PHPの配列処理

$a:
array (size=16)
0 => 文字列 '1' (長さ=1)
1 => 文字列 '3' (長さ=1)
2 => 文字列 '5' (長さ=1)
3 => 文字列「7」(長さ=1)
4 => 文字列 '10' (長さ=2)
5 => 文字列 '12' (長さ=2)
6 => 文字列 '14' (長さ=2)
7 => 文字列 '16' (長さ=2)
8 => 文字列 '17' (長さ=2)
9 => 文字列 '19' (長さ=2)
10 => 文字列 '21' (長さ=2)
11 => 文字列 '23' (長さ=2)
12 => 文字列 '26' (長さ=2)
13 => 文字列 '28' (長さ=2)
14 => 文字列 '30' (長さ=2)
15 => 文字列 '32' (長さ=2)

$b:
配列 (サイズ=16)
0 => 文字列 '2' (長さ=1)
1 => 文字列 '4' (長さ=1)
2 => 文字列 '6' (長さ=1)
3 => 文字列「8」(長さ=1)
4 => 文字列 '9' (長さ=1)
5 => 文字列 '11' (長さ=2)
6 => 文字列 '13' (長さ=2)
7 => 文字列 '15' (長さ=2)
8 => 文字列 '18' (長さ=2)
9 => 文字列 '20' (長さ=2)
10 => 文字列 '22' (長さ=2)
11 => 文字列 '24' (長さ=2)
12 => 文字列 '25' (長さ=2)
13 => 文字列 '27' (長さ=2)
14 => 文字列 '29' (長さ=2)
15 => 文字列 '31' (length=2)

上の数グループ。私は得られる [16, 1, 13, 4] [14, 3, 15, 2] [12, 5, 9, 8] [10,7, 9 、 8]
[32,17,29,20] [30,19,31,18] [28,21,25,24] [26,23,27,22]

回复讨论(解决方案)

看不出物么组装规律

/*规律$a = array(1,3,5,7,  10,12,14,16,  17,19,21,23,  26,28,30,32);			 a1	           a2	                    a3	        a4$b = array(2,4,6,8,  9,11,13,15,  18,20,22,24,  25,27,29,31);			 b1           b2                   b3                   b4[16, 1,13, 4][14, 3,15, 2][12, 5, 9, 8][10, 7, 9 ,8]a2 + a1 = b2 + b1 = 17[32,17,29,20][30,19,31,18][28,21,25,24][26,23,27,22]a4 + a3 = b4 + b3 = 49*/t(17);t(49);function t($n){	$a = array(1,3,5,7,  10,12,14,16,  17,19,21,23,  26,28,30,32);	$b = array(2,4,6,8,  9,11,13,15,  18,20,22,24,  25,27,29,31);	$a = array_chunk($a,4);	$b = array_chunk($b,4);	//15+16 = 31	if($n <= 31){		$useA = array_slice($a,0,2);		$useB = array_slice($b,0,2);	}else{		$useA = array_slice($a,2,4);		$useB = array_slice($b,2,4);	}	$data = array();	arsort($useA[1]);	foreach($useA[1] as $v1){		foreach($useA[0] as $v2){			if($n==$v1+$v2){				$tmp = array();				$tmp[] = $v1;				$tmp[] = $v2;				$data['a'][] = $tmp;			}		}	}	arsort($useB[1]);	foreach($useB[1] as $v1){		foreach($useB[0] as $v2){			if($n==$v1+$v2){				$tmp = array();				$tmp[] = $v1;				$tmp[] = $v2;				$data['b'][] = $tmp;			}		}	}	foreach($data['a'] as $k=>$v){		echo join(' , ',array_merge($v,$data['b'][$k])).'<br>';	}}/*16 , 1 , 15 , 214 , 3 , 13 , 412 , 5 , 11 , 610 , 7 , 9 , 832 , 17 , 31 , 1830 , 19 , 29 , 2028 , 21 , 27 , 2226 , 23 , 25 , 24*/

あれ吗?

[ 12, 5, 9, 8 ] 7, 9, 8 ]
怎么解释？

[ 10, 7, 9, 8 ] 应该是[10,7, 11 , 6]

最終結果は二数组の組み合わせ

a7 a0 b6 b1
a6 a1 b7 b0 a2 b4 b3
a4 a3 b5 b2
a15 a8 b14 b9
a14 a9 b15 b8
a13 a10 b12 b11
a12 a11 b13 b10

[ 12, 5, 9, 8 ]
[ 10, 7, 9, 8 ]

感觉是他写错了，不然就没就没没了

、就、就就就就他写错了了

他他[30、19、31、31、18]错了？


[ 12, 5, 9, 8 ]
[ 10, 7, 9, 8 ]
他感觉是写错了，不然没规律了

嗯，感觉这样就不是规律了、有限定性
すごい

我想应该这样做

$a = array(   1,  3,  5,  7,  10, 12, 14, 16,  17, 19, 21, 23,  26, 28, 30, 32);$b = array(   2,  4,  6,  8,   9, 11, 13, 15,  18, 20, 22, 24,  25, 27, 29, 31);function foo($a) {  $r = array();  foreach(array_chunk($a, 8) as $v) {    $t = array_chunk($v, 4);    $t = array_map(null, array_reverse($t[1]), $t[0]);    $r = array_merge($r, $t);  }  return $r;}$r = foo($a);foreach(foo($b) as $k=>$v) {  $k += $k % 2 ? -1 : 1;  $r[$k] = array_merge($r[$k], $v);}print_r($r);

Array(    [0] => Array        (            [0] => 16            [1] => 1            [2] => 13            [3] => 4        )    [1] => Array        (            [0] => 14            [1] => 3            [2] => 15            [3] => 2        )    [2] => Array        (            [0] => 12            [1] => 5            [2] => 9            [3] => 8        )    [3] => Array        (            [0] => 10            [1] => 7            [2] => 11            [3] => 6        )    [4] => Array        (            [0] => 32            [1] => 17            [2] => 29            [3] => 20        )    [5] => Array        (            [0] => 30            [1] => 19            [2] => 31            [3] => 18        )    [6] => Array        (            [0] => 28            [1] => 21            [2] => 25            [3] => 24        )    [7] => Array        (            [0] => 26            [1] => 23            [2] => 27            [3] => 22        ))

只是调动位置

确实是调换位置，这个有规律些

[12,  5,  9,  8]   [10,7, 9 ,     8] 是我写错，不好意思。宗各位所述已经解决谢谢你们。