ITPUB论坛-中国专业的IT技术社区

 找回密码
 注册
查看: 13963|回复: 45

[SQL] 这几年牛蛙写过的奇葩代码

[复制链接]
论坛徽章:
78
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋头
日期:2012-12-08 09:43:38鲜花蛋
日期:2012-11-17 12:02:07鲜花蛋
日期:2013-02-05 21:53:34复活蛋
日期:2012-11-17 12:02:07SQL极客
日期:2013-12-09 14:13:35SQL数据库编程大师
日期:2013-12-06 13:59:43SQL大赛参与纪念
日期:2013-12-06 14:10:50ITPUB季度 技术新星
日期:2012-11-27 10:16:10最佳人气徽章
日期:2013-03-19 17:24:25
发表于 2016-8-31 23:55 | 显示全部楼层 |阅读模式
本帖最后由 udfrog 于 2016-9-1 17:24 编辑

之前自己整理过一个文档,存着自己这几年在pub上写过的有趣代码。今天又追加了几个今年写的,一并分享出来,如果有缘人喜欢,那就很好。
我这个人不是个着调的开发,从来不爱写注释,所以看着痛苦的话,抱歉了。格式也不够好。
这几年写下来,我也总在体会,我自认为我的sql写得很不同寻常,的好,最根本的原因就是我对问题建模所产生的数据结构有相当的优越性,那优秀的算法自然也就不难形成。


巧的是,论坛新开了打赏功能,我觉得挺有意思的,如果喜欢,可以尽情打赏,让我尝个鲜。


1. Adjacent Digits


How many positive 10-digit integers are there with the property that for each digit N, the number of adjacent N's is at most N?


(Examples: 1223334444, 7777777822, 5999999999, 3434343434)


这个没去优化,在我的pc上大概要跑8分钟,也不清楚是否可以优化
with b as (
select level lv from dual connect by level<=9),
t1 (str) as (
select rpad(lv, level, lv)
from    b
connect by level<=lv and prior lv=lv and prior dbms_random.value()>0
),
t as (
select * from t1 where length(str)<=5
),
r (str) as (
select  str
from    t
union all
select  r.str||t.str
from    t, r
where   substr(r.str, -1, 1)!=substr(t.str, 1, 1)
and     length(r.str||t.str)<=5
),
r2 as (
select  str, substr(str, -1) trail, 5-nvl(length(rtrim(str, substr(str, -1))), 0) len
from    r
where   length(str)=5
),
r3 as (
select  str, substr(str, 1, 1) lead, 5-nvl(length(ltrim(str, substr(str, 1, 1))), 0) len
from    r
where   length(str)=5
)
select  56143*56143-count(*)
from    r2, r3
where   r2.trail=r3.lead
and     r2.len+r3.len>r2.trail;




2.Double Factored
Let''s call a number "double factored" if at least one of its prime factors repeats itself when factorised. What is the smallest positive integer that itself and its four neighbors (2 before, 1 before, the number itself, 1 after, 2 after) are double factored?
If the problem was asked for the number and its two neighbors (1 before, the number itself, 1 after) then the answer would be 49. (48=2x2x2x2x3 , 49=7x7, 50=2x5x5).


with df as (
select (rownum+1)*(rownum+1) df from dual connect by rownum<=20),
t as (
select rownum f from dual connect by rownum<=1000),
tmp as (
select  min(df*f) keep (dense_rank first order by df*f) r, row_number() over (order by df*f) rn
from  df, t
group by df*f)
select  min(a.r)-2
from  tmp a, tmp b
where a.r=b.r+4
and a.rn=b.rn+4;




3.Full Prime Number


All the numbers that can be read with two or more adjacent digits inside a prime number are also prime numbers. What is the greatest number having this property?


