puzzleup 2021

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37:258
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solomon_007

我回复的贴子被吞了

solomon_007

solomon_007 发表于 2021-9-2 20:09
我回复的贴子被吞了

#1


SQL> with t(n,x,y) as
  2  (select 1,0,0 from dual
  3   union all
  4   select 2,1,0 from dual
  5   union all
  6   select 3,2,0 from dual
  7   union all
  8   select 4,1/2,sqrt(3)/2 from dual
  9   union all
10   select 5,3/2,sqrt(3)/2 from dual
11   union all
12   select 6,1,sqrt(3) from dual
13   union all
14   select 7,0,sqrt(3) from dual
15  ),
16  s (n,x,y) as (
17  select rownum rn,x,y
18  from (
19        select distinct x,y
20        from (
21              select x,y from t
22              union all
23              select -x,y from t
24              union all
25              select -x,-y from t
26              union all
27              select x,-y from t
28             )
29         order by x,y
30       )
31  )
32  select count(*)
33    from (
34          select a.n n1,b.n n2,c.n n3,
35                 round(sqrt(power(a.x-b.x,2)+power(a.y-b.y,2)),2) ab,
36                 round(sqrt(power(a.x-c.x,2)+power(a.y-c.y,2)),2) ac,
37                 round(sqrt(power(b.x-c.x,2)+power(b.y-c.y,2)),2) bc
38            from s a,s b,s c
39           where
40                 a.n < b.n
41             and b.n < c.n
42         )
43   where ab=ac
44     and ac=bc
45  /

  COUNT(*)
----------
        66

newkid

给猫猫发一个章。
我的写法：

    *   *   *
  *   *   *   *
*   *   *   *   *   ------- 首先构造出这一行的坐标，Y=0, X=0,2,4,6,8，然后延申构造出上下各两行的坐标
  *   *   *   *     ------- 纵向点距设为1, 实际上应该为根号3, 在计算距离的时候补回来
    *   *   *


with m as (
SELECT (LEVEL-1)*2 AS X  ---- 构造出中间一行的坐标
      ,0 AS Y
  FROM DUAL
CONNECT BY LEVEL<=5
)
,d as (  --------- 延申构造出上下各两行的坐标
select m.x+abs(n) as x
      ,m.y+n as y
      ,rownum as id
  from m,(select level-3 n from dual connect by level<=5)
where abs(n)+m.x/2<=4
)
select *
from d d1,d d2,d d3
where d1.id<d2.id and d2.id<d3.id
and (d2.y-d1.y)*(d2.y-d1.y)*3+(d2.x-d1.x)*(d2.x-d1.x)   --------- 纵向乘以3补偿
    = (d3.y-d1.y)*(d3.y-d1.y)*3+(d3.x-d1.x)*(d3.x-d1.x)
and (d2.y-d1.y)*(d2.y-d1.y)*3+(d2.x-d1.x)*(d2.x-d1.x)
    = (d3.y-d2.y)*(d3.y-d2.y)*3+(d3.x-d2.x)*(d3.x-d2.x);

newkid

猫的回复被审核，我全放出来了。

solomon_007

newkid 发表于 2021-9-2 22:06
猫的回复被审核，我全放出来了。

还有其他贴子的回帖，也被河蟹了。。。WHAT THE FUCK!

你构造坐标的方法更好，我是手动构造第一象限，再复制3份，再去重

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极坐标版本

with
s as (select generate_series(1,6,1) i),
p as (select 0 d,0 r
union all select 1,pi()/3*i from s
union all select 2,pi()/3*i from s
union all select sqrt(3),pi()/6+pi()/3*i from s),
t as(select p.*,row_number()over() rn from p),
e as(
select row_number()over() n,p1.rn p1,p2.rn p2,(p1.d^2+p2.d^2-2*p1.d*p2.d*cos(p1.r-p2.r))::int d from t p1,t p2
where p1.rn>p2.rn),
tri as(
select e1.d from e e1,e e2
where e1.n>e2.n and e1.d=e2.d and e1.p1=e2.p1 and exists(select 1 from
e e3 where e3.d=e2.d and(e3.p1,e3.p2) in((e1.p2,e2.p2),(e2.p2,e1.p2))))
select count(*)cnt,d from tri group by rollup(d);

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n=4就很辛苦了

as (select 0 d,0 r
union all select 1,pi()/3*i from s
union all select 2,pi()/3*i from s
union all select sqrt(3),pi()/6+pi()/3*i from s
union all select 3,pi()/3*i from s
union all select 3*sqrt((1/6.0)^2+3/4.0),pi()/6-atan(1.0/6/(sqrt(3)/2))+pi()/3*i from s
union all select 3*sqrt((1/6.0)^2+3/4.0),pi()/6+atan(1.0/6/(sqrt(3)/2))+pi()/3*i from s
union all select 4,pi()/3*i from s
union all select 2*sqrt(3),pi()/6+pi()/3*i from s
union all select 4*sqrt((1/4.0)^2+3/4.0),pi()/6-atan(1.0/4/(sqrt(3)/2))+pi()/3*i from s
union all select 4*sqrt((1/4.0)^2+3/4.0),pi()/6+atan(1.0/4/(sqrt(3)/2))+pi()/3*i from s
),

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我想了一下，想还是用直角坐标，用到圆心的距离筛选不同的n

newkid

我全用整数坐标，都用不着根号。