puzzleup 2018

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10秒减01秒最小?
21-12
32-23
43-34
54-45
都一样

〇〇

newkid 发表于 2018-11-2 22:17
这也能扯上“用PLSQL的”？你的逻辑太强大。

我看了一下MMA的识别，竟然把0认成9，这识别出来的程序还 ...

训练不足

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alter session set nls_date_format ='mmddyyhh24mi';

with t as (select to_char(date'2018-11-08'+ interval '1' MINUTE * (level-1),'hhmi') d from dual connect by level<24*60),
a as(select t.d from t where not exists(select 1 from (select level-1 l from dual connect by level<=10) where instr(d,l,1,2)>0 ))
select min(a.d-a1.d) from a,a a1 where a.d>a1.d and translate('a'||a.d,'a'||a1.d,'a')='a';

9

lugionline

这也能叫puzzle？ 题目中1705892346 稍微变异一下1706892345， 1706892354，咱用mma的根本就不写代码

〇〇

lugionline 发表于 2018-11-8 10:03
这也能叫puzzle？ 题目中1705892346 稍微变异一下1706892345， 1706892354，咱用mma的根本就不写代码

改成最大的呢

solomon_007

# 15
好吧，你们都口算，我还是暴力一把吧

SQL> with t as (select level -1 n from dual connect by level <=100),
  2       d as (select lpad(n,2,'0') dd from t where n between 1 and 31 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  3       m as (select lpad(n,2,'0') mm from t where n between 1 and 12 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  4       y as (select lpad(n,2,'0') yy from t where n between 0 and 99 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  5       h as (select lpad(n,2,'0') hh from t where n between 0 and 23 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  6       i as (select lpad(n,2,'0') mi from t where n between 0 and 59 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  7       s (time_num) as (
  8          select dd||mm||yy||hh||mi time_num
  9            from d,m,y,h,i
10           where instr(dd,substr(mm,1,1))=0 and instr(dd,substr(mm,2,1))=0
11             and instr(dd||mm,substr(yy,1,1))=0 and instr(dd||mm,substr(yy,2,1))=0
12             and instr(dd||mm||yy,substr(hh,1,1))=0 and instr(dd||mm||yy,substr(hh,2,1))=0
13             and instr(dd||mm||yy||hh,substr(mi,1,1))=0 and instr(dd||mm||yy||hh,substr(mi,2,1))=0
14       )
15  select min(abs(a.time_num - b.time_num))
16    from s a,s b
17   where a.time_num <> b.time_num
18  /
MIN(ABS(A.TIME_NUM-B.TIME_NUM)
------------------------------
                             9

solomon_007

〇〇 发表于 2018-11-8 10:44
改成最大的呢

SQL> with t as (select level -1 n from dual connect by level <=100),
  2       d as (select lpad(n,2,'0') dd from t where n between 1 and 31 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  3       m as (select lpad(n,2,'0') mm from t where n between 1 and 12 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  4       y as (select lpad(n,2,'0') yy from t where n between 0 and 99 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  5       h as (select lpad(n,2,'0') hh from t where n between 0 and 23 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  6       i as (select lpad(n,2,'0') mi from t where n between 0 and 59 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
  7       s (time_num) as (
  8          select dd||mm||yy||hh||mi time_num
  9            from d,m,y,h,i
10           where instr(dd,substr(mm,1,1))=0 and instr(dd,substr(mm,2,1))=0
11             and instr(dd||mm,substr(yy,1,1))=0 and instr(dd||mm,substr(yy,2,1))=0
12             and instr(dd||mm||yy,substr(hh,1,1))=0 and instr(dd||mm||yy,substr(hh,2,1))=0
13             and instr(dd||mm||yy||hh,substr(mi,1,1))=0 and instr(dd||mm||yy||hh,substr(mi,2,1))=0
14       )
15  select min(abs(a.time_num - b.time_num)),max(abs(a.time_num - b.time_num))
16    from s a,s b
17   where a.time_num <> b.time_num
18  /
MIN(ABS(A.TIME_NUM-B.TIME_NUM) MAX(ABS(A.TIME_NUM-B.TIME_NUM)
------------------------------ ------------------------------
                             9                     1501979184

〇〇

solomon_007 发表于 2018-11-8 11:15
SQL> with t as (select level -1 n from dual connect by level

怎么保证是合法的日期

solomon_007

〇〇 发表于 2018-11-8 12:02
怎么保证是合法的日期

--加个日期格式验证                                                         
SQL>  with function fn_valid_date(p_str in varchar2)
  2        return int
  3        is
  4          l_date date;
  5        begin
  6          l_date := to_date(p_str,'ddmmyy');
  7          return 1;
  8        exception when others then
  9          return 0;
10        end;
11        t as (select level -1 n from dual connect by level <=100),
12        d as (select lpad(n,2,'0') dd from t where n between 1 and 31 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
13        m as (select lpad(n,2,'0') mm from t where n between 1 and 12 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
14        y as (select lpad(n,2,'0') yy from t where n between 0 and 99 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
15        h as (select lpad(n,2,'0') hh from t where n between 0 and 23 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
16        i as (select lpad(n,2,'0') mi from t where n between 0 and 59 and substr(n,1,1)<> nvl(substr(n,2,1),'x')),
17        s (time_num) as (
18           select dd||mm||yy||hh||mi time_num
19             from d,m,y,h,i
20            where instr(dd,substr(mm,1,1))=0 and instr(dd,substr(mm,2,1))=0
21              and instr(dd||mm,substr(yy,1,1))=0 and instr(dd||mm,substr(yy,2,1))=0
22              and instr(dd||mm||yy,substr(hh,1,1))=0 and instr(dd||mm||yy,substr(hh,2,1))=0
23              and instr(dd||mm||yy||hh,substr(mi,1,1))=0 and instr(dd||mm||yy||hh,substr(mi,2,1))=0
24              and fn_valid_date (dd||mm||yy) = 1
25        )
26   select min(abs(a.time_num - b.time_num)),max(abs(a.time_num - b.time_num))
27     from s a,s b
28    where a.time_num <> b.time_num
29  /
MIN(ABS(A.TIME_NUM-B.TIME_NUM) MAX(ABS(A.TIME_NUM-B.TIME_NUM)
------------------------------ ------------------------------
                             9                     1501979184               

newkid

〇〇 发表于 2018-11-8 10:44
改成最大的呢

因为日，月，小时的第一位都很有限，所以最大最小是很容易推理出来的。

我还是玩了一把SQL，BITAND真是好东西，速度比加菲猫快：
WITH n as (select to_char(level-1) n,power(2,level-1) b from dual connect by level<=10)
,ddmm as (
select dm,sum(b) b
   from (select to_char(date '2001-1-1'+level-1,'ddmm') dm from dual connect by level<=365)
       ,n
   where instr(dm,n)>0
group by dm having count(*)=4
)      
,hhmi as (
select hhmi,sum(b) b
  from (select to_char(date '2001-1-1'+level/(24*60),'hh24mi') hhmi from dual connect by level<24*60)
      ,n
   where instr(hhmi,n)>0
group by hhmi having count(*)=4
)
,ddmmhhmi as (
select dm,hhmi,trunc(log(2,b)) b1,round(log(2,b-power(2,trunc(log(2,b))))) b2
  from (select dm,hhmi,power(2,10)-1-ddmm.b-hhmi.b b from ddmm,hhmi
        where bitand(ddmm.b,hhmi.b)=0
       )
)
select min(str),max(str) from (
select dm||decode(n,1,b1||b2,b2||b1)||hhmi as str
  from ddmmhhmi,(select * from n where n<2)
);