出個(gè)題目

bskay

觀察規(guī)律,產(chǎn)生如下的序列(,不要用窮舉之類硬編碼的)
(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

規(guī)律呢... 原諒我的愚蠢.... 完全看不出規(guī)律  

bskay

Linux_manne 發(fā)表于 2016-02-16 18:09
規(guī)律呢... 原諒我的愚蠢.... 完全看不出規(guī)律

這就是題目的重點(diǎn),單純把算法翻譯為代碼的事情,是個(gè)人都可以做的

Linux_manne

回復(fù) 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
    

復(fù)制代碼

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. }
    

復(fù)制代碼

huazai0715

大神，快快解答吧！圍觀中

mswsg

生成的組合

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 ]

復(fù)制代碼

[(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

回復(fù) 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
    

復(fù)制代碼

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,這套題，我猜考python里的列表解析，map，filter 等函數(shù)，結(jié)合好了一行代碼搞定回復(fù) 8# rubyish