Puzzleup 2013挑战赛即将开始

udfrog

#15
bill@ORCL> with t as (
  2  select to_char(rownum-1) n from dual connect by rownum<=10),
  3  r(n, p, na) as (
  4  select     n, n, n
  5  from       t
  6  union all
  7  select     t.n, r.p||t.n, r.na||(r.p+t.n)
  8  from       r, t
  9  where      translate(r.p||t.n, (r.n+t.n)||'x', 'x')=r.p||t.n
10  and                instr(r.p, t.n)=0
11  and                instr(r.na, t.n)=0)
12  select     max(to_number(p)) from r;

MAX(TO_NUMBER(P))
-----------------
         92837465

已用时间:  00: 00: 00.19

〇〇

怎么老是概率题
No: 16       November 06, 2013  



Twelve Numbers

Twelve Numbers

There are twelve numbers from 1 to 12. Five of these numbers are given to you randomly, and the remaining seven numbers are given to your friend.

What is the probability that you have more even numbers than your friend has?

After simplification, enter your answer as a/b.

newkid

#No: 16

Twelve Numbers

There are twelve numbers from 1 to 12. Five of these numbers are given to you randomly, and the remaining seven numbers are given to your friend.

What is the probability that you have more even numbers than your friend has?

After simplification, enter your answer as a/b.

12个数
有从1到12的12个数。其中有五个随机给了你，剩下的七个给了你的朋友。
你的偶数比朋友多的概率是多少？
答案格式为简化后的分数 a/b。

这个也太简单了吧, CONNECT BY就搞定。

peter1166

#No: 16

WITH T AS (
          SELECT SYS_CONNECT_BY_PATH(L,',') P
                 ,REGEXP_COUNT(SYS_CONNECT_BY_PATH(L,','),'[^13579,]') CNT
          FROM(
                 SELECT LEVEL L FROM DUAL CONNECT BY LEVEL <= 12
              )
          WHERE LEVEL = 5
          CONNECT BY L < PRIOR L
          AND  LEVEL <= 5
          )
SELECT COUNT(CASE WHEN CNT>3 THEN 1 ELSE NULL END)
       /COUNT(*)  PB
FROM T;

        PB
----------
0.12121212

peter1166

#No: 16
方法2：

WITH T AS (SELECT LEVEL N  FROM   DUAL  CONNECT BY LEVEL <= 12)
     ,T1 AS (
            SELECT T1.N N1
                  ,T2.N N2
                  ,T3.N N3
                  ,T4.N N4
                  ,T5.N N5
                  ,5 - (MOD(T1.N,2) + MOD(T2.N,2) + MOD(T3.N,2) + MOD(T4.N,2) + MOD(T5.N,2)) EVEN_NUM
            FROM   T T1
                  ,T T2
                  ,T T3
                  ,T T4
                  ,T T5
            WHERE  T2.N > T1.N
            AND    T3.N > T2.N
            AND    T4.N > T3.N
            AND    T5.N > T4.N )
SELECT COUNT(CASE WHEN EVEN_NUM > 3 THEN 1 ELSE NULL END) / COUNT(*)  RATE
FROM   T1;

newkid

#17
Order Of The Cards

You and your friend are told that there are N cards, each are numbered from 1 to N. You will randomly select 6 cards. Using the “perfect coding” system which you had set with your friend previously, you will select one card and transfer the other five to your friend. By just examining the order of these numbered five cards, your friend will be able to know the card you’ve selected.

What can be the maximum value for N?

你和你的朋友被告知总共有N张卡片，编号分别从1到N。你随机选出六章卡。利用你和朋友事先约定好的“完美编码”系统，你挑出一张卡并且把另外五张传给你的朋友。仅靠着检查这无张带编号的卡的顺序，你的朋友就能够你选择了哪张卡。

N的最大值是多少？

newkid

朋友要完成的任务是从剩下的N-5中确定一张。“我”要利用五张卡的顺序把这个信息透露出来。5!=120，总共可以表达120个数，所以N=120+5=125

这个“完美编码”是这样运作的：

朋友看到五张卡，就知道剩下的N-5张是什么。把剩下的N-5张卡编号，从1~120

把传递信息的5张卡从小到大编为ABCDE

确定排列以及编号之间的映射关系, 比如：
ABCDE=1
ABCED=2
.....
EDCBA=120

那么，朋友根据排列就能知道编号，根据编号就能把卡挑选出来。

udfrog

newkid 发表于 2013-11-14 01:07
朋友要完成的任务是从剩下的N-5中确定一张。“我”要利用五张卡的顺序把这个信息透露出来。5!=120，总共可以 ...

这。。。虎
只是这不是只能表示120个么，这+5怎么来的

udfrog

哦，是125

newkid

你到底看明白没有？我这方法有问题吗？