急.数字重组计算问题

问题描述:
现在有一组数字需要计算重组后总数,以最快的速度计算出总数.
数字不一定就是给出的这些,也有可能是其它数字
数字:
例一
1->0123
2->01234
3->678
总数:60


例二
1->0123456789
2->0123456789
3->01234
总数:500


写一个方法计算出重组后的总数和重组后的号码

回复讨论(解决方案)

不明白你的意思
最好给个样例

看不懂~~~~

样例:

1->0
2->01
3->01

重组后的号码为000,001,010,011,总数为4

这个意思？

$ar = array('0123','01234','678',);print_r(foo($ar));function foo(&$ar, $i=0) {  $res = array();  if($i+1 Array  
 (  
     [0] => 006  
     [1] => 007  
     [2] => 008  
     [3] => 016  
     [4] => 017  
     [5] => 018  
     [6] => 026  
     [7] => 027  
     [8] => 028  
     [9] => 036  
     [10] => 037  
     [11] => 038  
     [12] => 046  
     [13] => 047  
     [14] => 048  
     [15] => 106  
     [16] => 107  
     [17] => 108  
     [18] => 116  
     [19] => 117  
     [20] => 118  
     [21] => 126  
     [22] => 127  
     [23] => 128  
     [24] => 136  
     [25] => 137  
     [26] => 138  
     [27] => 146  
     [28] => 147  
     [29] => 148  
     [30] => 206  
     [31] => 207  
     [32] => 208  
     [33] => 216  
     [34] => 217  
     [35] => 218  
     [36] => 226  
     [37] => 227  
     [38] => 228  
     [39] => 236  
     [40] => 237  
     [41] => 238  
     [42] => 246  
     [43] => 247  
     [44] => 248  
     [45] => 306  
     [46] => 307  
     [47] => 308  
     [48] => 316  
     [49] => 317  
     [50] => 318  
     [51] => 326  
     [52] => 327  
     [53] => 328  
     [54] => 336  
     [55] => 337  
     [56] => 338  
     [57] => 346  
     [58] => 347  
     [59] => 348  
 )  
  如果不考虑重复问题，这个不就是排列组合了？  
 1->0123  
 2->01234  
 3->678  
 总数 = A(4,1)*A(5,1)*A(3,1) = 4*5*3 = 60 
 
  如果不考虑重复问题，这个不就是排列组合了？  
 1->0123  
 2->01234  
 3->678  
 总数 = A(4,1)*A(5,1)*A(3,1) = 4*5*3 = 60  
 代码？  

echo strlen('0123')*strlen('01234')*strlen('678');//??这个算最快不？
  
  
  

  噢！只要计算组合数啊？！我还以为要求所有的组合呢  

$ar = array('0123456789', '0123456789', '01234');echo array_product(array_map('strlen', $ar));
  
  其实你们的方法都是对的..只是我没有想到最快的...一时脑浆糊了..谢谢了 ... 
 
  看来我还是要多看PHP手册，又被xu版大人秒杀了。 
 
  发现数组中一个奇怪的问题,不知道是不是我写的有问题.  


$strArr = array(    1 => 0, 3 => 1, 6 => 2, 10 => 3, 15 => 4, 21 => 5, 28 => 6, 36 => 7, 45 => 8,    55 => 9, 63 => 10, 69 => 11, 73 => 12, 75 => 13, 75 => 14, 73 => 15, 69 => 16,    63 => 17, 55 => 18, 45 => 19, 36 => 20, 28 => 21, 21 => 22, 15 => 23, 10 => 24, 6 => 25,    3 => 26, 1 => 27);echo $str = '1,2,3,6,7,25,23,15,18'."\n";$sArr = explode(',',$str);foreach($sArr as $v){    echo array_search($v,$strArr)."\n";}
  

 结果:  
1,2,3,6,7,25,23,15,186157355Array(    [1] => 27    [3] => 26    [6] => 25    [10] => 24    [15] => 23    [21] => 22    [28] => 21    [36] => 20    [45] => 19    [55] => 18    [63] => 17    [69] => 16    [73] => 15    [75] => 14)
  

 我的0-13的这个怎么都没有了???  
  不明白这个 0-13 是什么意思 
 
  你的 $strArr 数组键名重复了，因此后面的覆盖前面的。print_r($strArr); 就知道了。 
 
  $strArr = array(  
    1 => 0,  
    3 => 1,  
    6 => 2,  
   10 => 3,  
   15 => 4,  
   21 => 5,  
   28 => 6,  
   36 => 7,  
   45 => 8,  
   55 => 9,  
   63 => 10,  
   69 => 11,  
   73 => 12,  
   75 => 13,  
    75 => 14,  
    73 => 15,  
    69 => 16,  
    63 => 17,  
    55 => 18,  
    45 => 19,  
    36 => 20,  
    28 => 21,  
    21 => 22,  
    15 => 23,  
    10 => 24,  
     6 => 25,  
     3 => 26,  
     1 => 27  
 );  
 套红的下标将覆盖上面已有的下标