Croatian Open Competition in Informatics
The 1st Croatian Open Competition in Informatics results (Indonesian contestants) :
| Rank | Position | Score | Name |
5 7 186 Brian Marshal
15 21 96 Timotius Nugroho Chandra
22 34 71 Anggakara Hendra
25 41 65 Ricky Winata
26 45 62 James Edward
28 47 59 Aldrian Obaja -> OSN 2007 math gold medalist
28 47 59 Kristanto S
30 52 53 Reinardus Surya P. -> Retry for 123
31 53 50 Muhammad Irfan Alfarabbi
40 74 20 Sam Aryasa
41 89 8 Felix Halim -> Disappointed with the tasks (too easy).
42 90 0 Angelina Veni -> Retry for 61
There were only 9 contestants among they who will go to the 1st national training camp, why? And I’m a little bit disappointed with the result. Well, hoping better performance later…
November 1, 2007
9 responses to Croatian Open Competition in Informatics
beuh jadi malu kak
left it in the middle of the competition, too exhausted after CC@G 
anyway, i’ve made it yesterday w/out trial and error, i scored 61. not really good though, but not 0 
hmm… sebenernya, gw srednji uda slese, tapi pas tinggal 12 DETIK… Ga sempet submit. Gw coba di analysis, trnyata bener skali tembak… Jadi harusnya 53+70 = 123… Eh, trnyata srednji itu O(n) ya?
Well, it’s great Reinardus (you got the gold in OSN 2007, right?)…
Yes, you can solve “srednji” at linear time, judge’s solution and mine were same. Here’s my Pascal code:
Program srednji;
var hasil,tmp,h,l,i,n,b:longint;
data:array[1..100001]of longint;
xa,xb:array[-100001..100001]of longint;
begin
readln(n,b);
for i:=1 to n do
begin
read(data[i]);
if data[i]=b then tmp:=i;
end;
fillchar(xa,sizeof(xa),0);
fillchar(xb,sizeof(xb),0);
xa[0]:=1;h:=0;l:=0;
for i:=tmp-1 downto 1 do
begin
if (data[i]>b) then inc(h) else inc(l);
inc(xa[h-l]);
end;
xb[0]:=1;h:=0;l:=0;
for i:=tmp+1 to n do
begin
if (data[i]>b) then inc(h) else inc(l);
inc(xb[h-l]);
end;
hasil:=0;
for i:=-100001 to 100001 do hasil:=hasil+xa[i]*xb[-i];
writeln(hasil);
end.
Duh salah
var
ar:array[0..100001]of longint;
bnyk:array[-100001..100001]of longint;
tot,n,m,i,z,y:longint;
med:boolean;
begin
readln(n,m);
med:=false;
z:=0;
for i:=1 to n do
begin
read(ar[i]);
if ar[i]=m then
begin
y:=i;
med:=true;
end;
if med then
begin
if ar[i]>m then inc(z);
if ar[i]m then dec(z);
if ar[i]<m then inc(z);
tot:=tot+bnyk[z];
end;
writeln(tot);
end.
Good, such a smart code…
Well, it’s ok to post a long comment here.
hmm… koq agak sulit ya menulis potongan kode…
Pokoknya gini
1. baca dulu smua, sambil baca, yg ada di sbelah kanan median dikasi nilai (klo lbh dari +1, klo kurang -1)…
2. diliat dari median downto 1, trus dikasi nilai (kebalikannya dari yg prtama, klo lbh dari -1, klo kurang +1). Trus panggil nilai di array indeks ke sekian
nilai2 ar[z] (ar[-100001..100001] itu menyimpan ad berapa banyak kasus yg (bnyk angka yg > median – bnyk angka yg < median)=z
yah, kira2 gitu deh… T_T
Oh, admin itu kk Brian ya?
coba lagi ya? terakhir deh nih…
[code]
var
ar:array[0..100001]of longint;
bnyk:array[-100001..100001]of longint;
tot,n,m,i,z,y:longint;
med:boolean;
begin
readln(n,m);
med:=false;
z:=0;
for i:=1 to n do
begin
read(ar[i]);
if ar[i]=m then
begin
y:=i;
med:=true;
end;
if med then
begin
if ar[i]>m then inc(z);
if ar[i]m then dec(z);
if ar[i]<m then inc(z);
tot:=tot+bnyk[z];
end;
writeln(tot);
end.
[/code]
Yes, I am the admin here…
Well I must say that you are playing a nice host… You really know how to welcome people warmly. Such an interactive blog you we have here…