Oracle SQL group by cube with distinct ID - oracle

Question

Example data (complete table has more columns and millions of rows):
invoice_number |year |department |euros
-------------------------------------------------------------
1234 |2010 |1 | 200
1234 |2011 |1 | 200
1234 |2011 |2 | 200
4567 |2010 |1 | 450
4567 |2010 |2 | 450
4567 |2010 |3 | 450
My Objective:
I want to sum the euros for every year and every department in every possible combination.
How result should look:
year |department |euros
--------------------------------------------
2010 |1 |650
2010 |2 |450
2010 |3 |450
2010 |(null) |650
2011 |1 |200
2011 |2 |200
(null) |1 |650
(null) |2 |650
(null) |3 |450
(null) |(null) |650
My query:
select year
, department
, sum(euros)
from table1
group by cube (
year
, department
)
Problem:
One invoice number can occur in several categories. For example, one invoice can have items from 2010 and 2011. This is no problem when I want to show the data per year. However, when I want the total over all years the euros will be summed twice, one time for each year. I want the functionality of 'group by cube' but I want to sum only distinct invoice numbers for aggregations.
Problem table:
year |department |euros
--------------------------------------------
2010 |1 |650
2010 |2 |450
2010 |3 |450
2010 |(null) |1550
2011 |1 |200
2011 |2 |200
(null) |1 |850
(null) |2 |650
(null) |3 |450
(null) |(null) |1950
Is it possible to do what I want? So far my search has yielded no results. I have created a SQL Fiddle, I hope it works

Answer 1

[Removed previous "solution"]
New attempt: here is quite an ugly solution, but it seems to work, even when two invoices have the same amount. With two table accesses, you should check if performance is acceptable.
SQL> with table1_cubed as
2 ( select year
3 , department
4 , grouping_id(year,department) gid
5 from table1
6 group by cube(year,department)
7 )
8 , join_distinct_invoices as
9 ( select distinct x.*
10 , r.invoice_number
11 , r.euros
12 from table1_cubed x
13 inner join table1 r on (nvl(x.year,r.year) = r.year and nvl(x.department,r.department) = r.department)
14 )
15 select year
16 , department
17 , sum(euros)
18 from join_distinct_invoices
19 group by year
20 , department
21 , gid
22 order by year
23 , department
24 /
YEAR DEPARTMENT SUM(EUROS)
---------- -------------------- ----------
2010 1 650
2010 2 450
2010 3 450
2010 650
2011 1 200
2011 2 200
2011 200
1 650
2 650
3 450
650
11 rows selected.

Answer 2

select year
,department
,case when GROUPING_id(year,department) in (3) then sum(dist_euro) else sum(euros) end sums
,decode(GROUPING_id(year,department),0,'NO GROUP',1,'DEPARTMENT IS NULL',2,'YEAR IS NULL',3,'TOTAL OVER ALL YEARS') info
from (
select year
, department
, euros
,case when row_number() over(partition by year order by year) = 1 then euros else 0 end dist_euro
from table1)
group by cube (
year
, department
)
order by GROUPING_id(year,department)