PLSQL算24點(diǎn)
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??? 很早以前就想用PLSQL寫(xiě)一個(gè)算24點(diǎn)的程序了，后來(lái)想了一下沒(méi)有研究下去，最近上ITPUB的時(shí)候正好大家都在討論，在大牛的基礎(chǔ)上修改了一下，基本上思想和算法都是沒(méi)有變的，就是學(xué)習(xí)了一下。還有兩個(gè)大問(wèn)題沒(méi)有處理好，一是只用了枚舉法，用PLSQL應(yīng)該可以換個(gè)更好的遞歸算法的；二是沒(méi)有濾重；反正最后的結(jié)果只是最簡(jiǎn)單的得到答案而已。
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??? 具體的可以參見(jiàn)：
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create or replace function f_op (p_n1 number, p_op varchar2, p_n2 number) return number
as
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;
/
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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
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?????? r_result := f_op(f_op(f_op(r.a,r.o1,r.b),r.o2,r.c),r.o3,r.d);
?????? if r_result between 23.9 and 24.1 then
?????? dbms_output.put_line('(('||r.a||r.o1||r.b||')'||r.o2||r.c||')'||r.o3||r.d);
?????? end if; --((a b) c) d
??????
?????? r_result := f_op(f_op(r.a,r.o1,f_op(r.b,r.o2,r.c)),r.o3,r.d);
?????? if r_result between 23.9 and 24.1 then
?????? dbms_output.put_line('('||r.a||r.o1||'('||r.b||r.o2||r.c||'))'||r.o3||r.d);
?????? end if; --(a (b c)) d
??????
?????? r_result := f_op(f_op(r.a,r.o1,r.b),r.o2,f_op(r.c,r.o3,r.d));
?????? if r_result between 23.9 and 24.1 then
?????? dbms_output.put_line('('||r.a||r.o1||r.b||')'||r.o2||'('||r.c||r.o3||r.d||')');
?????? end if; --(a b) (c d)
??????
?????? r_result := f_op(r.a,r.o1,f_op(f_op(r.b,r.o2,r.c),r.o3,r.d));
?????? if r_result between 23.9 and 24.1 then
?????? dbms_output.put_line(r.a||r.o1||'(('||r.b||r.o2||r.c||')'||r.o3||r.d||')');
?????? end if; --a ((b c) d)
???????????
?????? r_result := f_op(r.a,r.o1,f_op(r.b,r.o2,f_op(r.c,r.o3,r.d)));
?????? if r_result between 23.9 and 24.1 then
?????? dbms_output.put_line(r.a||r.o1||'('||r.b||r.o2||'('||r.c||r.o3||r.d||'))');
?????? end if; --a (b (c d))
? end loop;
end;
/
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exec pro_24(3,3,8,8);
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