Select unique

We have a transaction record that, for the sake of brevity, I will use
a simple paradigm to convey my need:

A sales clerk sells several pairs of shoes over the period of a day,
each & every day the clerk works. Each transaction is recorded in
a database. All clerks use a single physical register but login to
capture unique ID.  As each clerk logs in, it automatically logs out
the prior user and writes a zero-dollar record.

Each record logs the RcdNbr, clerk's ID, date-timestamp and sales
amount. (Other fields not pertinent to this discussion)

Example:
Rcd...CID...yyyy-mmdd.hhmm...Sale
001...ABC...2005-0101.0850...10.00
002...ABC...2005-0101.0930...00.00
003...DEF...2005-0101.1000...15.51
004...DEF...2005-0101.1200...00.00
005...ABC...2005-0101.1300...12.83
006...ABC...2005-0101.1530...00.00

The above is unsorted. I would first sort by CID (ClerkID) to get:

Rcd...CID...yyyy-mmdd.hhmm...Sale
001...ABC...2005-0101.0850...10.00
002...ABC...2005-0101.0930...00.00
005...ABC...2005-0101.1300...12.83
006...ABC...2005-0101.1530...00.00
003...DEF...2005-0101.1000...15.51
004...DEF...2005-0101.1200...00.00

As a double-check of a clerk's actual worked hours we want to extract
the last record for each clerk for each day. In the above sample, I'd
need records 004 & 006.

The transaction file covers 6 months worth of data.

Is there a statement that I can construct that would extract the last
(greatest?) time for each clerk for each day?

bump one more (last) time.

Something like this?

select
CID,
convert(char(8), tmstamp, 112) as 'Day',
max(tmstamp) as 'Last Logout'
from
dbo.SomeTable
group by
CID,
convert(char(8), tmstamp, 112)
order by
'CID', 'Day'

Simon