PUZZLEUP 2014

newkid

lastwinner 发表于 2014-11-12 14:09
嗯，对，4人的话，分组可能性：C(4,2)*C(2,2)/P(2,2)=3
最终是两个队，两个队中任意一队是同校的就行，所 ...

这就是我的方法，你再仔细看我的公式，虽然繁琐，但有章可循。

lastwinner

newkid 发表于 2014-11-13 05:43
这就是我的方法，你再仔细看我的公式，虽然繁琐，但有章可循。

对，感觉出来了，一加一减，一加一减……

lastwinner

newkid 发表于 2014-11-13 05:41
#16 Colored Squares

如果没有最后的同色必须相邻的条件，结果就会多出很多了
这是一个有额外限制条件的概率问题

piliskys

newkid 发表于 2014-11-8 05:15
配对方法总数：（分母）
C(24,2)*C(22,2)*C(20,2)*C(18,2)*C(16,2)*C(14,2)*C(12,2)*C(10,2)*C(8,2)*C(6,2 ...

真牛B，头痛

peter1166

# 16

这个是不是可以简化成 9块儿分4组块儿，每组块儿之多3，有多少种可能？ 再乘以 4! = 24 就是最终答案？
9块儿分4组好像有 104种可能 ， 那答案是不是 104 *24 = 2496 ?

piliskys

#15
--24个人
with a as (select power(2, rownum-1) rn, rownum rnc from dual connect by rownum <=24)
, b as (select rownum rn,  a.rn+b.rn  txx ,  
--flag为1是同一队
case when b.rnc=a.rnc+1 and b.rnc<=16 and mod(b.rnc,2)=0  then 1 else 0 end  flag
  from a, a b where b.rn>a.rn)
,d (idx, rn, txx,flag) as
(  select 1 idx, b.rn,  b.txx ,  b.flag   from b
   union all
   select d.idx+1, b.rn , d.txx +b.txx txx  ,d.flag+b.flag
    from d,b where b.rn>d.rn  and  bitand(d.txx,b.txx)=0
   )
--4代表至少4队
select sum(case when  d.flag>=4 then 1 else 0 end) ff,   count(1) alll
,sum(case when  d.flag>=1 then 1 else 0 end) /  count(1)
--12是配成12队的
from d where d.idx=12
---------------------------------写的效率很低，运行不出来

peter1166

piliskys 发表于 2014-11-14 13:18
#15
--24个人
with a as (select power(2, rownum-1) rn, rownum rnc from dual connect by rownum d.rn  ...

  我的SQL也运行不出来，没好意识贴出来。
看看你的能运行到几个人。

peter1166

16#
9块儿分4组块儿，每组块儿之多3的所有可能如下；

,1,23,456,789
,1,23,457,689
,1,23,478,569
,1,235,47,689
,1,235,478,69
,1,236,45,789
,1,236,457,89
,1,236,47,589
,1,245,36,789
,1,245,369,78
,1,25,369,478
,1,258,369,47
,12,3,456,789
,12,3,457,689
,12,3,478,569
,12,356,4,789
,12,356,47,89
,12,356,478,9
,12,36,45,789
,12,36,457,89
,12,36,47,589
,12,369,4,578
,12,369,45,78
,12,369,457,8
,12,369,458,7
,12,369,47,58
,12,369,478,5
,123,4,56,789
,123,4,569,78
,123,4,578,69
,123,45,6,789
,123,45,689,7
,123,45,69,78
,123,456,7,89
,123,456,78,9
,123,457,6,89
,123,457,69,8
,123,458,69,7
,123,47,5,689
,123,47,56,89
,123,47,568,9
,123,47,569,8
,123,47,58,69
,123,47,589,6
,123,478,5,69
,123,478,56,9
,124,3,56,789
,124,3,569,78
,124,3,578,69
,124,356,7,89
,124,356,78,9
,124,36,5,789
,124,36,578,9
,124,36,589,7
,124,369,5,78
,124,369,58,7
,125,3,47,689
,125,3,478,69
,125,36,4,789
,125,36,47,89
,125,36,478,9
,125,369,4,78
,125,369,47,8
,14,2,356,789
,14,2,369,578
,14,23,56,789
,14,23,569,78
,14,23,578,69
,14,235,6,789
,14,235,689,7
,14,235,69,78
,14,236,5,789
,14,236,578,9
,14,236,589,7
,14,25,36,789
,14,25,369,78
,14,256,3,789
,14,258,369,7
,145,2,36,789
,145,2,369,78
,145,23,6,789
,145,23,689,7
,145,23,69,78
,145,236,7,89
,145,236,78,9
,147,2,356,89
,147,2,36,589
,147,2,369,58
,147,23,5,689
,147,23,56,89
,147,23,568,9
,147,23,569,8
,147,23,58,69
,147,23,589,6
,147,235,6,89
,147,235,69,8
,147,236,5,89
,147,236,58,9
,147,25,3,689
,147,25,36,89
,147,25,369,8
,147,256,3,89
,147,258,3,69
,147,258,36,9

