I think we can do it this way.

Sum(all boards length) / K = A ( tentative avg. lenght to be painted by each
painter)

Now start from B1, and keep going further till Bi, till Sum(B1-Bi) is less
than A. So this goes to painter P1.

Same way for P2, start from i+1 till j.

Condition: If  i+1 itself is > A. Then new A = length(B i+1).

Keep going like this till Bn


On Mon, Apr 4, 2011 at 3:20 PM, rajat ahuja <[email protected]>wrote:

> like u hav boards of length of length
> 7 2 6 9 4 and u hav 3 painters who can work ||ly
> so now
> one way to distribute is
> (7 )(2 6 9) (4) so time in ths case is 17
> suppose we do (7 2)(6)(9 4) time  in ths case is 13
> or i can do (7 2)(6 9 )(4) time in ths case is 15
>  i m takin 1 unit time to paint one meter so it is directly equal to length
>
>
>
> so we hav to make time and ans is 13
>
> On Mon, Apr 4, 2011 at 3:11 PM, Rakib Ansary Saikot <
> [email protected]> wrote:
>
>> I didnt quite get this problem. Sample case?
>>
>> On 4/4/11, rajat ahuja <[email protected]> wrote:
>> > You have to paint N boards of length {B1, B2, B3… BN}. There are K
>> painters
>> > available and you are also given how much time a painter takes to paint
>> 1
>> > unit of board. You have to get this job done as soon as possible under
>> the
>> > constraints that any painter will only paint continuous sections of
>> board,
>> > say board {2, 3, 4} or only board {1} or nothing but not board {2, 4,
>> 5}.
>> >
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