{-
Recursive solution using sqrt(n) subsets. Runs in O(n).

Recursively computes the solution on sqrt(n) subsets of size sqrt(n). 
Then recurses on the product sum of each subset.
Then for each element in each subset, it computes the product with
the product sum of all other products.
Then flattens all subsets.

Recurrence on the run time is T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n

Suppose that T(n) ≤ cn in O(n).

T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
    ≤ sqrt(n)*c*sqrt(n) + c*sqrt(n) + n
    ≤ c*n + c*sqrt(n) + n
    ≤ (2c+1)*n
    ∈ O(n)

Note that ceiling(sqrt(n)) can be computed using a binary search 
and O(logn) iterations, if the sqrt instruction is not permitted.
-}

otherProducts [] = []
otherProducts [x] = [1]
otherProducts [x,y] = [y,x]
otherProducts a = foldl' (++) [] $ zipWith (\s p -> map (*p) s) solvedSubsets subsetOtherProducts
    where 
      n = length a

      -- Subset size. Require that 1 < s < n.
      s = ceiling $ sqrt $ fromIntegral n

      solvedSubsets = map otherProducts subsets
      subsetOtherProducts = otherProducts $ map product subsets

      subsets = reverse $ loop a []
          where loop [] acc = acc
                loop a acc = loop (drop s a) ((take s a):acc)

//这是Java中的递归解决方案 //从main product(a,1,0)调用如下;

public static double product(double[] a, double fwdprod, int index){
    double revprod = 1;
    if (index < a.length){
        revprod = product2(a, fwdprod*a[index], index+1);
        double cur = a[index];
        a[index] = fwdprod * revprod;
        revprod *= cur;
    }
    return revprod;
}

左旅行->右和保持保存产品。称之为过去。- > O (n) 旅行右->左保持产品。称之为未来。- > O (n) 结果[i] =过去[i-1] *将来[i+1] -> O(n) 过去[-1]= 1;和未来(n + 1) = 1;

O(n)

还有一个O(N^(3/2))非最优解。不过，这很有趣。

首先预处理大小为N^0.5的每个部分乘法(这在O(N)时间复杂度中完成)。然后，计算每个数字的其他值的倍数可以在2*O(N^0.5)时间内完成(为什么?因为您只需要将其他((N^0.5) - 1)数字的最后一个元素相乘，并将结果与属于当前数字组的((N^0.5) - 1)数字相乘。对每一个数都这样做，可以得到O(N^(3/2))时间。

例子:

4, 6, 7, 2, 3, 1, 9, 5, 8

部分结果: 4*6*7 = 168 2*3*1 = 6 9*5*8 = 360

要计算3的值，需要将其他组的值乘以168*360，然后乘以2*1。

php版本 使用不除法的array_product函数。 如果我们将i的值临时设为1，那么数组product将完全满足我们的需要

<?php
function product($key, $arr)
{
    $arr[$key] = 1;
    return array_product($arr);
};
$arr = [1, 2, 3, 4, 5];
$newarr = array();


foreach ($arr as $key => $value) {

    $newarr[$key] = product($key, $arr);
}
print_r($newarr);

int[] b = new int[] { 1, 2, 3, 4, 5 };            
int j;
for(int i=0;i<b.Length;i++)
{
  int prod = 1;
  int s = b[i];
  for(j=i;j<b.Length-1;j++)
  {
    prod = prod * b[j + 1];
  }
int pos = i;    
while(pos!=-1)
{
  pos--;
  if(pos!=-1)
     prod = prod * b[pos];                    
}
Console.WriteLine("\n Output is {0}",prod);
}