* 定義下面的集合允許重複的元素，索引由0開始，另外
使用交集(∩)、聯集(∪)結果不包含重複元素。
* 定義二個集合的比對函數，取得a數量和b數量的結果。
{count_a, count_b} := guestResult(s1, s2)
* 代入的例子:
s1 = {2,6,5,2}
s2 = {2,3,5,1}
* 以下為guestResult函數的運算步驟
1. 取交集
定義s1、s2的交集結果
int = s1 ∩ s2 = {2, 5}
^ (intersection)
2. 反查位置
定義 table(i, s) 為查詢 int[i] 在 s 中其索引的集合
table(0, s1) = {0, 3}
table(0, s2) = {0}
table(1, s1) = {2}
table(1, s2) = {2}
table(2, s1) = null
table(2, s2) = null
table(3, s1) = null
table(3, s2) = null
3. 統計ab
定義 count_a(i) = #(table(i, s1) ∩ table(i, s2))
定義 count_b(i) = max{#table(i, s1), #table(i, s2)} - count_a(i)
=> count_a = count_a(0) + count_a(1) + ... + count_a(3)
count_b = count_b(0) + count_b(1) + ... + count_b(3)
PS: 排除null相加
count_a(0) = #{0} = 1
count_a(1) = #{2} = 1
count_a(2) = #null = null
count_a(3) = #null = null
count_b(0) = max{#{0, 3}, #{0}} - 1 = max{2, 1} - 1 = 2 - 1 = 1
count_b(1) = max{#{2}, #{2}} - 1 = max{1, 1} - 1 = 1 - 1 = 0
count_b(2) = max{#null, #null} - null = null
count_b(3) = max{#null, #null} - null = null
count_a = 1 + 1 = 2
count_b = 1 + 0 = 1
~~the end~~ By s3748679 |
