puzzleup 2018

newkid

#15

TIME DIFFERENCES

17 May 1989 - 23:46 (17/05/89 - 23:46) can be numerically written as 1705892346, where each digit from 0 to 9 is used exactly once. What is the minimum numerical difference between two different time information having this property.

Note: 24-hour format is used therefore the time part is between 00:00 and 23:59.         


1989年5月17日 - 23:46（17/05/89 - 23:46）可以用数字写成1705892346，其中0到9之间的每个数字恰好只使用一次。具有这种属性的两个不同时间信息之间的最小数值差异是多少？

注意：使用24小时格式，因此时间部分在00:00和23:59之间。

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这题的答案应该就是把最后两位数字颠倒一下。

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答案不跨天？

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数字把日写在年前，不可能跨天

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

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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

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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