出个题目

bskay

观察规律,产生如下的序列(,不要用穷举之类硬编码的)
(2, 1, 2, 1)
(2, 1, 2, 0)
(2, 1, 3, 1)
(2, 1, 3, 0)
(2, 2, 2, 2)
(2, 2, 2, 1)
(2, 2, 3, 2)
(2, 2, 3, 1)
(2, 3, 3, 2)
(2, 3, 3, 1)
(2, 3, 4, 2)
(2, 3, 4, 1)

Linux_manne

规律呢... 原谅我的愚蠢.... 完全看不出规律  

bskay

Linux_manne 发表于 2016-02-16 18:09
规律呢... 原谅我的愚蠢.... 完全看不出规律

这就是题目的重点,单纯把算法翻译为代码的事情,是个人都可以做的

Linux_manne

回复 3# bskay


    你真棒！！

rubyish

maybe ~~

1. 2 1 2 1
   
2. 2 1 2 0
   
3. 2 1 3 1
   
4. 2 1 3 0
   
5. 2 2 2 2
   
6. 2 2 2 1
   
7. 2 2 3 2
   
8. 2 2 3 1
   
9. 2 3 3 2
   
10. 2 3 3 1
    
11. 2 3 4 2
    
12. 2 3 4 1
    

复制代码

1. #!/usr/bin/perl
   
2. use 5.010;
   
3. 
   
4. my ( $this, $that ) = ( 0, 1 );
   
5. 
   
6. sub make {
   
7.     $a = [ $this - $that, $that - $this ];
   
8.     for my $it ( $this .. $that ) {
   
9.         push @_, map {
   
10.             $b = [@$_];
    
11.             $b->[ $#{ $_[ $it - $it ] } - $it ] += $a->[$it];
    
12.             $b
    
13.         } @_;
    
14.     }
    
15.     @_;
    
16. }
    
17. 
    
18. my @it = make [ map @$_, make [$that] ];
    
19. 
    
20. for my $it ( $this + $this .. $that + $that ) {
    
21.     $_->[ $that - $this ]++, $_->[ -$it ]++, say "@$_" for @it;
    
22. }
    

复制代码

huazai0715

大神，快快解答吧！围观中

mswsg

生成的组合

1. a =(1,2,3)
   
2. b = (2,3,4)
   
3. c = (0,1,2)
   
4. [(2,x,y,z) for x in a for y in b for z in c ]

复制代码

[(2, 1, 2, 0),
(2, 1, 2, 1),
(2, 1, 2, 2),
(2, 1, 3, 0),
(2, 1, 3, 1),
(2, 1, 3, 2),
(2, 1, 4, 0),
(2, 1, 4, 1),
(2, 1, 4, 2),
(2, 2, 2, 0),
(2, 2, 2, 1),
(2, 2, 2, 2),
(2, 2, 3, 0),
(2, 2, 3, 1),
(2, 2, 3, 2),
(2, 2, 4, 0),
(2, 2, 4, 1),
(2, 2, 4, 2),
(2, 3, 2, 0),
(2, 3, 2, 1),
(2, 3, 2, 2),
(2, 3, 3, 0),
(2, 3, 3, 1),
(2, 3, 3, 2),
(2, 3, 4, 0),
(2, 3, 4, 1),
(2, 3, 4, 2)]
然后再filter？

rubyish

回复 7# mswsg

wode guancha...shi
what is begin ~

1. 
   
2.     b e g i n
   
3.     b e g i n
   
4.     b e g i n
   
5.     b e g i n
   
6.    
   
7.     2 1 2 1
   
8.     2 1 2 0
   
9.     2 1 3 1
   
10.     2 1 3 0
    
11.    
    
12.     2 2 2 2
    
13.     2 2 2 1
    
14.     2 2 3 2
    
15.     2 2 3 1
    
16.    
    
17.     2 3 3 2
    
18.     2 3 3 1
    
19.     2 3 4 2
    
20.     2 3 4 1
    

复制代码

col2 ? 1 -> 2 -> 3 => begin = [?, 0, ?, ?]... 0->1->2->3
col3 [2233] + 0 -> [2233] + 1 -> [3344]
Gindex 2 => + 1 => begin = [?, 0, 2, ?]...
col4 [1010] + 1 -> [2121] + 0 -> [2121]
Gindex 1 => + 1 => begin = [?, 0, 2, 1]...

col1 => (index -0 = 0 = -4)

col1(-0): Gindex  0 -> + 1 => begin block = [1021][1020][1031][1030]
col4(-1): Gindex  1 -> + 1
col3(-2): Gindex  2 -> + 1
col2(-3): ++


ok: now

    1 0 2 1
    1 0 2 0
    1 0 3 1
    1 0 3 0
_______________

    2 1 2 1
    2 1 2 0
    2 1 3 1
    2 1 3 0
   
    2 2 2 2
    2 2 2 1
    2 2 3 2
    2 2 3 1
   
    2 3 3 2
    2 3 3 1
    2 3 4 2
    2 3 4 1

begin block:

[1, 0, 2, 1],
[1, 0, 2, 0],
[1, 0, 3, 1],
[1, 0, 3, 0]

=

[1021][1020]+[1031][1030] => [1031][1030] = [10(2+1=3)1][10(2+1=3)0]] // + 1         // + 1
[1021][1020] = [1021] + [1020] => [1020] = [102(1-1=0)]               // - 1


[ - 1 and + 1 ]: self + (self - 1) and self + (self + 1)

[1 0 2 1]

[1021] = [10] + [21] => [21] = [10]+1 (1+1, 0+1) self + (self+1) // +1
[10] = [1] + [0] => [0] = [1]-1 (1-1) self + (self-1)             // -1

RESULT => [1021][1020][1031][1030] => [1021] => 1
1 (-1+1) => [1021] (-1+1) => [1021][1020][1031][1030] => RESULT

begin = 1 is ok!!

RESULT = fun(1)

write code for fun ~~

mswsg

哈哈,这套题，我猜考python里的列表解析，map，filter 等函数，结合好了一行代码搞定回复 8# rubyish