Exampe:6173 is such a number. Because 61, 17, 73, 617, 173 and 6173 are prime numbers


WITH t0 AS (
    SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (100000)/2-1-1
    ),
t as(SELECT rn from t0
      where mod(rn,3)<>0
      and mod(rn,5)<>0
      and mod(rn,7)<>0
    )
,p AS (
SELECT TO_CHAR(rn) n,LENGTH(rn) len
   FROM (SELECT rn from t
         MINUS
         SELECT t1.rn * t2.rn
           FROM t t1, t t2
         WHERE t1.rn <= t2.rn
            AND t1.rn BETWEEN 11 AND SQRT(100000)
            AND t1.rn * t2.rn <100000
       )
WHERE rn>10
),
tmp (n, digit) as (
select  n, 2 digit
from  p
where len=2
union all
select  a.n||substr(b.n, -1) n, a.digit+1 digit
from  tmp a, tmp b, p
where a.digit=b.digit
and substr(a.n, 2)=substr(b.n, 1, b.digit-1)
and a.n||substr(b.n, -1)=p.n)
select max(to_number(n)) from tmp;




4.Prime Sums


All of the digits of a number are different. The sums of all neighboring three digits of this number are prime numbers. What is the largest number having these properties?


with b as (
select rownum-1 n from dual connect by rownum<=10),
p as (
select level n from dual where (level>=2 and mod(level, 2)<>0 and mod(level, 3)<>0 and mod(level, 5)<>0) or level in (2, 3, 5) connect by level<=23),
tmp as (
select     d1.n||d2.n||d3.n d
from       b d1, b d2, b d3, p
where      d1.n<>d2.n
and        d1.n<>d3.n
and        d2.n<>d3.n
and        d1.n+d2.n+d3.n=p.n),
t (d) as (
select     d
from       tmp
where      not d like '0%'
union all
select     t.d||substr(tmp.d, -1)
from       t, tmp
where      substr(t.d, -2)=substr(tmp.d, 1, 2)
and        instr(t.d, substr(tmp.d, -1))=0)
select max(to_number(d)) from t;




5.Test


35 students took a test of 100 problems.


-Each problem is solved by exactly one girl and one boy.
-There are at least one girl who solved exactly one problem, and at least one girl who solved exactly two problems.
-There are at least one boy who solved exactly four problems, and at least one boy who solved exactly five problems.
-The number of the problems solved by the girl who has solved the maximum number of problems is X.
-The number of the problems solved by the boy who has solved the maximum number of problems is Y.


If the minimum values for both X and Y are the same, find the number of girls who have taken the test.


select  n
from  (select rownum+2 n from dual connect by rownum<=35-2-2-1)
where ceil((100-9)/(35-n-2))=ceil((100-3)/(n-2));




6.Solve Problem: Complex Sequence Generation
One of the questions on the Roundtable discussion was: "On one of the forums (SQL.ru) sometimes one of the members challenge everybody by posting a task (Friday challenge) and seeking for the best solution. I remember we had something like that about year ago here as well. Can we bring that back?"
Our "Solve Problem" feature never went away - we just hadn't posted one for a while. And Faig's question spurred me to take action. So we now have up on the website the following coding challenge:
Oracle sequences generate the next integer value available in the sequence. But sometimes applications rely on keys with a more complex structure. In this challenge, a player from Islamabad asks for some help to generate an alpha-numeric value in a sequence that follows a very interesting set of rules (explained in the question). He writes:
I need to write a function with this header:
FUNCTION plch_next_key (from_this_key_in in varchar2) RETURN VARCHAR2in which the key is a 5 digit alpha-numeric sequence of values, each of which has two parts:


1. Letter part: letters from A-Z, a-z. The number of letters increase from left to right
2. Digit part: left filled with 0s to length 5, after letters.
The rightmost letter stays fixed, and the remainder (a number starting with 1) is incremented, is then left padded with 0s to length 5 (including letters), up to all 9s.
When you have all 9s and the rightmost letter is not z, then increment the rightmost letter through the sequence A-Z, a-z and the remaining digits are reset to 1 (padded left with 0s)s.
When the rightmost letter is z, but there are non-z letters to the left, the rightmost z and the other letters are incremented A-Z, a-z as needed, with the rightmost z's becoming A's and the 9s are reset to 1 (padded left with 0s).
When all letters are z and all digits are 9s, then replace each z with A, replace the first 9 with A, and the remaining digits are reset to 1 (padded left with 0s).
Here's a partial set of key ranges to illustrate these rules:
A0001...A9999B0001...B9999....Z0001...Z9999a0001...z9999AA001...AA999AB001...AB999....AZ001...AZ999Aa001...Az999BA001...BZ999Ba001...Bz999CA001...zz999AAA01...zzz99AAAA1...zzzz9AAAAA...zzzzzYour mission, if you choose to accept it, is to implement this function so that it performs efficiently and is easy to understand.


