发现这种题对sql不成问题

Question

Answer 1

如果不要求 a,b,c 不同， 就可以像上面那样推理。

Answer 2

喔，忘了个条件，a,b,c 要求不相同

SQL> with t as (select level n from dual connect by level <=9)

  2  select a,b,c,max_value

  3    from (

  4  select a.n a,b.n b,c.n c,

  5         max(1/(a.n+2019/(b.n+1/c.n))) over() max_value,

  6         rank () over( order by 1/(a.n+2019/(b.n+1/c.n)) desc) rk

  7    from t a,t b,t c

  8   where a.n <> b.n and b.n <> c.n and c.n <> a.n    --增加条件

  9  )

10  where rk = 1;

         A          B          C  MAX_VALUE

---------- ---------- ---------- ----------

         2          9          1 0.00490436

Answer 3

怎么知道a,c=1,2还是2,1分母更小？

Answer 4

with a as (select level l from dual connect by level<=9)

select * from (

select  max(1/(a.l+2019/(b.l+1/c.l))) keep(dense_rank first order by 1/(a.l+2019/(b.l+1/c.l)) desc) value,

max(a.l||','||b.l||','||c.l) keep(dense_rank first order by 1/(a.l+2019/(b.l+1/c.l)) desc) key from a a,  a b, a c where

a.l<>b.l and b.l<>c.l

and a.l<>c.l) ;

value                                                                         key

0.004904364884747425208435507601765571358509        2,9,1

Answer 5

口算：1/(a+2019/(b+1/c))，要使得表达式的值最大，那分母要最小 a+2019/(b+1/c)，a最小取1， 2019/(b+1/c)最小，那么 (b+1/c)要求最大，b最大9，1/c最大，C取最小，c=1

Answer 6

小菜一碟

with t as (select level n from dual connect by level <=9)

select a,b,c,max_value

  from (

select a.n a,b.n b,c.n c,

       max(1/(a.n+2019/(b.n+1/c.n))) over() max_value,

       rank () over( order by 1/(a.n+2019/(b.n+1/c.n)) desc) rk

  from t a,t b,t c

)

where rk = 1

；

A     B       C      MAX_VALUE

1        9        1        0.00492853622474125