puzzleup 2018

阿吉2009 · 发表于 2018-11-26 16:07

收藏了。

newkid · 发表于 2018-11-29 03:35

#18 PRIME NUMBER

Starting from the rightmost digit, all digits of a prime number are removed and after removal of each digit the resulting number is still prime. What is the largest such number?

For example: 3797 has this property because 3797, 379, 37 and 3 are prime numbers.

从一个素数的最右边的数字开始，删除素数的所有数字，并且在移除每个数字之后，得到的数字仍然是素数。这样的素数最大是多少？

例如：3797具有此属性，因为3797,379,37和3是素数。
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除了搞素数表好像也没其他办法。

〇〇 · 发表于 2018-11-29 05:51

第一位2357找中间都是奇数数字的cte

solomon_007 · 发表于 2018-11-29 09:53

〇〇发表于 2018-11-29 05:51
第一位2357找中间都是奇数数字的cte

按你的：

SQL> set timing on;
SQL>
SQL>
SQL> with
  2  function fn_prime (p_prime in integer) return int
  3  is
  4 i integer :=2;
  5  begin
  6 while (i<=sqrt(p_prime)) loop
  7    if mod(p_prime,i) = 0 then
  8       return 0;
  9    else
10       i := i+1;
11    end if;
12 end loop;
13 return 1;
14  end;
15  t as (select level n from dual where mod(level,2) <> 0 connect by level <=10),
16  s(lvl,nlist) as (select 1,decode(n,1,2,n) from t where n < 9
17                      union all
18                      select lvl + 1,
19                            to_number(s.nlist||t.n)
20                         from s,t
21                      where lvl < 10
22                         and fn_prime (to_number(s.nlist||t.n)) = 1 )
23  select max(nlist) from s
24  /
MAX(NLIST)
----------
  73939133
Executed in 0.063 seconds

SQL>
SQL> with
  2  function fn_prime (p_prime in integer) return int
  3  is
  4 i integer :=2;
  5  begin
  6 while (i<=sqrt(p_prime)) loop
  7    if mod(p_prime,i) = 0 then
  8       return 0;
  9    else
10       i := i+1;
11    end if;
12 end loop;
13 return 1;
14  end;
15  t as (select level n from dual where mod(level,2) <> 0 connect by level <=10),
16  s(lvl,nlist) as (select 1,decode(n,1,2,n) from t where n < 9
17                      union all
18                      select lvl + 1,
19                            to_number(s.nlist||t.n)
20                         from s,t
21                      where lvl < 15
22                         and fn_prime (to_number(s.nlist||t.n)) = 1 )
23  select max(nlist) from s
24  /
MAX(NLIST)
----------
  73939133
Executed in 0.062 seconds

所以 73939133 这就应该是最大的了。

solomon_007 · 发表于 2018-11-29 10:25

可以不要位数判断：

SQL> with
  2  function fn_prime (p_prime in integer) return int
  3  is
  4 i integer :=2;
  5  begin
  6 while (i<=sqrt(p_prime)) loop
  7    if mod(p_prime,i) = 0 then
  8       return 0;
  9    else
10       i := i+1;
11    end if;
12 end loop;
13 return 1;
14  end;
15  t as (select level n from dual where mod(level,2) <> 0 connect by level <=10),
16  s(nlist) as (select decode(n,1,2,n) from t where n < 9
17             union all
18             select to_number(s.nlist||t.n)
19             from s,t
20             where fn_prime (to_number(s.nlist||t.n)) = 1 )
21  select max(nlist) from s
22  /
MAX(NLIST)
----------
  73939133