newkid

peter1166 发表于 2014-11-15 10:55
16#
9块儿分4组块儿，每组块儿之多3的所有可能如下；

你把SQL写出来就发章。
我写的是这样，感觉还是有点啰嗦：

WITH T (s,cnt,flag) AS (
SELECT CAST(c AS VARCHAR2(10)),1,0
FROM (SELECT TO_CHAR(LEVEL) c FROM DUAL CONNECT BY LEVEL<=4)
UNION ALL
SELECT s||c
      ,cnt+1
      ,CASE WHEN FLAG=1 THEN 0 WHEN REGEXP_COUNT(s,c) =1 AND MOD(cnt+1,3)<>0 AND cnt+1>3 AND SUBSTR(s,cnt-1,1)=c THEN 1 ELSE 0 END
  FROM t,(SELECT TO_CHAR(LEVEL) c FROM DUAL CONNECT BY LEVEL<=4)
WHERE cnt<9
       AND (flag=1 AND SUBSTR(s,cnt,1)=c
            OR flag=0 AND INSTR(s,c)=0
            OR flag=0 AND REGEXP_COUNT(s,c) IN (1,2)
                AND (cnt+1>3 AND MOD(cnt+1,3)<>1 AND (SUBSTR(s,cnt+1-3,1)=c OR SUBSTR(s,cnt,1)=c)
                     OR cnt+1>3 AND SUBSTR(s,cnt+1-3,1)=c
                     OR MOD(cnt+1,3)<>1 AND  SUBSTR(s,cnt,1)=c
                     OR REGEXP_COUNT(s,c) =1 AND MOD(cnt+1,3)<>0 AND cnt+1>3 AND SUBSTR(s,cnt-1,1)=c
                    )
            )
)
SELECT COUNT(*) FROM t
WHERE cnt=9
       AND TRANSLATE('1234','A'||s,'A') IS NULL;

newkid

peter1166 发表于 2014-11-14 13:24
我的SQL也运行不出来，没好意识贴出来。
看看你的能运行到几个人。

我也写了一个，只试了一半的数据：

假设总共N个人，P对来自同校，要求至少有K对来自同校的配对方法。

找出所有的配对，求出 K~P 对同校的组合，求出剩下非同校配对的组合。

VAR N NUMBER;
VAR P NUMBER;
VAR K NUMBER;
EXEC :N:=12;
EXEC :P:=4;
EXEC :K:=2;


WITH all_pairs AS
(SELECT N1,N2,CASE WHEN CEIL(N1/2)=CEIL(N2/2) AND N2/2<=:P THEN 1 ELSE 0 END AS SAME_SCHOOL
       ,BIT1+BIT2 bits
  FROM (SELECT LEVEL N1,POWER(2,LEVEL-1) BIT1 FROM DUAL CONNECT BY LEVEL<=:N)
      ,(SELECT LEVEL N2,POWER(2,LEVEL-1) BIT2 FROM DUAL CONNECT BY LEVEL<=:N)
WHERE N1<N2
)
,same_pairs as (SELECT * FROM all_pairs WHERE SAME_SCHOOL=1)
,diff_pairs as (SELECT * FROM all_pairs WHERE SAME_SCHOOL=0)
,same_comb(s,bits,cnt,last_n1) AS (
SELECT CAST(N1||','||N2 AS VARCHAR2(100)),bits,1,N1
  FROM same_pairs
UNION ALL
SELECT s||'|'||N1||','||N2,same_comb.bits+same_pairs.bits,cnt+1,same_pairs.N1
  FROM same_comb,same_pairs
WHERE N1>LAST_N1 AND CNT<:P
)
,diff_comb(s,bits,cnt,last_n1) AS (
SELECT CAST(N1||','||N2 AS VARCHAR2(100)),bits,1,N1
  FROM diff_pairs
UNION ALL
SELECT s||'|'||N1||','||N2,diff_comb.bits+diff_pairs.bits,cnt+1,diff_pairs.N1
  FROM diff_comb,diff_pairs
WHERE N1>LAST_N1 AND CNT<:N/2-:K
       AND BITAND(diff_comb.bits,diff_pairs.bits)=0
)
SELECT COUNT(*)
       --s.s||'|'||d.s,s.cnt,d.cnt
  FROM (SELECT s.*,ROWNUM FROM same_comb s WHERE cnt>=:K) s,diff_comb d
WHERE s.cnt+d.cnt=:N/2 AND BITAND(s.bits,d.bits)=0
       ;