} i \\
=(n+1)(m+1)-\frac{1}{2}\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+3}{2}\right]- \\
\frac{1}{2}\left[\frac{m+n-p}{2}\right]\left[\frac{m+n-p+2}{2}\right] \\
=(m+1)(n+1)-\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+2}{2}\right] \text {, } \\
\end{array}
$$
$$
\begin{array}{l}
q_{0}-(m+1-s)(n+1-s) \\
\text { points belong to } R=T_{1} \cup T_{2} \cup \cdots \cup T_{0} \text {. } \\
\text { By the pigeonhole principle, at least } \\
{\left[\frac{q_{0}-(m+1-s)(n+1-s)-1}{s}\right]+1} \\
=-\left[\frac{m+n-p+2}{2}\right]+(m+n+2)-s+1 \\
=-\left[\frac{m+n-p+2}{2}\right]+(m+n+2)-\left[\frac{m+n-p+1}{2}\right]+1 \\
=p+2 \\
\end{array}
$$

students corresponding to points belong to some subset $T_{i}$ among $T_{1}, T_{2}, \cdots, T_{1}$. These $p+2$ students have at most $0,1, \cdots, p$ different scores in the third exam, so there must be two students with the same score in the third exam. By the construction of subset $T_{i}$, one of these two students (denoted as $A$) has scores in the first and second exams that are not lower than the other student (denoted as $B$). Therefore, for each exam, $A$'s score is not lower than $B$'s score.
In summary, the minimum value of $q$ is
$$
q_{0}=(m+1)(n+1)-\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+2}{2}\right]+1.
$$
[Note] In particular, when $m=p=n$, we get the conclusion of problem 2: the minimum value of $q$ is
$$
\begin{array}{l}
q_{0}=(n+1)^{2}-\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]+1 \\
=\left[\frac{3(n+1)^{2}+1}{4}\right]+1 .
\end{array}
$$

Let $n=7$, then $q_{0}=49$. This not only gives the conclusion of problem 3, but also shows that 49 is the value that makes the conclusion hold.
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