create or replace function plch_next_key(from_this_key_in in varchar2)
return varchar2
is
len int:=5;
v_digit varchar2(32);
v_char varchar2(32);
begin
v_char:=regexp_substr(from_this_key_in, '\D+');
v_digit:=regexp_substr(from_this_key_in, '\d+');
return case when replace(v_digit, '9') is not null then
v_char||lpad((v_digit+1), length(v_digit),'0')
else
substr(
regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 1)||
nvl(replace(chr(ascii(regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 2))+1), '[', 'a'), 'A')||
replace(regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 3), 'z', 'A')||
rpad('0', length(v_digit)+nvl(sign(length(replace(v_char, 'z'))), 0)-2, '0')||'1',
1, len
)
end;
end;
/




7.Four Digits


You have a number where any digit appears at most twice. The sum of all neighboring four digits in this number is a square number.


What is the maximum possible value for this number?


Example: 205290 is such a number because no digit appears more than twice and 2+0+5+2, 0+5+2+9, and 5+2+9+0 are square numbers.




with t as (
  select to_char(level-1) l from dual connect by level<=10),
t2 as(
select
a.l||b.l||c.l||d.l v4 --??
from t a,t b,t c,t d
where a.l+b.l+c.l+d.l in(4,9,16,25)
and a.l<>0
),
r(vx,lv)as(
select cast(v4 as varchar(40)),1 from t2
union all
select vx||l,lv+1 from r,t
where
instr(vx,l,1,2)=0 --????2?,????????????
and substr(vx,-1,1)+substr(vx,-2,1)+substr(vx,-3,1)+l in(4,9,16,25)
and lv<=16)
select to_char(max(0+vx)) from r where regexp_substr(vx, '([[:digit:]])(\1){2}') is not null;




8.Ten Numbers


_
|_|_ _
|_|_|_|
|_|_|_|
|_|_|_|


Place each of the ten numbers between 1 and 10 in the squares so that no two adjacent squares (horizontally or vertically) should contain;


consecutive numbers,
numbers with the same parity (odd or even).


In how many different ways can this task be accomplished?


with t as (
select rownum n from dual connect by rownum<=8)
select count(*)*2
from t
where  level=8 and abs(connect_by_root(n)-n) in (3, 5, 7)
start with n in (2, 4, 6)
connect by nocycle abs(prior n -n) in (3, 5, 7);




9.Plastic Digits


There are four sets of plastic digits. Each set has four digits (1, 2, 3, 4) and each set has a different colour (red, blue, green, yellow). You will place these 16 digits into 4x4 table such that every two adjacent squares (vertically or horizontally) should have either the same color or the same number.>

In how many different ways can this placement be done?>

If the problem was asked for 2 digits, 2 colors and a 2x2 table, then the answer would be 8.
  
  
WITH d AS (
SELECT n,c,POWER(2,ROWNUM-1) AS id
   FROM (SELECT LEVEL n FROM DUAL CONNECT BY LEVEL<=4)
       ,(SELECT 'R' c FROM DUAL UNION ALL SELECT 'B' c FROM DUAL UNION ALL SELECT 'G' c FROM DUAL UNION ALL SELECT 'Y' c FROM DUAL)
)
,t(cnt,bits,n0,c0,n1,c1,n2,c2,n3,c3) AS (
SELECT 1
       ,id
       ,null
      ,null
       ,null
      ,null
       ,null
      ,null
       ,n
       ,c
   FROM d
where        rownum=1
UNION ALL
SELECT t.cnt+1
       ,t.bits+d.id
       ,nvl(t.n1, d.n)
       ,nvl(t.c1, d.c)
       ,t.n2
       ,t.c2
       ,t.n3
       ,t.c3
       ,d.n
       ,d.c
FROM t,d
WHERE cnt<16
        AND BITAND(t.bits,d.id)=0
        AND (MOD(t.cnt,4)=0 OR d.n=t.n3 OR d.c=t.c3)
        AND (t.cnt<4 OR d.n = t.n0 OR d.c = t.c0)
)
SELECT COUNT(*)*16 FROM t WHERE cnt=16;




10.Balls And Boxes

What is the number of different ways to distribute different colored ten balls into four different boxes?

-All boxes will have different number of balls.
-An empty box is allowed.

