Puzzleup 2013挑战赛即将开始

〇〇

newkid 发表于 2013-9-5 00:01
#7 Ten Numbers

_

In how many different ways
对称的也算不同的？

newkid

没啥对称，上面凸了个角。

〇〇

--7
with t as (select level-1 c from dual connect by level<=10)
,s(lv,c,s) as(select 1,c,cast(c as varchar(10)) from t union all
select lv+1, t.c ,s||t.c from s,t where instr(s.s,t.c)=0 and abs(t.c-s.c)<>1 and mod(t.c-s.c,2)<>0)
select count(*) from s where lv=2;

〇〇

/*
a
bcd
gfe
hij
用0代替10，不知对不对*/
with t as (select level-1 c from dual connect by level<=10)
,s(lv,c,s) as(select 1,c,cast(c as varchar(10)) from t union all
select lv+1, t.c ,s||t.c from s,t where instr(s.s,t.c)=0 and abs(t.c-s.c)<>1 and mod(t.c-s.c,2)<>0)
,u as(select substr(s,1,1)a,
substr(s,2,1)b,
substr(s,3,1)c,
substr(s,4,1)d,
substr(s,5,1)e,
substr(s,6,1)f,
substr(s,7,1)g,
substr(s,8,1)h,
substr(s,9,1)i,
substr(s,10,1)j
from s where lv=10 )
--select count(*) from u
select cast(a||chr(10)||b||c||d||chr(10)||g||f||e||chr(10)||h||i||j as varchar(20))x from u
where
abs(b-g)<>1 and mod(b-g,2)<>0 and
abs(c-f)<>1 and mod(c-f,2)<>0 and
abs(f-i)<>1 and mod(f-i,2)<>0 and
abs(e-j)<>1 and mod(e-j,2)<>0;
X
--------------------
1
472
905
638

1
496
703
258

1
638
905
472

1
694
307
852

1
836
509
274

1
852
307
694

8
147
692
305

8
163
490
725

8
305
692
147

8
361
094
527

8
503
296
741

8
527
094
361

〇〇

逐个替换的结果应该是符合的，但不确定有没有遗漏
17* select translate(cast(a||chr(10)||b||c||d||chr(10)||g||f||e||chr(10)||h||i||j as varchar(20)),'0123456789','123456789a')x from u
SQL> /

X
----------------------------------------
2
583
a16
749

2
5a7
814
369

2
749
a16
583

2
7a5
418
963

2
947
61a
385

2
963
418
7a5

9
258
7a3
416

9
274
5a1
836

9
416
7a3
258

9
472
1a5
638

9
614
3a7
852

9
638
1a5
472

〇〇

〇〇 发表于 2013-9-5 12:26
逐个替换的结果应该是符合的，但不确定有没有遗漏
17* select translate(cast(a||chr(10)||b||c||d||chr( ...

发现怪事，第3排第2个不是1就是10

newkid

我直接用十六进制，但排列上没想到用你那种Z字形的方法。

WITH t (cnt,c1,c2,c3,l,r) AS (
SELECT 1 cnt, LEVEL c1,TO_NUMBER(NULL) c2,TO_NUMBER(NULL) c3,TO_NUMBER(NULL) l,CAST(TO_CHAR(LEVEL,'X') AS VARCHAR2(20)) FROM DUAL CONNECT BY LEVEL<=10
UNION ALL
SELECT t.cnt+1
      ,DECODE(MOD(t.cnt,3),1,n.n,t.c1)
      ,DECODE(MOD(t.cnt,3),2,n.n,t.c2)
      ,DECODE(MOD(t.cnt,3),0,n.n,t.c3)
      ,DECODE(MOD(t.cnt,3),0,TO_NUMBER(NULL),n.n)
      ,t.r||TO_CHAR(n.n,'X')
  FROM t,(SELECT LEVEL n FROM DUAL CONNECT BY LEVEL<=10) n      
WHERE t.cnt<10
       AND INSTR(t.r,TO_CHAR(n.n,'X'))=0
       AND NVL(DECODE(MOD(t.cnt,3),1,t.c1,2,t.c2,t.c3)-n.n,3) IN (-9,-7,-5,-3,3,5,7,9)
       AND NVL(t.l-n.n,3) IN (-9,-7,-5,-3,3,5,7,9)
)
SELECT r FROM t WHERE cnt=10;

R
----------------------
2 5 8 3 A 1 6 7 4 9
2 5 A 7 8 1 4 3 6 9
2 7 4 9 A 1 6 5 8 3
2 7 A 5 4 1 8 9 6 3
2 9 4 7 6 1 A 3 8 5
2 9 6 3 4 1 8 7 A 5
9 2 5 8 7 A 3 4 1 6
9 2 7 4 5 A 1 8 3 6
9 4 1 6 7 A 3 2 5 8
9 4 7 2 1 A 5 6 3 8
9 6 1 4 3 A 7 8 5 2
9 6 3 8 1 A 5 4 7 2

12 rows selected.

newkid

其实，如果事后过滤的话CONNECT BY就足够用了：
WITH u AS (
select TO_NUMBER(substr(s,1,1) ,'X') a,
       TO_NUMBER(substr(s,2,1) ,'X') b,
       TO_NUMBER(substr(s,3,1) ,'X') c,
       TO_NUMBER(substr(s,4,1) ,'X') d,
       TO_NUMBER(substr(s,5,1) ,'X') e,
       TO_NUMBER(substr(s,6,1) ,'X') f,
       TO_NUMBER(substr(s,7,1) ,'X') g,
       TO_NUMBER(substr(s,8,1) ,'X') h,
       TO_NUMBER(substr(s,9,1) ,'X') i,
       TO_NUMBER(substr(s,10,1),'X') j,
       s
  FROM (SELECT REPLACE(SYS_CONNECT_BY_PATH(TRIM(TO_CHAR(n,'X')),'/'),'/') s
          FROM (SELECT LEVEL n FROM DUAL CONNECT BY LEVEL<=10)
         WHERE LEVEL=10
         CONNECT BY NOCYCLE abs(n-PRIOR n)<>1 and mod(n-PRIOR n,2)<>0
       )
)
SELECT s FROM u
where abs(b-g)<>1 and mod(b-g,2)<>0 and
       abs(c-f)<>1 and mod(c-f,2)<>0 and
       abs(f-i)<>1 and mod(f-i,2)<>0 and
       abs(e-j)<>1 and mod(e-j,2)<>0;

〇〇

newkid 发表于 2013-9-6 04:43
其实，如果事后过滤的话CONNECT BY就足够用了：
WITH u AS (
select TO_NUMBER(substr(s,1,1) ,'X') a,

n-PRIOR n good

newkid

#8 Unique Codes

Using each of the digits from 0 to 9 once, you are going to make codes of 10 digits length. Any 4 adjacent digits used in one code, can’t be used samely in another code.

How many codes can be obtained according to this rule?

Example:
If you have the code 0792435861, then you can’t have a code containing 0792, 7924, 9243, 2435, 4358, 3586 or 5861.

唯一编码
用数字0-9构成10位编码，每个数字只用一次。在一个编码中的任意相邻四位数字不能被另外一个编码使用。
根据这个规则总共能构造出多少个编码？
例子：
如果你有一个编码0792435861，那么你就不能在其他编码中包含 0792, 7924, 9243, 2435, 4358, 3586 或 5861.

最先答出来的人可以向我索取一个徽章！