回復 1# mathca 的帖子
第 11 題
p = bc - ad = bc - a(-a - b - c) = (a + b)(a + c)
由於 a + b ≧ a + c
故 a + b = p，a + c = 1
b = p - a，c = 1 - a，d = -(a + b + c) = -(p + 1 - a) = a - p - 1
a ≧ b = p - a
p ≦ 2a
c ≧ d
1- a ≧ a - p - 1
p ≧ 2a - 2
p = 2a - 1
a = (p + 1)/2