If the problem was asked for 3 balls (white, black, red) and 2 boxes, then the answer would be 8.
1. (empty) - (w,b,r), 2. (w,b,r) - (empty), 3. (w) - (b,r), 4. (b) - (w,r), 5. (r) - (b,w), 6. (b,r) - (w)
7. (w,r) - (b), 8. (b,w) - (r)


此解法刻意追求代码的趣味性,性能远非最优
with t as (
select  sign(bitand((level-1), 1))+sign(bitand((level-1), 2))+sign(bitand((level-1), 4))+sign(bitand((level-1), 8))+
  sign(bitand((level-1), 16))+sign(bitand((level-1), 32))+sign(bitand((level-1), 64))+sign(bitand((level-1), 128))+
  sign(bitand((level-1), 256))+sign(bitand((level-1), 512)) s, rownum-1 n from dual connect by level<=1024)
select  count(*)*24
from  t t1, t t2, t t3, t t4
where bitand(t1.n, t2.n)=0
and bitand(t1.n, t3.n)=0
and bitand(t1.n, t4.n)=0
and bitand(t2.n, t3.n)=0
and bitand(t2.n, t4.n)=0
and bitand(t3.n, t4.n)=0
and t1.n+t2.n+t3.n+t4.n=1023
and t1.s>t2.s
and t2.s>t3.s
and t3.s>t4.s;




11.Two Pawns

Five pieces are randomly selected from a standard chess set. If it is known that at least one of these pieces is a pawn, what is the probability that at least two of these five pieces are white pawns?

Note:In a standard chess set there are 8 pawns, 2 rooks, 2 bishops, 2 knights, 1 queen, and 1 king for each color (white and black) making a total of 32 pieces.

After simplification, enter your answer as a/b.


with t as (
select case when rownum<=8 then 0 when rownum<=16 then 1 else 2 end s, rownum n from dual connect by rownum<=32)
select  count(case when regexp_count(sys_connect_by_path(s, ','), '0')>=2 then 1 else null end)||'/'||count(*)
from  t
where level=5
connect by prior n<n and level<=5
start with s in (0, 1);




12.Digits

A number has different digits, and the digits of the sum of each neighboring two digits are not used in this number.

What is the largest number satisfying these conditions?

Example:
7692
7+6=13, 6+9=15, 9+2=11 (The digits 1, 3 and 5 are not used.)


with t as (
select to_char(rownum-1) n from dual connect by rownum<=10),
r(n, p, na) as (
select     n, n, n
from       t
union all
select     t.n, r.p||t.n, r.na||(r.p+t.n)
from       r, t
where      translate(r.p||t.n, (r.n+t.n)||'x', 'x')=r.p||t.n
and instr(r.p, t.n)=0
and instr(r.na, t.n)=0)
select     max(to_number(p)) from r;




13.        千王之王
52张牌,四张A,随机打乱后问,从左到右一张一张翻直到出现第一张A,请问平均要翻几张牌?


with b as (
select rownum n from dual connect by rownum<=51),
f (fac, n, lv) as (
select b.n, b.n, b.n from b where n>=2
union all
select     f.fac*b.n, f.n, f.lv-1
from       f, b
where      f.lv-1=b.n),
p (p, n, lv) as (
select     48, 48, 1
from       dual
union all
select     p.p*b.n, p.n-1, p.lv+1
from       p, b
where      p.n-1=b.n)
select     sum(b.n*nvl(p.p, 1)*f.fac)/sum(nvl(p.p, 1)*f.fac)
from       f, p, b
where      f.lv=1
and (52-f.n)=b.n
and b.n-1=p.lv(+)
and b.n<=49;




14.        构造最大数
给定只包含正数的数组,给出一个方法,将数组中的数拼接起来,得到的数,是最大的。


with a  as (
select 4 x from dual union all
select 94 x from dual union all
select 14 x from dual union all
select 9 x from dual union all
select 1 x from dual),
b as (
select  max(length(x)) len
from    a)
select  listagg(x, '') within group (order by rpad(x, len*2, x) desc)
from    a, b;


注:关于此算法rpad(x, len*2, x)的证明,在101楼
http://www.itpub.net/forum.php?mod=viewthread&tid=1804279&page=11#pid21599534




