$36 \cdot 64$ In a math competition, the sum of the scores of $B$ and $D$ is equal to the sum of the scores of $A$ and $C$. If the scores of $B$ and $C$ are swapped, then the sum of the scores of $A$ and $C$ will exceed the sum of the scores of $B$ and $D$. In addition, it is known that the score of $D$ is more than the sum of the scores of $B$ and $C$. Assuming all scores are non-negative, the order from highest to lowest score is
(A) $D, A, C, B$.
(B) $D, A, B, C$.
(C) $D, C, B, A$.
(D) $A, D, C, B$.
(E) $A, D, B, C$.
(38th American High School Mathematics Examination, 1987)
[Solution] Let $a, b, c, d$ be the scores of $A, B, C, D$ respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
b+d=a+c, \\
a+b>c+d, \\
d>b+c .
\end{array}\right.
$$

From (1), we have $d=a+c-b$ and substituting into (3) gives
$$
a>2 b,
$$

From (1), we have $a=b+d-c$ and substituting into (2) gives
$$
b>c,
$$

From (1), we have $b=a+c-d$ and substituting into (2) gives
$$
a>d,
$$

From (3), it is clear that $d>b$ and $d>c$.
In summary, we have $a>d>b>c$.
Therefore, the answer is $(E)$.