I had made the substituition but I don't get to go back to 2i.
n-2i=2m
Then
SUM[lg(2m)]= SUM [lg2+lgm]=SUM[lg2]+SUM[lgm]= SUM [1] +SUM[lgm]
OBS: SUM = SUMMATION.
I stop here. I dont know how to continue. Can you help me.
Thanks in advance
Allysson
Ajinkya Kale escreveu:
substitute
let n - 2i = 2mie 2i = n-2m
hence
SUM { lg (n-2i) } = SUM { lg (2m) }
no the limits
upper limit = i = n/2-1 ie 2i = n-2 ie n-2m = n-2 ie m=1
lower limit = i = 0ie 2i = 0 ie n-2m = 0ie m=n/2
therefore the summation is
SUM { lg((2m) }
On 10/16/07, Muntasir Azam Khan [EMAIL PROTECTED] wrote:
h am terrible sorry dude. it is subsequence!!!
On Oct 13, 11:39 pm, kannan [EMAIL PROTECTED] wrote:
hellow!
here is the problem statement.
you have to find the subset having the maximum sum in the given array
of
