无聊，蛋帖，来做题，有兴趣进！

[AcHerat]

数学题 A * B = C 用SQL求解
A、B、C 由（ 1、2、3、4、5、6、7、8、9 ） 9 个数字中组成，且每个数字只能取一次，9 个数字必须全部取到，不得多取、重取
比方说，1738 * 4 = 6952

个人求得：

138         42          5796        138425796
157         28          4396        157284396
159         48          7632        159487632
186         39          7254        186397254
198         27          5346        198275346
297         18          5346        297185346
483         12          5796        483125796
1738        4           6952        173846952
1963        4           7852        196347852

不知答案是否对，等楼下结果。

[AcHerat]

--try

),t1 as
(
    select id from #tb a 
	where id between 1 and 99
		and not exists (select 1 from t0 where len(ltrim(a.id))-len(replace(ltrim(a.id),ltrim(id),'')) >= 2)
)

[litter_man]

路过，帮顶

[airsethh]

我们云南要干旱死了，没水洗脸了

[叶子]

--看着挺笨重的，我本地运行250-300毫秒之间。
;with m as ( 
select number as id from master..spt_values where type = 'p' and number between 1 and 9),
m1 as (select ltrim(t1.id) as c1 ,ltrim(t2.id) as c2 ,ltrim(t3.id) as c3 ,ltrim(t4.id) as c4
       from  m t1 cross join m t2 cross join m t3 cross join m t4
       where t1.id <> t2.id and t1.id <> t3.id and t1.id <> t4.id
       and t2.id <> t3.id and t2.id <> t4.id and t3.id <> t4.id
      ),
m2 as (select ltrim(t1.id) as c1 ,ltrim(t2.id) as c2 ,ltrim(t3.id) as c3
       from  m t1 cross join m t2 cross join m t3
       where t1.id <> t2.id and t1.id <> t3.id and t2.id <> t3.id
      ),
m3 as (select ltrim(t1.id) as c1 ,ltrim(t2.id) as c2
       from  m t1 cross join m t2 where t1.id <> t2.id
      ),
m4 as ( select (a.c1 + a.c2 + a.c3) as c1 ,( b.c1 + b.c2 ) as c2 ,
        ltrim(( a.c1 + a.c2 + a.c3 ) + ( b.c1 + b.c2 )+ ltrim(( a.c1 + a.c2 + a.c3 ) * 1 * ( b.c1+ b.c2 ))) as c3
        from  m2 a ,m3 b
        where ( a.c1 + a.c2 + a.c3 ) * 1 * ( b.c1 + b.c2 ) < 9876 union all
        select ( a.id ) ,( b.c1 + b.c2 + b.c3 + b.c4 ) ,
        ltrim(ltrim(a.id) + ( b.c1 + b.c2 + b.c3 + b.c4 )+ ltrim(a.id * 1 * ( b.c1 + b.c2 + b.c3 + b.c4 )))
        from  m a ,m1 b
        where    a.id * 1 * ( b.c1 + b.c2 + b.c3 + b.c4 ) < 9876
      )
    select  c1 ,c2 ,c3 from m4
    where   len(c3) = 9
            and charindex('1', c3) > 0
            and charindex('2', c3) > 0
            and charindex('3', c3) > 0
            and charindex('4', c3) > 0
            and charindex('5', c3) > 0
            and charindex('6', c3) > 0
            and charindex('7', c3) > 0
            and charindex('8', c3) > 0
            and charindex('9', c3) > 0 

[arnold]

学习一下知识支持一个。

[sunylf]

有點難度.

[daerter]

不会。

[zhangtieqiang]

楼上说的对呀

[zhangtieqiang]

楼上说的对呀

[追风筝的孩子]

omg。。。我只能路过

[天亮后说晚安]

omg。。。我只能路过

[lmkzzs]

不会 是新手 有那个哥哥姐姐教教俺捏 QQ1040634262

[傅琦鸿]

[淘淘大师]

路过。。。

[ssccbb1]

小三V5

[纳须弥]

现在才知道什么叫做人才。。。。各位都是大神啊

[弘恩]

看一看我的代码！ 这个问题，我有N种解法。
下面的代码是个人觉得最快的,
如果再改进，可以改用 CHARINDEX

with a1 as
(select '1' as a1
union all select '2'
union all select '3'
union all select '4'
union all select '5'
union all select '6'
union all select '7'
union all select '8'
union all select '9'),
a2 as ( select a.a1 + b.a1 as a2 from a1 a ,a1 b where a.a1 <> b.a1 ),
b3 as ( select a.a1 + b.a1 + c.a1 as b3 from a1 a ,a1 b ,a1 c where a.a1 <> b.a1 and a.a1 <> c.a1 and b.a1 <> c.a1 ),
b4 as ( select a.a1 + b.a1 + c.a1+ d.a1 as b4 from a1 a ,a1 b ,a1 c ,a1 d where a.a1 <> b.a1 and a.a1 <> c.a1 and a.a1 <> d.a1 and b.a1 <> c.a1 and b.a1 <> d.a1 and c.a1 <> d.a1 ),
c4 as ( select a.a1 + b.a1 + c.a1+ d.a1 as c4 from a1 a ,a1 b ,a1 c ,a1 d where a.a1 <> b.a1 and a.a1 <> c.a1 and a.a1 <> d.a1 and b.a1 <> c.a1 and b.a1 <> d.a1 and c.a1 <> d.a1 )


select a1.a1 a, b4.b4 b, c4.c4 c from a1 ,b4 ,c4 where cast ( a1.a1 as int ) * cast ( b4.b4 as int ) = cast ( c4.c4 as int )
and a1.a1 + b4.b4 + c4.c4 like '%1%'
and a1.a1 + b4.b4 + c4.c4 like '%2%'
and a1.a1 + b4.b4 + c4.c4 like '%3%'
and a1.a1 + b4.b4 + c4.c4 like '%4%'
and a1.a1 + b4.b4 + c4.c4 like '%5%'
and a1.a1 + b4.b4 + c4.c4 like '%6%'
and a1.a1 + b4.b4 + c4.c4 like '%7%'
and a1.a1 + b4.b4 + c4.c4 like '%8%'
and a1.a1 + b4.b4 + c4.c4 like '%9%'

union all
select a2.a2 ,b3.b3 ,c4.c4 from a2 ,b3 ,c4 where cast ( a2.a2 as int ) * cast ( b3.b3 as int ) = cast ( c4.c4 as int )
and a2.a2 + b3.b3 + c4.c4 like '%1%'
and a2.a2 + b3.b3 + c4.c4 like '%2%'
and a2.a2 + b3.b3 + c4.c4 like '%3%'
and a2.a2 + b3.b3 + c4.c4 like '%4%'
and a2.a2 + b3.b3 + c4.c4 like '%5%'
and a2.a2 + b3.b3 + c4.c4 like '%6%'
and a2.a2 + b3.b3 + c4.c4 like '%7%'
and a2.a2 + b3.b3 + c4.c4 like '%8%'
and a2.a2 + b3.b3 + c4.c4 like '%9%'

[无尘阁]

大牛们，求收留

[524929657雯]

[benbenLY]

留个脚印