15.        编码任务
假设有一个池塘,里面有无穷多的水。现有2个空水壶,容积分别为5升和6升。如何只用这2个水壶从池塘里取得3升的水(最后,这三升水,在其中一个壶里)


with
t(n1, n2, n3, n4, lv) as (
select 5, 0, 6, 0, 1
from DUAL
union all
select decode(t.n1, 0, 5, decode(t.n2, 6, t.n1, greatest(t.n1-(6-t.n2), 0))) n1,
decode(t.n1, 0, t.n2, decode(t.n2, 6, 0, least(t.n1+t.n2, 6))) n2,
decode(t.n3, 0, 6, decode(t.n4, 5, t.n3, greatest(t.n3-(5-t.n4), 0))) n3,
decode(t.n3, 0, t.n4, decode(t.n4, 5, 0, least(t.n3+t.n4, 5))) n4,
t.lv+1
from t
where --t.lv+1 <=42
t.n1!=3
and t.n2!=3
and t.n3!=3
and t.n4!=3
)
select *
from t;




16. 三只青蛙的换位游戏
百度一个可用的链接 http://www.3366.com/flash/32099.shtml


with t as (
select round(2.5-level) pos, to_char(ceil(level/2)) c
from dual
  connect by level<=4
),
r (s, x_pos, path) as (
  select '222x111' s, 4, ''
  from dual
  union all
  select substr(replace(r.s, 'x', t.c), 1, x_pos+t.pos-1)||'x'||substr(replace(r.s, 'x', t.c), x_pos+t.pos+1), x_pos+t.pos, path||to_char(x_pos+t.pos)||'#'
  from t, r
  where substr(r.s, x_pos+t.pos, 1)=t.c
  and x_pos+t.pos between 1 and 7
)
select *
from r
where s='111x222';




17. 第三届sql大赛第一题
题目链接 http://www.itpub.net/thread-1943911-1-1.html


相当得意的一段代码
insert into TICTACTOE
with b as (
select level||'' n, decode(level, 1,111,2,1001,3,10100001,4,10010,5,111100,6,10010000,7,1100010,8,1001000,11000100) v from dual connect by level<=9
),
r(p, xp, op, i, w) as (
select n, v, 0, 0, ''
from b
union all
select p||n, decode(i, 1, v, 0)+xp, decode(i, 1, 0, v)+op, mod(i+1, 2),
case when instr(decode(i, 1, v, 0)+xp, '3')>0 then 'X' else case when instr(decode(i, 1, 0, v)+op, '3')>0 then 'O' else null end end
from r, b
where instr(r.p, n)=0
and w is null
)
select p, translate(translate('123456789', p, 'XOXOXOXOX'), '123456789', '---------'), nvl(w, 'D')
from r
where w is not null or length(p)=9;




18. 12个人带7种颜色帽子,出现不少于3个相同的概率
qingyun搞来的一个题目 http://www.itpub.net/thread-1905121-1-1.html


直接贴newkid不厌其烦帮我改两步来加速的版本
with t as (
select  power(10, level-1) n
from    dual
connect by level<=7
),
t2 (comb, lv) as (
select  n, 1
from    t
union all
select  t.n+t2.comb, t2.lv+1
from    t, t2
where   instr(t.n+t2.comb, '3')=0
and     t2.lv<=2
)
,t3 as (
select  comb, count(*) cnt
from    t2
where   lv=3
group by comb
)
,t4 AS (
SELECT a.comb+b.comb comb,sum(a.cnt*b.cnt) as cnt
  from t3 a,t3 b
where  translate(a.comb+b.comb, 'x012', 'x') is null
group by a.comb+b.comb
)
select  TO_CHAR(power(7, 12)-sum(a.cnt*b.cnt))
from    t4 a, t4 b
where   translate(a.comb+b.comb, 'x012', 'x') is null;




oracle sql programming art.sql

16.49 KB, 下载次数: 47

打赏鼓励一下!

