MS SQL Server cycle calculation

Dears, I have an problems with a calculation of this formula:

Column “A” has always the same value for a group of SubIDs.
In the “Result” column I need formula (function) like in the example columns “Result Formula Description”.
I tried many approaches but each of them help me with a half of the problem.
Please, help me!

with cte as (
  select 1 as subId, 100 as A, 0 as X, 10 as Y, 0 as Result   union all
  select 2,100,10,0,0   union all
  select 3,100,20,0,0   union all
  select 4,100,0,50,0   union all
  select 5,100,60,0,0   union all
  select 1,1000,0,100,0   union all
  select 2,1000,100,0,0   union all
  select 3,1000,200,0,0   union all
  select 4,1000,0,500,0 
)
select 
    subid, 
    a, 
    x, 
    y, 
    a+x-y
    +ISNULL((select sum(x) from cte c2 where c2.a=c1.a and c2.subid<c1.subid),0)
    -ISNULL((select sum(y) from cte c2 where c2.a=c1.a and c2.subid<c1.subid),0)
        as Result
from cte c1
order by a,subid

output:

subid       a           x           y           Result
----------- ----------- ----------- ----------- -----------
1           100         0           10          90
2           100         10          0           100
3           100         20          0           120
4           100         0           50          70
5           100         60          0           130
1           1000        0           100         900
2           1000        100         0           1000
3           1000        200         0           1200
4           1000        0           500         700

IMHO you should deal with recursion.

In this case I’ve used an UDF inline function to do the recursive job:

CREATE FUNCTION dbo.getResult(@A int)
RETURNS TABLE AS
RETURN
(
    WITH tree AS
    (
        SELECT   t1.subID, t1.A, t1.X, t1.Y, 
                 Result = (A + X - Y)
        FROM     t t1
        WHERE    t1.A = @A AND t1.subID = 1

      UNION ALL

        SELECT  t2.subID, t2.A, t2.X, t2.Y, 
                Result = tree.Result + t2.X - t2.Y
        FROM    tree
        JOIN    t t2
                ON t2.A = tree.A and t2.subID = tree.subID + 1
    )
    SELECT   *
    FROM     tree
)

Then you can use a CROSS APPLY join with every distinct value of A column.

SELECT DISTINCT
    r.subID, t.A, r.X, r.Y, r.Result
FROM
    t
CROSS APPLY 
    dbo.getResult(A) r
ORDER BY
    t.A, r.subID;

subID |    A |   X |   Y | Result
----: | ---: | --: | --: | -----:
    1 |  100 |   0 |  10 |     90
    2 |  100 |  10 |   0 |    100
    3 |  100 |  20 |   0 |    120
    4 |  100 |   0 |  50 |     70
    5 |  100 |  60 |   0 |    130
    1 | 1000 |   0 | 100 |    900
    2 | 1000 | 100 |   0 |   1000
    3 | 1000 | 200 |   0 |   1200
    4 | 1000 |   0 | 500 |    700

db<>fiddle here

Take care when using a recursive function, you should set OPTION (MAXRECURSION XXX) just to ensure it can hold all returned records.