PUZZLEUP 2015

lastwinner · 发表于 2015-7-30 17:44

人肉一下一位数
一位数：0~9，和的范围是0~18，共计19个不同的结果
任取两个数，说明取的先后没有顺序关系，所以总共是100种取法，而不需要去除诸如1,1/4,4/5,5等等看似重复的取法
下面开始验证：

  1  with t as (select level-1 n from dual where level >0 connect by level <=10) --0 也是一位数
  2  , u as (select a.n+b.n s, count(*) cnt from t a, t b group by a.n+b.n) --和的个数
  3  , v as (select sum(power(cnt,2)) fz, power(power(10,2),2) fm from u )
  4* select fz, fm, fz/fm, 29/405 from v
SCOTT@lw.lw2> /

      FZ       FM    FZ/FM    29/405
---------- ---------- ---------- ----------
   670    10000    .067 .071604938

  1  with t as (select level n from dual where level >0 connect by level <10) --0 难道不是一位数？
  2  , u as (select a.n+b.n s, count(*) cnt from t a, t b group by a.n+b.n) --和的个数
  3  , v as (select sum(power(cnt,2)) fz, power(power(10,2),2) fm from u )
  4* select fz, fm, fz/fm, 29/405 from v
SCOTT@lw.lw2> /

      FZ       FM    FZ/FM    29/405
---------- ---------- ---------- ----------
   489    10000    .0489 .071604938

我去，看来拿3,4和拿4,3视作同一种取法
  1  with t as (select level-1 n from dual where level >0 connect by level <=10) --看来0应该是一位数
  2  , u as (select a.n+b.n s, count(*) cnt from t a, t b where a.n<=b.n group by a.n+b.n) --和的个数，3,4和4,3视作同一种取法
  3  , v as (select sum(power(cnt,2)) fz, power(55,2) fm from u )
  4* select fz, fm, fz/fm, 29/405 from v
SCOTT@lw.lw2> /

      FZ       FM    FZ/FM    29/405
---------- ---------- ---------- ----------
   195    3025  .06446281 .071604938

  1  with t as (select level n from dual where level >0 connect by level <10) --0真的不是一位数，FT
  2  , u as (select a.n+b.n s, count(*) cnt from t a, t b where a.n<=b.n group by a.n+b.n) --和的个数，3,4和4,3视作同一种取法
  3  , v as (select sum(power(cnt,2)) fz, power(45,2) fm from u )
  4* select fz, fm, fz/fm, 29/405 from v
SCOTT@lw.lw2> /

      FZ       FM    FZ/FM    29/405
---------- ---------- ---------- ----------
   145    2025 .071604938 .071604938

lastwinner · 发表于 2015-7-30 17:46

本帖最后由 lastwinner 于 2015-7-30 17:47 编辑

solomon_007 发表于 2015-7-30 10:05
SQL>
SQL> with t as (select level n from dual where level >0 connect by level with t as (select ...

(9*8/2+9)*(9*8/2+9) q

——————————————————————
这部分我是这样算的1+2+……+9=9*(9+1)/2=45

咱俩思路不太一样

〇〇 · 发表于 2015-7-30 20:58

where level >=0没必要

newkid · 发表于 2015-7-30 21:28

piliskys 发表于 2015-7-30 13:16
我感觉这问题是讨论概率，而题中的2个1位数的答案是组合比例
首先，[1,1]这两个数字，和[1,2]这两个数字是 ...

那你认为1位数的 29/405应该是怎么算出来的？
老头显然认为[1,1]和[1,2]算同等概率。

newkid · 发表于 2015-7-30 21:36

改题目了：

Dear PuzzleUp Contenders,

A change has been made to the text. The text "...two 2-digit numbers..." has been changed as follows: "...two different 2-digit numbers..."
Please take notice and submit your answers accordingly.

For clarification; 2-digit numbers are (10, 11,...,99), 1-digit numbers are (0, 1,...,9).

Regards,
The PuzzleUp Team

0算1位数，两个数不相同。