[ Ice-D @ 22.12.2008. 16:06 ] @
A group of students are members of a club that travels annually to different
locations.
Their destinations in the past have included Indianapolis, Phoenix,
Nashville,Philadelphia, San Jose, and Atlanta. This spring they are planning a trip to Eindhoven.
The group agrees in advance to share expenses equally, but it is not practical to shareevery expense as it occurs.
Thus individuals in the group pay for particular things, such as meals, hotels, taxi rides, and plane tickets.
After the trip, each student's expenses are tallied and money is exchanged so that the net cost to each is the same, to within one cent.
In the past, this money exchange has been tedious and time consuming.

Your job is to compute, from a list of expenses, the minimum amount of money that must change hands in order to equalize (within one cent) all the students' costs.

Input

Standard input will contain the information for several trips. Each trip
consists of a

line containing a positive integer n denoting the number of students on the
trip. This is

followed by n lines of input, each containing the amount spent by a student
in dollars

and cents. There are no more than 1000 students and no student spent more
than

$10,000.00. A single line containing 0 follows the information for the last
trip.

Output

For each trip, output a line stating the total amount of money, in dollars
and cents,

that must be exchanged to equalize the students' costs.

Sample Input

3

10.00

20.00

30.00

4

15.00

15.01

3.00

3.01

0

Sample Output

$10.00

$11.99

Kod drugo inputa ( 4 studenta ) output je 11.99. Aritmetička sredina je
9,005. Da bi svi jednako potrošili student 3.00 mora dodati studentu 15.01
6,005, a student 3.01 mora studentu 15,00 dati 5,995. Dakle ukupna razmjena
novca iznosi 5,995 + 6,005 = 12,00 a ne 11, 99.
[ Eurora3D Team @ 24.12.2008. 01:20 ] @
Nisi uzeo u obzir da je najmanja jedinica cent
Mislim da su zaokruzili na prvu vecu vrednost (zato sto se broj zavrsava sa 5) i da im kalkulacija ide ovako
(15.00 + 15.01 + 3.00 + 3.01) / 4 = 36.02 / 4 = 9.01 (a ne 9.005)
i onda
3.00 mora da doda 15.01 iznos $6.00
a 3.01 mora da doda 15,00 iznos $5.99
Sto zajedno daje $11.99
[ jablan @ 24.12.2008. 09:59 ] @
Ice, ignoriši decimalnu tačku i radi sa celim brojevima, biće mnogo jasnije... ;)
[ Ice-D @ 24.12.2008. 14:38 ] @
thx, sad mi je jasno