谁帮我写个函数方法

[qq4473169]

最近在研究彩票的问题，有一个关于复式数的分解问题始终在困扰我。

例子

/**
比如4个号码分解成3个号码和2个号码
4码：[1,2,3,4] 一组号码
分解成3码后:[1,2,3] [1,2,4] [1,3,4] [2,3,4] 四组三位数的号码
分解成2码后:[1,2] [1,3] [1,4] [2,3] [2,4] [3,4] 六组二位数的号码
不考虑位置

又如5个号码分解成4个号码
4码：[1,2,3,4,5] 一组号码
分解成4码后:[1,2,3,4] [1,2,3,5] [1,2,4,5] [1,3,4,5] [2,3,4,5]四组四位数的号码
分解成3码后:[1,2,3] [1,2,4] [1,2,5] [1,3,4] [1,3,5] [1,4,5] [2,3,4] [2,3,5] [2,4,5] [3,4,5] 十组三位数的号码
分解成2码后:[1,2] [1,3] [1,4] [1,5] [2,3] [2,4] [2,5] [3,4] [3,5] [4,5] 十组二位数的
-------------------------------------------------
大家应该看出规律了吧
谁能帮我写个方法
array lottery(array $num, int $i){
}
@param $num 是指一组数字
@param $i 是指分解的码数
返回结果数组

如
$arr1 = lottery(array(1,2,3,4,5), 4);
$arr2 = lottery(array(1,2,3,4,5), 3);
此时
$arr1 的值应该是一个数组
[1,2,3,4] [1,2,3,5] [1,2,4,5] [1,3,4,5] [2,3,4,5]
而$arr2 的值也是一个数组
分解成3码后:[1,2,3] [1,2,4] [1,2,5] [1,3,4] [1,3,5] [1,4,5] [2,3,4] [2,3,5] [2,4,5] [3,4,5]




这个问题我反复研究过，貌似挺难的。。过年了，给66分，祝答题者六六大顺
*/

[ZT_King]

我复制贴过来吧, 这个实现还是比较高效的算法.

function Combination($arr, $size = 1) { 
    $len = count($arr); 
    $max = pow(2,$len) - pow(2,$len-$size);
    $min = pow(2,$size)-1; 
 
    $r_arr = array(); 
    for ($i=$min; $i<=$max; $i++){
        $count = 0; 
        $t_arr = array(); 
        for ($j=0; $j<$len; $j++){
            $a = pow(2, $j); 
            $t = $i&$a; 
            if($t == $a){ 
                $t_arr[] = $arr[$j]; 
                $count++; 
            } 
        }      
        if($count == $size){ 
            $r_arr[] = $t_arr;         
        } 
    } 
    return $r_arr; 
} 
$r = Combination(array(1,2,3,4,5),3); 
echo '<pre>';
print_r($r); 
echo '</pre>';
/*
Array
(
    [0] => Array
        (
            [0] => 1
            [1] => 2
            [2] => 3
        )

    [1] => Array
        (
            [0] => 1
            [1] => 2
            [2] => 4
        )

    [2] => Array
        (
            [0] => 1
            [1] => 3
            [2] => 4
        )

    [3] => Array
        (
            [0] => 2
            [1] => 3
            [2] => 4
        )

    [4] => Array
        (
            [0] => 1
            [1] => 2
            [2] => 5
        )

    [5] => Array
        (
            [0] => 1
            [1] => 3
            [2] => 5
        )

    [6] => Array
        (
            [0] => 2
            [1] => 3
            [2] => 5
        )

    [7] => Array
        (
            [0] => 1
            [1] => 4
            [2] => 5
        )

    [8] => Array
        (
            [0] => 2
            [1] => 4
            [2] => 5
        )

    [9] => Array
        (
            [0] => 3
            [1] => 4
            [2] => 5
        )

)
*/

[ematphp]

2楼的效率还是蛮高的！

[xuzuning]

这是求 m 取 n （n<=m）的组合，现成的代码很多