DECLARE @AGE TABLE(AGE INT)
INSERT INTO @AGE
SELECT NUMBER FROM MASTER..SPT_VALUES WHERE TYPE='P' AND NUMBER BETWEEN 0 AND 36
DECLARE @HOUSE TABLE(WINDOW INT)
INSERT INTO @HOUSE
SELECT T1.NUMBER*T2.NUMBER 
FROM (SELECT NUMBER FROM MASTER..SPT_VALUES WHERE TYPE='P' AND NUMBER BETWEEN 1 AND 100) T1
inner JOIN (SELECT NUMBER FROM MASTER..SPT_VALUES WHERE TYPE='P' AND NUMBER BETWEEN 1 AND 100) T2
ON T1.NUMBER>=T2.NUMBER
;WITH MU AS (
SELECT T1.AGE AS A1,T2.AGE AS A2,T3.AGE AS A3,T1.AGE+T2.AGE+T3.AGE AS HE
FROM @AGE T1
INNER JOIN @AGE T2 ON T1.AGE>=T2.AGE 
INNER JOIN @AGE T3 ON T2.AGE>=T3.AGE
AND T1.AGE+T2.AGE+T3.AGE IN (
SELECT WINDOW FROM @HOUSE
) AND T1.AGE*T2.AGE*T3.AGE=36
)
SELECT A1,A2,A3 FROM MU T1 WHERE EXISTS(
SELECT 1 FROM MU T2 WHERE T2.HE=T1.HE AND T2.A1<>T1.A1
) AND A1>A2
--9    2    2

------解决方案--------------------

怎么算出两个答案？
SQL code
DECLARE @number TABLE(n INT)
INSERT INTO @number
SELECT NUMBER FROM MASTER..SPT_VALUES WHERE TYPE='P' AND NUMBER BETWEEN 2 AND 9
;with tb as(
select a.n an,b.n bn,a.n+b.n [a+b],a.n*b.n [a*b] from @number a, @number b where a.n <= b.n )
,tb1 as --甲猜不到，说明两个数的和不唯一
(select * from tb where [a+b] in(select [a+b] from tb group by [a+b] having COUNT(1) > 1))
,tb2 as --乙猜不到，说明两个数的积不唯一
(select * from tb1 where [a*b] in(select [a*b] from tb1 group by [a*b] having COUNT(1) > 1))
,tb3 as --甲猜到了说明两个数的和已经唯一了
(select * from tb2 where [a+b] in(select [a+b] from tb2 group by [a+b] having COUNT(1) = 1))
,tb4 as --乙猜到了说明两个数的和已经唯一了
(select * from tb3 where [a*b] in(select [a*b] from tb3 group by [a*b] having COUNT(1) = 1))
select * from tb4

------解决方案--------------------

SQL code

--情况一：可以相等
select a.number as an,b.number bn,a.number+b.number sm,a.number*b.number m into #tb from master..spt_values a
,master..spt_values b
where a.type='p' and b.type='p'
and a.number between 2 and 9
and b.number between 2 and 9
and a.number<=b.number


;with ct1 as(
select * from #tb a where exists
(select 1 from #tb where a.sm=sm group by sm having COUNT(1)>=2)
),
ct2 as
(select * from ct1 a where exists(
select 1 from ct1 where a.m=m group by m having COUNT(1)>=2)
)
,ct3 as
(select * from ct2 a where exists(
select 1 from ct2 where a.sm=sm group by sm having COUNT(1)=1)
)
,ct4 as(
select * from ct3 a where exists(
select 1 from ct3 where a.m=m group by m having COUNT(1)=1)
)
select * from ct4

drop table #tb
go
--情况一：不可以相等


select a.number as an,b.number bn,a.number+b.number sm,a.number*b.number m into #tb from master..spt_values a
,master..spt_values b
where a.type='p' and b.type='p'
and a.number between 2 and 9
and b.number between 2 and 9
and a.number<b.number


;with ct1 as(
select * from #tb a where exists
(select 1 from #tb where a.sm=sm group by sm having COUNT(1)>=2)
),
ct2 as
(select * from ct1 a where exists(
select 1 from ct1 where a.m=m group by m having COUNT(1)>=2)
)
,ct3 as
(select * from ct2 a where exists(
select 1 from ct2 where a.sm=sm group by sm having COUNT(1)=1)
)
,ct4 as(