First I marked the trivial empty squares (rows where the clue is 0), and I marked the same clues next to the edge or empty rows.
Then let's check the biggest clues. The are two 9s. C10 seems better, becase there are two marked empty cells while in R3 there is only one.
So C10 contain10 unmarked cells and 9 are filled. We can make the simple deductions.
Now we can use the trick. R1C10 and R9C10 are also filled.
So in C10 there remains 2 unfilled cells, R4C10 and R12C10.
Because of the 7 in R4 R4C10 cannot be empty.
We can fill the whole C10, then easy to fill the whole right side.
Now lets check the biggest clues again. Try to find rows/columns where there are N empty cells and N-1 are filled.
These are R3 and R11-R12.
Now we can use again the trick. And mark some cells in C1.
Then there are some simple deductions again such as R3C6 cannot be empty because then it is not possible to finish R2.
So finishing this puzzle isn't too hard.
But I want to show one more thing.
In C1-C2 there are 7s. It is an odd clue so there is at least one rectangle with odd width in this column. It can be just 3 width.
And clues in R9 and R10 are same, so it cannot be in the bottom part. Just in the up part.
Clues in R1 and R2 are also same. So it can be just in R1-R2-R3.
This is the way how it is possible to use the same clues trick.