PLSQL算24点

PLSQL算24点

很早以前就想用PLSQL写一个算24点的程序了，后来想了一下没有研究下去，最近上ITPUB的时候正好大家都在讨论，在大牛的基础上修改了一下，基本上思想和算法都是没有变的，就是学习了一下。还有两个大问题没有处理好，一是只用了枚举法，用PLSQL应该可以换个更好的递归算法的；二是没有滤重；反正最后的结果只是最简单的得到答案而已。

具体的可以参见：

http://www.itpub.net/viewthread.php?tid=1061129&extra=page%3D1%26amp%3Bfilter%3Ddigest&page=1

http://www.itpub.net/viewthread.php?tid=1077244&extra=&page=1

create or replace function f_op (p_n1 number, p_op varchar2, p_n2 number) return number

begin

return case when p_op = '+' then p_n1 + p_n2

when p_op = '-' then p_n1 - p_n2

when p_op = '*' then p_n1 * p_n2

when p_op = '/' and p_n2<>0 then p_n1 / p_n2

else null

end;

end f_op;

create or replace procedure pro_24(p1 number, p2 number, p3 number, p4 number) as

r_result number default 0;

begin

for r in (

with t_num as

(select 1 id,p1 as n from dual

union

select 2 id,p2 as n from dual

union

select 3 id,p3 as n from dual

union

select 4 id,p4 as n from dual),

t_op as

(select '+' as o from dual

union

select '-' as o from dual

union

select '*' as o from dual

union

select '/' as o from dual)

select distinct

a.n as a,

o1.o as o1,

b.n as b,

o2.o as o2,

c.n as c,

o3.o as o3,

d.n as d

from t_num a, t_num b, t_num c, t_num d,

t_op o1, t_op o2, t_op o3

where a.id not in (b.id, c.id, d.id)

and b.id not in (c.id, d.id)

and c.id <> d.id) loop

r_result := f_op(f_op(f_op(r.a,r.o1,r.b),r.o2,r.c),r.o3,r.d);