5人打赏

论坛徽章:
496
紫蜘蛛
日期:2007-09-26 17:05:56奥运会纪念徽章:垒球
日期:2008-09-15 01:28:12生肖徽章2007版:鸡
日期:2008-11-17 23:40:58生肖徽章2007版:马
日期:2008-11-18 05:09:48数据库板块每日发贴之星
日期:2008-11-29 01:01:02数据库板块每日发贴之星
日期:2008-12-05 01:01:03生肖徽章2007版:虎
日期:2008-12-10 07:47:462009新春纪念徽章
日期:2009-01-04 14:52:28数据库板块每日发贴之星
日期:2009-02-08 01:01:03生肖徽章2007版:蛇
日期:2009-03-09 22:18:53
发表于 2016-9-1 00:31 | 显示全部楼层
打赏按钮怎么看不到了?要是我有钱就给你五毛。

使用道具 举报

回复
论坛徽章:
78
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋头
日期:2012-12-08 09:43:38鲜花蛋
日期:2012-11-17 12:02:07鲜花蛋
日期:2013-02-05 21:53:34复活蛋
日期:2012-11-17 12:02:07SQL极客
日期:2013-12-09 14:13:35SQL数据库编程大师
日期:2013-12-06 13:59:43SQL大赛参与纪念
日期:2013-12-06 14:10:50ITPUB季度 技术新星
日期:2012-11-27 10:16:10最佳人气徽章
日期:2013-03-19 17:24:25
 楼主| 发表于 2016-9-1 08:00 | 显示全部楼层
newkid 发表于 2016-9-1 00:31
打赏按钮怎么看不到了?要是我有钱就给你五毛。

我靠,这功能有bug,我发新帖的时候忘了选开启打赏,等再编辑选开启,保存不上,我去@喵喵。
5毛?你也做得出

使用道具 举报

回复
论坛徽章:
399
紫蛋头
日期:2012-05-21 10:19:41迷宫蛋
日期:2012-06-06 16:02:49奥运会纪念徽章:足球
日期:2012-06-29 15:30:06奥运会纪念徽章:排球
日期:2012-07-10 21:24:24鲜花蛋
日期:2012-07-16 15:24:59奥运会纪念徽章:拳击
日期:2012-08-07 10:54:50奥运会纪念徽章:羽毛球
日期:2012-08-21 15:55:33奥运会纪念徽章:蹦床
日期:2012-08-21 21:09:51奥运会纪念徽章:篮球
日期:2012-08-24 10:29:11奥运会纪念徽章:体操
日期:2012-09-07 16:40:00
发表于 2016-9-1 08:12 | 显示全部楼层
udfrog 发表于 2016-9-1 08:00
我靠,这功能有bug,我发新帖的时候忘了选开启打赏,等再编辑选开启,保存不上,我去@喵喵。
5毛?你也 ...

这可是newkid的5毛,不能和一般人的比
就像波尔特给你的0.01秒

使用道具 举报

回复
论坛徽章:
22
林肯
日期:2013-10-25 08:53:562014年世界杯参赛球队: 澳大利亚
日期:2014-05-26 09:13:482014年世界杯参赛球队: 阿尔及利亚
日期:2014-05-28 10:46:58喜羊羊
日期:2015-03-04 14:54:422015年新春福章
日期:2015-03-06 11:59:47懒羊羊
日期:2015-06-16 10:25:42双鱼座
日期:2015-07-30 08:48:43白羊座
日期:2015-08-05 20:05:55天蝎座
日期:2015-08-07 22:32:03马上有车
日期:2014-05-07 11:14:31
发表于 2016-9-1 08:35 | 显示全部楼层
我原来想赏的,一看没有那个红色的打赏按钮,那就算了吧

使用道具 举报

回复
论坛徽章:
289
生肖徽章2007版:猴
日期:2008-05-16 11:28:59生肖徽章2007版:马
日期:2008-10-08 17:01:01SQL大赛参与纪念
日期:2011-04-13 12:08:17授权会员
日期:2011-06-17 16:14:53ITPUB元老
日期:2011-06-21 11:47:01ITPUB官方微博粉丝徽章
日期:2011-07-01 09:45:27ITPUB十周年纪念徽章
日期:2011-09-27 16:30:472012新春纪念徽章
日期:2012-01-04 11:51:22海蓝宝石
日期:2012-02-20 19:24:27铁扇公主
日期:2012-02-21 15:03:13
发表于 2016-9-1 09:29 | 显示全部楼层
我觉得 newkid 大师的也应该像LZ一样整理一份PDF(就这个论坛中解过的题)!

