In addition to these assumptions, you have also assumed that numbers are
greater than 1 else * will lower the result.

Regards,
Akash Agrawal
http://tech-queries.blogspot.com/


On Thu, Nov 25, 2010 at 11:18 AM, Algoose chase <harishp...@gmail.com>wrote:

> For this specific case since only 2 operators are used : + , *      and we
> know that * is the operator that maximizes the value(provided both the
> operands are not equal to one / none of the operand is zero and also given
> that operands are +ve ).
> Doing * operation as late as possible should suffice right ?
>
> For Eg: Do all additions in the first pass and do all multiplications in
> 2nd pass.
>
> is there be any case where the above mentioned logic fails ?
>
> On Wed, Nov 24, 2010 at 4:07 PM, Amir hossein Shahriari <
> amir.hossein.shahri...@gmail.com> wrote:
>
>> you can  use an algorithm similar to matrix chain multiplication i.e. if
>> dp[i][j] is the maximum value that you can get with the numbers v_i to v_j
>> and in order to maximize it find k that maximizes ( dp[i][k]  op_k  dp[k][j]
>> )
>> v_i is the ith value and op_k is the kth operator
>> obviously if i==j : dp[i][j] = v_i
>>
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