Euler Project 挨个做 - 之一(Q1-50)

tree_new_bee

这个题不难。 只是边界需要试探。

with todd as (select rownum*2+1 n from dual connect by rownum<=10000/6)
,toddcomp as (select  distinct t1.n*t2.n o from todd t1,todd t2 where t1.n<=t2.n and t1.n*t2.n<=10000)
, tprim as (select  column_value prim from table(pkg_prim2.seive_numlist(10000)))
select min(o) from toddcomp
where not exists (select 1 from tprim
where prim<toddcomp.o
and sqrt((toddcomp.o - prim)/2) = trunc(sqrt((toddcomp.o - prim)/2))
)

    MIN(O)
----------
      5777

Executed in 114.921 seconds

数据量并不大，这么慢有点出乎预料之外。
估计应该又是优化器惹的祸。 用上面的MATERIALIZE试试



with todd as (select rownum*2+1 n from dual connect by rownum<=10000/6)
,toddcomp as (select /*+ MATERIALIZE */ distinct t1.n*t2.n o from todd t1,todd t2 where t1.n<=t2.n and t1.n*t2.n<=10000)
, tprim as (select /*+ MATERIALIZE */ column_value prim from table(pkg_prim2.seive_numlist(10000)))
select min(o) from toddcomp
where not exists (select 1 from tprim
where prim<toddcomp.o
and sqrt((toddcomp.o - prim)/2) = trunc(sqrt((toddcomp.o - prim)/2))
)

    MIN(O)
----------
      5777

Executed in 1.594 seconds

这下可以接受了。

jboracle1981

楼主11g的递归with都用得那么如火纯清了。。
高等数据也那么牛。。。
偶个高中毕业的， IQ也不高，想写也写不出来。。

tree_new_bee

上面是用奇合数-质数，看是否是平方数。
下面改成用奇合数-平方数， 看是否是质数。

with todd as (select rownum*2+1 n from dual connect by rownum<=10000/6)
,toddcomp as (select  distinct t1.n*t2.n o from todd t1,todd t2 where t1.n<=t2.n and t1.n*t2.n<=10000)
,tsqr as (select rownum*rownum s from dual connect by rownum*rownum<=10000)
select min(o) from toddcomp
where not exists (select 1 from tsqr
where 2*s<toddcomp.o
and pkg_prim2.isprim(toddcomp.o-2*s) = 0
)

    MIN(O)
----------
      5777

Executed in 1.656 seconds

这次不加提示也很快。

tree_new_bee

原帖由 jboracle1981 于 2010-12-28 11:21 发表
楼主11g的递归with都用得那么如火纯清了。。
高等数据也那么牛。。。
偶个高中毕业的， IQ也不高，想写也写不出来。。

做这些题用的数学知识里，没有超过高中程度的内容。

再说了， 超过高中数学程度的东西， 我也早忘光了， 想用也用不上。

linyisen1985

以NB哥为榜样，好好学习开发知识，年前就研究这篇帖子了

〇〇

问题55：
How many Lychrel numbers are there below ten-thousand?
题目简介：首先介绍一个数学名词：利克瑞尔数（Lychrel Number）：指的是将该数与将该数各数位逆序翻转
后形成的新数相加、并将此过程反复迭代后，结果永远无法是一个回文数的自然数。
　　现在已知对于10000以内的自然数，要么能够在50步（指将这个数与其逆序后的数相加的过程）内得到一
个回文数；要么该数是一个利克瑞尔数。
　　求：10000以内利克瑞尔数的个数
答　　案：249

〇〇

问题56：Considering natural numbers of the form, a^b, finding the maximum digital sum.
题目简介：记f(x)表示自然数x的各位数字之和，a^b表示a的b次方。对1≤a,b <100，求f(a^b)的最大值。
答　　案：972

tree_new_bee

原帖由 〇〇 于 2010-12-28 21:24 发表
问题55：

怎么一下子跳到55了？
这不是逼着我赶进度吗？

tree_new_bee

Q47  Find the first four consecutive integers to have four distinct primes factors.

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.

Find the first four consecutive integers to have four distinct primes factors. What is the first of these numbers?

tree_new_bee

又要质因数分解。
这种题用SQL总是有种无力的感觉。

SQL:

with targ as (select 200000 arg from dual)
,t1 as (select /*+ MATERIALIZE */ rownum n from targ connect by rownum<=arg)
,t2 as (select /*+ MATERIALIZE */ column_value d from targ, table(pkg_prim2.seive_numlist((arg/2))))
,t3 as (select n, count(d)  dn
from  t1, t2 where n>d and mod(n, d) = 0 group by n having count(d)=4 )
--select * from t3 where dn=3
select connect_by_root(n) from t3 where  level=4 connect by n=prior n+1  


放到那里不管它， 看看多久能出来结果。

CONNECT_BY_ROOT(N)
------------------
            134043

Executed in 1835.078 seconds

SQL>

吼吼， 跑了半个小时！