使用道具 举报

回复
认证徽章
论坛徽章:
168
SQL数据库编程大师
日期:2016-01-13 10:30:43SQL极客
日期:2013-12-09 14:13:35SQL大赛参与纪念
日期:2013-12-06 14:03:45最佳人气徽章
日期:2015-03-19 09:44:03现任管理团队成员
日期:2015-08-26 02:10:00秀才
日期:2015-07-28 09:12:12举人
日期:2015-07-13 15:30:15进士
日期:2015-07-28 09:12:58探花
日期:2015-07-28 09:12:58榜眼
日期:2015-08-18 09:48:03
发表于 2016-9-1 10:59 | 显示全部楼层

打赏按钮怎么看不到了?要是我有钱就给你五毛。

使用道具 举报

回复
论坛徽章:
78
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋头
日期:2012-12-08 09:43:38鲜花蛋
日期:2012-11-17 12:02:07鲜花蛋
日期:2013-02-05 21:53:34复活蛋
日期:2012-11-17 12:02:07SQL极客
日期:2013-12-09 14:13:35SQL数据库编程大师
日期:2013-12-06 13:59:43SQL大赛参与纪念
日期:2013-12-06 14:10:50ITPUB季度 技术新星
日期:2012-11-27 10:16:10最佳人气徽章
日期:2013-03-19 17:24:25
 楼主| 发表于 2016-9-1 11:04 | 显示全部楼层
Naldonado 发表于 2016-9-1 10:59
打赏按钮怎么看不到了?要是我有钱就给你五毛。

凸(艹皿艹 )

使用道具 举报

回复
认证徽章
论坛徽章:
168
SQL数据库编程大师
日期:2016-01-13 10:30:43SQL极客
日期:2013-12-09 14:13:35SQL大赛参与纪念
日期:2013-12-06 14:03:45最佳人气徽章
日期:2015-03-19 09:44:03现任管理团队成员
日期:2015-08-26 02:10:00秀才
日期:2015-07-28 09:12:12举人
日期:2015-07-13 15:30:15进士
日期:2015-07-28 09:12:58探花
日期:2015-07-28 09:12:58榜眼
日期:2015-08-18 09:48:03
发表于 2016-9-1 11:09 | 显示全部楼层
〇〇 发表于 2016-9-1 08:12
这可是newkid的5毛,不能和一般人的比
就像波尔特给你的0.01秒

0.01s够青蛙生一堆蝌蚪。。

使用道具 举报

回复
论坛徽章:
399
紫蛋头
日期:2012-05-21 10:19:41迷宫蛋
日期:2012-06-06 16:02:49奥运会纪念徽章:足球
日期:2012-06-29 15:30:06奥运会纪念徽章:排球
日期:2012-07-10 21:24:24鲜花蛋
日期:2012-07-16 15:24:59奥运会纪念徽章:拳击
日期:2012-08-07 10:54:50奥运会纪念徽章:羽毛球
日期:2012-08-21 15:55:33奥运会纪念徽章:蹦床
日期:2012-08-21 21:09:51奥运会纪念徽章:篮球
日期:2012-08-24 10:29:11奥运会纪念徽章:体操
日期:2012-09-07 16:40:00
发表于 2016-9-1 13:26 | 显示全部楼层
solomon_007 发表于 2016-9-1 09:29
我觉得 newkid 大师的也应该像LZ一样整理一份PDF(就这个论坛中解过的题)!

应该出一本书《苏神记》

使用道具 举报

回复

您需要登录后才可以回帖 登录 | 注册

本版积分规则

TOP技术积分榜 社区积分榜 徽章 电子杂志 团队 统计 虎吧 老博客 知识索引树 读书频道 积分竞拍 文本模式 帮助
  ITPUB首页 | ITPUB论坛 | 数据库技术 | 企业信息化 | 开发技术 | 微软技术 | 软件工程与项目管理 | IBM技术园地 | 行业纵向讨论 | IT招聘 | IT文档 | IT博客
  ChinaUnix | ChinaUnix博客 | ChinaUnix论坛 | SAP ERP系统
CopyRight 1999-2011 itpub.net All Right Reserved. 北京盛拓优讯信息技术有限公司版权所有 联系我们 网站律师 隐私政策 知识产权声明
京ICP备16024965号 北京市公安局海淀分局网监中心备案编号:11010802021510 广播电视节目制作经营许可证:编号(京)字第1149号
  
快速回复 返回顶部 返回列表