puzzleup 2020 11月开始

dhhb

105个，1256种.

newkid

dhhb 发表于 2020-12-13 16:34
105个，1256种.

我答案和你一样，你的代码是怎么写的？
等有空来验证一下猫猫的写法。

solomon_007

修改了 67 楼，规定了一下向量的方向，当考察正方形的某个顶点时，以这个顶点为中心，方向指向其它顶点和被考察的外面的点

SQL>
SQL>  with t as (select level n from dual connect by level <= 6),
  2        p as (select row_number() over(order by t1.n,t2.n) rn,t1.n x,t2.n y from t t1,t t2),
  3        s as (
  4  select p1.x x1,p1.y y1,
  5         p2.x x2,p2.y y2,
  6         p3.x x3,p3.y y3,
  7         p4.x x4,p4.y y4
  8    from p p1,p p2,p p3,p p4
  9   where p1.rn < p2.rn
10     and p2.rn < p3.rn
11     and p3.rn < p4.rn
12     and (p2.x-p1.x)*(p3.x-p1.x)+(p2.y-p1.y)*(p3.y-p1.y) = 0
13     and (p2.x-p1.x)*(p4.x-p2.x)+(p2.y-p1.y)*(p4.y-p2.y) = 0 and (p3.x-p1.x)*(p4.x-p3.x)+(p3.y-p1.y)*(p4.y-p3.y) = 0
14     and power((p2.x-p1.x),2)+power((p2.y-p1.y),2) = power((p3.x-p1.x),2)+power((p3.y-p1.y),2)
15  )
16  select
17         /*s1.x1 x1,s1.y1 y1,
18         s1.x2 x2,s1.y2 y2,
19         s1.x3 x3,s1.y3 y3,
20         s1.x4 x4,s1.y4 y4,
21         s2.x1 x5,s2.y1 y5,
22         s2.x2 x6,s2.y2 y6,
23         s2.x3 x7,s2.y3 y7,
24         s2.x4 x8,s2.y4 y8 */
25         count(*)
26    from s s1,s s2
27   where ( (s2.x1-s1.x1)*(s1.x3-s1.x1)+(s2.y1-s1.y1)*(s1.y3-s1.y1) < 0
28        or (s2.x1-s1.x1)*(s1.x2-s1.x1)+(s2.y1-s1.y1)*(s1.y2-s1.y1) < 0
29        or (s2.x1-s1.x2)*(s1.x4-s1.x2)+(s2.y1-s1.y2)*(s1.y4-s1.y2) < 0
30        or (s2.x1-s1.x2)*(s1.x1-s1.x2)+(s2.y1-s1.y2)*(s1.y1-s1.y2) < 0
31        or (s2.x1-s1.x3)*(s1.x1-s1.x3)+(s2.y1-s1.y3)*(s1.y1-s1.y3) < 0
32        or (s2.x1-s1.x3)*(s1.x4-s1.x3)+(s2.y1-s1.y3)*(s1.y4-s1.y3) < 0
33        or (s2.x1-s1.x4)*(s1.x2-s1.x4)+(s2.y1-s1.y4)*(s1.y2-s1.y4) < 0
34        or (s2.x1-s1.x4)*(s1.x3-s1.x4)+(s2.y1-s1.y4)*(s1.y3-s1.y4) < 0 )
35    and  ( (s2.x2-s1.x1)*(s1.x3-s1.x1)+(s2.y2-s1.y1)*(s1.y3-s1.y1) < 0
36        or (s2.x2-s1.x1)*(s1.x2-s1.x1)+(s2.y2-s1.y1)*(s1.y2-s1.y1) < 0
37        or (s2.x2-s1.x2)*(s1.x4-s1.x2)+(s2.y2-s1.y2)*(s1.y4-s1.y2) < 0
38        or (s2.x2-s1.x2)*(s1.x1-s1.x2)+(s2.y2-s1.y2)*(s1.y1-s1.y2) < 0
39        or (s2.x2-s1.x3)*(s1.x1-s1.x3)+(s2.y2-s1.y3)*(s1.y1-s1.y3) < 0
40        or (s2.x2-s1.x3)*(s1.x4-s1.x3)+(s2.y2-s1.y3)*(s1.y4-s1.y3) < 0
41        or (s2.x2-s1.x4)*(s1.x2-s1.x4)+(s2.y2-s1.y4)*(s1.y2-s1.y4) < 0
42        or (s2.x2-s1.x4)*(s1.x3-s1.x4)+(s2.y2-s1.y4)*(s1.y3-s1.y4) < 0 )
43    and  ( (s2.x3-s1.x1)*(s1.x3-s1.x1)+(s2.y3-s1.y1)*(s1.y3-s1.y1) < 0
44        or (s2.x3-s1.x1)*(s1.x2-s1.x1)+(s2.y3-s1.y1)*(s1.y2-s1.y1) < 0
45        or (s2.x3-s1.x2)*(s1.x4-s1.x2)+(s2.y3-s1.y2)*(s1.y4-s1.y2) < 0
46        or (s2.x3-s1.x2)*(s1.x1-s1.x2)+(s2.y3-s1.y2)*(s1.y1-s1.y2) < 0
47        or (s2.x3-s1.x3)*(s1.x1-s1.x3)+(s2.y3-s1.y3)*(s1.y1-s1.y3) < 0
48        or (s2.x3-s1.x3)*(s1.x4-s1.x3)+(s2.y3-s1.y3)*(s1.y4-s1.y3) < 0
49        or (s2.x3-s1.x4)*(s1.x2-s1.x4)+(s2.y3-s1.y4)*(s1.y2-s1.y4) < 0
50        or (s2.x3-s1.x4)*(s1.x3-s1.x4)+(s2.y3-s1.y4)*(s1.y3-s1.y4) < 0 )
51    and  ( (s2.x4-s1.x1)*(s1.x3-s1.x1)+(s2.y4-s1.y1)*(s1.y3-s1.y1) < 0
52        or (s2.x4-s1.x1)*(s1.x2-s1.x1)+(s2.y4-s1.y1)*(s1.y2-s1.y1) < 0
53        or (s2.x4-s1.x2)*(s1.x4-s1.x2)+(s2.y4-s1.y2)*(s1.y4-s1.y2) < 0
54        or (s2.x4-s1.x2)*(s1.x1-s1.x2)+(s2.y4-s1.y2)*(s1.y1-s1.y2) < 0
55        or (s2.x4-s1.x3)*(s1.x1-s1.x3)+(s2.y4-s1.y3)*(s1.y1-s1.y3) < 0
56        or (s2.x4-s1.x3)*(s1.x4-s1.x3)+(s2.y4-s1.y3)*(s1.y4-s1.y3) < 0
57        or (s2.x4-s1.x4)*(s1.x2-s1.x4)+(s2.y4-s1.y4)*(s1.y2-s1.y4) < 0
58        or (s2.x4-s1.x4)*(s1.x3-s1.x4)+(s2.y4-s1.y4)*(s1.y3-s1.y4) < 0 )


