PUZZLEUP 2015

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newkid 发表于 2015-8-14 08:43
全加上ROWNUM就老实了：

with t as(select level-1 l from dual connect by levelb.l*c.l

和解决谓词推进的一个办法
上次有个例子人工展开比递归快很多是怎么回事

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newkid 发表于 2015-8-14 08:43
全加上ROWNUM就老实了：

with t as(select level-1 l from dual connect by levelb.l*c.l

真的快了
  COUNT(L)
----------
         0

已用时间:  00: 00: 00.05

solomon_007

newkid 发表于 2015-8-14 08:51
你这WHERE LEVEL

因为组成一个10位数,最多只要8个3位数每两个3位数两位首尾相接

jixch

又开始了啊

newkid

#4 SUM OF PALINDROMES

Three different numbers between 10 and 1000 are palindromes and their sum is also a palindrome. What is the maximum possible value of the multiplication of these three numbers?

-The numbers cannot start with 0, for example 010 is not a valid number.
-A palindrome is a number which reads the same backward or forward. Examples: 11, 101, 292.


在10和1000之间有三个不同的回文数字，并且它们的和也是回文。这三个数相乘的最大可能值是多少？

数字不能以0开始，例如010不是一个有效的数字。
回文数是这样一个数字，它从左或者右读取都相同，例如：11，101，292。

又是一个没意思的题目。

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找出回文数表自连接

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newkid 发表于 2015-8-20 02:34
#4 SUM OF PALINDROMES

Three different numbers between 10 and 1000 are palindromes and their sum i ...

444+222+333=999,最大值只能比这个更大

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〇〇 发表于 2015-8-20 08:38
444+555=999,最大值只能比这个更大

with t as(select level l from dual connect by level<=999)
,a as (select l from t where to_char(l)=reverse(to_char(l))and l>10)
select max(a.l*b.l*c.l*(a.l+b.l+c.l))
from a,a b,a c where a.l<b.l and b.l<c.l and to_char(a.l+b.l+c.l)=reverse(to_char(a.l+b.l+c.l));  2    3    4  

MAX(A.L*B.L*C.L*(A.L+B.L+C.L))
------------------------------
                    1.3142E+12

SQL> set num 14
SQL> /

MAX(A.L*B.L*C.L*(A.L+B.L+C.L))
------------------------------
                 1314213733404

乘多了，以楼下为准

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〇〇 发表于 2015-8-20 08:48
with t as(select level l from dual connect by level10)
select max(a.l*b.l*c.l*(a.l+b.l+c.l))
fr ...

with t as(select level l from dual connect by level<=999)
,a as (select l from t where to_char(l)=reverse(to_char(l))and l>10)
,d as(select a.l a,b.l b,c.l c,a.l+b.l+c.l e,a.l*b.l*c.l d
from a,a b,a c where a.l<b.l and b.l<c.l and to_char(a.l+b.l+c.l)=reverse(to_char(a.l+b.l+c.l)))
select * from d where d=(select max(d) from d);  2    3    4    5  

             A              B              C              E              D
-------------- -------------- -------------- -------------- --------------
           787            797            858           2442      538171062

lugionline

In[1]:= Last@Sort@Map[{Apply[Times, #], #} &, Select[
    Tuples[11*Range[9] \[Union] Flatten[Array[101 #1 + 10 #2 - 10 &, {9, 10}]], 3],
    Apply[Greater, #] && ((StringReverse[#] == #) & @IntegerString@Total@#) &]]

Out[1]= {538171062, {858, 797, 787}}

改进了一下