PUZZLEUP 2015

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〇〇 发表于 2015-8-13 08:34
把思路用程序体现

with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t9 as(select a.l||b.l||c.l l from t3 a,t3 b,t3 c)
select max(l) from t9 where substr(l,2,3) in(select l from t3)
and substr(l,3,3) in(select l from t3)
and substr(l,5,3) in(select l from t3)
and substr(l,6,3) in(select l from t3)
;
太慢

lastwinner

〇〇 发表于 2015-8-13 08:58
with t as(select level-1 l from dual connect by levelb.l*c.l
and a.lb.l and b.lc.l and c.la.l)
, ...

慢在造数环节，找了半天找到了
http://www.itpub.net/thread-1796279-1-1.html
居然不在精华区，白翻了半天

lastwinner

〇〇 发表于 2015-8-13 08:34
把思路用程序体现

先求出所有的符合条件的3位数
然后做connect by，尽可能的以9开头，尽可能的以较大的数衔接在右边（左边数的后两位与右边数的前两位匹配），尽可能的使结果长度为10（level=8）

〇〇

lastwinner 发表于 2015-8-13 09:18
慢在造数环节，找了半天找到了
http://www.itpub.net/thread-1796279-1-1.html
居然不在精华区，白翻了 ...

t9还是很快的，最后一步慢

〇〇

〇〇 发表于 2015-8-13 09:36
t9还是很快的，最后一步慢

SQL> with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t9 as(select a.l||b.l||c.l l from t3 a,t3 b,t3 c
where a.l<>b.l and b.l<>c.l and c.l<>a.l)
select count(*) from t9;  2    3    4    5    6  

  COUNT(*)
----------
   2924064

已用时间:  00: 00: 00.40

〇〇

lastwinner 发表于 2015-8-13 09:32
先求出所有的符合条件的3位数
然后做connect by，尽可能的以9开头，尽可能的以较大的数衔接在右边（左边 ...

4位还有，5位就没了？
SQL> with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t5 as(select distinct a.l||substr(b.l,-1) l from t4 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select max(l) from t5;  2    3    4    5    6    7    8  

MAX(L)
--------------------------------------------------------------------------------


已用时间:  00: 00: 00.07
SQL> with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select max(l) from t4;  2    3    4    5    6  

MAX(L)
--------------------------------------------------------------------------------
9807

已用时间:  00: 00: 00.02
SQL> with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select count(l) from t4;  2    3    4    5    6  

  COUNT(L)
----------
       504

已用时间:  00: 00: 00.02
SQL> with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where sqrt(a.l)/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t5 as(select distinct a.l||substr(b.l,-1) l from t4 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select count(l) from t5;  2    3    4    5    6    7    8  

  COUNT(L)
----------
         0

已用时间:  00: 00: 00.07

〇〇

〇〇 发表于 2015-8-13 09:56
4位还有，5位就没了？
SQL> with t as(select level-1 l from dual connect by levelb.l*c.l
and a.lb. ...

看错了，平方不是平方根
with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where a.l*a.l/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t5 as(select distinct a.l||substr(b.l,-1) l from t4 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select count(l) from t5;

〇〇

〇〇 发表于 2015-8-13 09:58
看错了，平方不是平方根
with t as(select level-1 l from dual connect by levelb.l*c.l
and a.lb.l a ...

10位还是很慢
with t as(select level-1 l from dual connect by level<=10)
,t3 as(select distinct a.l||b.l||c.l l from t a,t b,t c where a.l*a.l/2>b.l*c.l
and a.l<>b.l and b.l<>c.l and c.l<>a.l)
,t4 as(select distinct a.l||substr(b.l,-1) l from t3 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t5 as(select distinct a.l||substr(b.l,-1) l from t4 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t6 as(select distinct a.l||substr(b.l,-1) l from t5 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t7 as(select distinct a.l||substr(b.l,-1) l from t6 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t8 as(select distinct a.l||substr(b.l,-1) l from t7 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t9 as(select distinct a.l||substr(b.l,-1) l from t8 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
,t10 as(select distinct a.l||substr(b.l,-1) l from t9 a,t3 b
where substr(a.l,-2,2)=substr(b.l,1,2) and instr(a.l,substr(b.l,-1))=0)
select count(l) from t10;

lastwinner

lastwinner 发表于 2015-8-13 09:32
先求出所有的符合条件的3位数
然后做connect by，尽可能的以9开头，尽可能的以较大的数衔接在右边（左边 ...

with t as (select rownum-1 n from dual connect by rownum<=10)
,u as (select cast(a.n||b.n||c.n as char(3)) n3, cast(a.n||b.n as char(2)) n2l, cast(b.n||c.n as char(2)) n2r, cast(a.n as char(1)) n1l, cast(c.n as char(1)) n1r from t a, t b, t c where a.n<>b.n and b.n<>c.n and c.n<>a.n
         and power(a.n,2)>2*b.n*c.n)
,v as(select u.n3, level lvl, replace(sys_connect_by_path(case when level=1 then n3 else n1r end, ','),',') p, u.n1r from u --start with to_number(n3)>800
         connect by n2l=prior n2r
                and n1r<>prior n1l/*初步过滤*/)
,w as (select n3, lvl, to_number(p) pn from v where instr(substr(p,1,length(p)-3),n1r)=0/*终极过滤，不允许重复数字出现*/ order by 3 desc)
select * from w where rownum<2;

N3           LVL               PN
--- ---------- ----------
107             7        985432107

lastwinner

〇〇 发表于 2015-8-13 10:06
10位还是很慢
with t as(select level-1 l from dual connect by levelb.l*c.l
and a.lb.l and b.lc.l  ...

看89#，十来毫秒出