  COUNT(*)
----------
      3565

solomon_007

p1.rn < p2.rn
10     and p2.rn < p3.rn
11     and p3.rn < p4.rn  

这个条件已经确定了顶点的位置，所以后面的条件要依据这个来改。。。。不能瞎定。。。

newkid

你用4x4 验证一下你的代码，应该有20种画法。

jihuyao

If there exist more directions other than 0 and 45 degrees for 6x6 grid it is hard to imagine  how many directions there could be for a relatively large grid NxN.  I would finger count those extreme squares pairing with those “normal” squares since 6x6 grid is small enough (lol).  As to using where clause to check each pair I would prefer using a function which take rid, cid, and length as inputs and returns flag value 0 or 1.  Basically it should be able to process the check like “logical calculation”.  As to the pairing of abnormal and abnormal (say 45 and 45 ), I realized the condition should be the same as that for the pair of normal and normal.  This is because if rotating the grid clockwise wise (but keep squares unmoved) then it becomes the pair of normal and normal except the length unit is different (but it has no impact on the sharing condition).  So for any pair of abnormal 1 and abnormal 2 it can be treated as a pair of normal and a modified abnormal.  Hopefully this transformation can speed up the validation process in part two.

newkid

jihuyao 发表于 2020-12-14 13:10
If there exist more directions other than 0 and 45 degrees for 6x6 grid it is hard to imagine  how m ...

还finger count呢，你有几个手指头？你的function是要检查“一对”正方形的关系吧，那么看你的参数设计已经败了，你以为两个正方形永远都是同样角度的？还是再仔细看看56楼的图，里面两个正方形一个正的一个斜的。

newkid

#7 COUNTERFEIT COINS

There are 16 coins, 3 of which are slightly heavier or lighter than the others. The counterfeit coins both have the same weight but it is not known whether they are heavier or lighter than the genuine ones.

Using a balance scale, what is the minimum number of weighings in order to determine whether the counterfeit coins are heavier or lighter?

Notes:

All the coins look identical.
The genuine coins have the same weight.
You can compare any number of coins on the balance scale.

又有不良信息，不让我发布中文译文。

newkid

有16枚硬币，其中3枚比其他硬币略重或略轻。假 币 的重量都一样，但不知道比真币重还是轻。用天枰称，要确定假 币是重还是轻，最少要称多少次？

注意：
所有的硬币看起来都是一样的。
真币的重量是一样的。
你可以在天枰上比较任何数量的硬币。

-------这道题没有说要把三枚假 币找出来，只是想知道轻重

newkid

"假 币"是特么的敏感词，但是中间加了个空格就可以了！