mysql 实现 五个海盗分金币

海盗分金问题：有五个海盗，劫掠了100两金子，需要分 赃。办法是抓阄，盗亦有道。抓到第一个阄的人，可以先提出一个分 配方案，如果他的方案被一半以上的人同意， 就照他的方案分金子，否则，第一个人就要 被杀掉。余下的人也照此办理。我们的问题是：如果你是第一个人，你 会提出怎样的分配方案？为了分析问题更确定，我们假定每个人 都是追求自己利益极大化的人。

with  a as
(
  select cast(help_topic_id as SIGNED) as n from mysql.help_topic   where  help_topic_id <101  order by help_topic_id desc
)
, max1  as
(
select  max(n) as max1 from a 
)
,max2 as 
(
  select
  p2.n as max2,p1.n as max1  from a p1,a p2
  where p1.n + p2.n = 100
  and p1.n >=(select max1 from max1)
)
,max3 as
(
  select/*+ JOIN_ORDER(P3,P2,P1) 
        JOIN_SUFFIX(P1) 
        NO_BNL() */
  p3.n as max3,p2.n as max2,p1.n as max1 from a p1,a p2,a p3,max2
  where p1.n+p2.n+p3.n = 100 
   and p3.n +p2.n <= 100
  and sign(p2.n- max2) + sign(p1.n - max1) >=0
limit 1
)
,max4 as
(
 select 
 /*+ JOIN_ORDER(p4,P3,P2,P1) 
        JOIN_SUFFIX(P1) 
        NO_BNL() */
 p4.n as max4,p3.n as max3,p2.n as max2,p1.n as max1       
 from a p1,a p2 ,a p3,a p4,max3
 where p1.n+p2.n+p3.n + p4.n = 100 
  and p4.n + p3.n +p2.n <= 100
 and sign(p3.n - max3) + sign(p2.n - max2) + sign(p1.n - max1) >0
 limit 1
)
,max5 as
(
 select 
 /*+ JOIN_ORDER(p5,p4,P3,P2,P1) 
        JOIN_SUFFIX(P1) 
        NO_BNL() */
 p5.n as max5,p4.n as max4,p3.n as max3,p2.n as max2,p1.n as max1       
 from a p1,a p2 ,a p3,a p4,a p5,max4
 where p1.n+p2.n+p3.n + p4.n  + p5.n = 100 
  and p5.n + p4.n + p3.n +p2.n <=100
 and sign(p4.n - max4) + sign(p3.n - max3) + sign(p2.n - max2) + sign(p1.n - max1) >=0
 limit 1
)
SELECT * FROM MAX5 

max5 max4 max3 max2 max1
97 0 1 0 2