Have you attempted to solve it or are you just baiting me

I think the grid shading problem is sodoku-esque.
For any line (row or column) let N be the number of numbers shown for that line. Let Xi be each of the numbers (i=1 to N).
Then if sigma(Xi,i=1..N) + N - 1 = 25 (25 is the length of each line, I think) then that line is completely defined - meaning there is only one possible solution.
sigma(Xi,i=1..N) means add up all the numbers.
For example, there is a row that is 1 3 1 3 10 2
N = 6
sigma(Xi,i=1..N) = 20
20 + 6 - 1 = 25.
So there is only one way of laying that row out (# means a shaded cell, _ means a white cell)
#_###_#_###_##########_##
For any line where that adds up to 24 there is one floating white cell to insert at the start or end or somewhere next to another white cell.
For lines that add up to 23 there are two floating white cells.
Etc.
So fill in all the 25s first. Then try as many of the 24s as you can. Then 23s. Etc. Each time you fill in a line you can go back and attempt to complete any intersecting higher order lines (i.e. if you've got some 24 columns left and you fill in a row then that might solve one or more of the columns).
Note also that lines with smaller N are easier to solve than lines with larger N because there are fewer spaces to insert the floating white cells.
It might help to mark white cells that are known to be white (as opposed to undecided) with a dot or something.
Or write some code that fills it in for you ... it's actually more fun than filling it in by hand unless you're very bored
Cheers,
Robin