Program Pascal Linked List

program linkedlist2;
uses crt;
type data = record
    nama,npm,tgl : array[1..10]of string;
    alamat,link: array[1..10]of integer;
    end;
var list: array[1..10]of integer;
    asc,desc : array[1..10]of string;
    mhs : data;
    avail,i,isi,opsi,del,start,temp,temp2,ujung: integer;

procedure Ascending;
var x, y, Imax : integer;
Smax, temp3 : string;
begin
for x := 1 to 10 do
begin
Smax := asc[x];
for y := x to 10 do
begin
if (asc[y] <= Smax) then
begin
Smax := asc[y];
Imax := y;
end;
end;
temp3 := asc[x];
asc[x] := asc[Imax];
asc[Imax] := temp3;
end;
end;

 procedure Descending;
 var x, y, Imin : integer;
 Smin, temp4 : string;
 begin
 for x := 1 to 10 do
 begin
 Smin := desc[x];
 for y := x to 10 do
 begin
 if (desc[y] >= Smin) then
 begin
 Smin := desc[y];
 Imin := y;
 end;
 end;
 temp4 := desc[x];
 desc[x] := desc[Imin];
 desc[Imin] := temp4;
 end;
 end;

procedure cekavail;
var j : integer;
begin
for j:=1 to 10 do
    begin
    if mhs.link[avail]=0 then
        begin
        temp:=0;
        break;
        end
    else if mhs.link[avail]=mhs.alamat[j] then
        begin
        if mhs.nama[j]='' then
            begin
            temp:=j;
            break;
            end;
        end;
    end;
end;

procedure menu;
begin
clrscr;
writeln('           PROGRAM LINKED LIST (MENGGUNAKAN AVAIL)');
writeln('         |Alamat| Nama |  NPM  | Tgl.Lahir | Link |');
for i := 1 to 10 do
    begin
    asc[i]:=mhs.nama[i];
    desc[i]:=mhs.nama[i];
    mhs.alamat[i]:=i;
    gotoxy(10,i+2);write('|',mhs.alamat[i]);
    gotoxy(17,i+2);write('|',mhs.nama[i]);
    gotoxy(24,i+2);writeln('|',mhs.npm[i]);
    gotoxy(32,i+2);writeln('|',mhs.tgl[i]);
    gotoxy(44,i+2);writeln('| ',mhs.link[i],' |');
    end;
if isi=0 then
    begin
    avail:=3;
    start:=3;
    end
else
    begin
    avail:=temp;
    for i:=1 to isi do
        begin
        if list[i]<>0 then
            begin
            start:=list[i];
            break;
            end;
        end;
    end;
if avail<>0 then
    begin
    gotoxy(52,avail+2);write(' << AVAIL ');
    end;
gotoxy(1,start+2);write('START >> ');
list[isi+1]:=avail;
gotoxy(1,13);writeln('AVAIL : ',avail);
gotoxy(1,14);writeln('1. List ');
gotoxy(1,15);writeln('2. Insert ');
gotoxy(1,16);writeln('3. Delete ');
gotoxy(14,14);writeln('4. Ascending Sort ');
gotoxy(14,15);writeln('5. Descending Sort ');
gotoxy(14,16);writeln('6. Exit ');
write('Select your choice : ');readln(opsi);
if opsi =1 then
    begin
    if isi=0 then
        begin
        writeln('THERE IS NO DATA !');
        readln;
        menu;
        end
    else
        begin
        for i:=1 to isi do
            begin
            if list[i]<>0 then
                begin
                write('[',mhs.nama[list[i]],']');
                if i<>isi then write(' >> ');
                end;
            end;
        readln;
        menu;
        end;
    end
else if opsi =2 then
    begin
    if (avail=0) then
        begin
        writeln('ERROR : DATA OVERFLOW');
        readln;
        menu;
        end
    else
        begin
        inc(isi);
        write('Masukkan nama : ');readln(mhs.nama[avail]);
        write('Masukkan npm : ');readln(mhs.npm[avail]);
        write('Masukkan tanggal lahir : ');readln(mhs.tgl[avail]);
        temp2:=avail;
        if isi<>1 then mhs.link[list[isi-1]]:=temp2;
        cekavail;
        mhs.link[avail]:=0;
        menu;
        end;
    end
else if opsi =3 then
    begin
    if isi=0 then
        begin
        writeln('ERROR: DATA UNDERFLOW');
        readln;
        menu;
        end
    else
        begin
        write('Data pada alamat keberapa yang akan dihapus ? ');readln(del);
        mhs.nama[del]:='';mhs.npm[del]:='';mhs.tgl[del]:='';
        for i:=1 to 10 do
            begin
            if mhs.link[i]=del then
                begin
                mhs.link[i]:=mhs.link[del];
                break;
                end;
            end;
        mhs.link[del]:=0;
        mhs.link[ujung]:=del;
        ujung:=del;
        for i:=1 to 10 do
            begin
            if list[i]=del then
                begin
                list[i]:=0;
                break;
                end;
            end;
        menu;
        end;
    end
else if opsi=4 then
    begin
    ascending;
    for i:= 1 to 10 do
        begin
        if asc[i]<>'' then
            begin
            write('[',asc[i],']');
            if i<>10 then write(' >> ');
            end;
        end;
    readln;
    menu;
    end
else if opsi=5 then
    begin
    descending;
    for i:= 1 to 10 do
        begin
        if desc[i]<>'' then
            begin
            write('[',desc[i],']');
            if i<>10 then write(' >> ');
            end;
        end;
    readln;
    menu;
    end;
end;

begin {program utama}
isi:=0;
mhs.link[1]:=7;
mhs.link[2]:=4;
mhs.link[3]:=5;
mhs.link[4]:=10;
mhs.link[5]:=1;
mhs.link[6]:=8;
mhs.link[7]:=9;
mhs.link[8]:=0;
mhs.link[9]:=2;
mhs.link[10]:=6;
ujung:=8;
menu;
end.

PROGRAM PASCAL SELECTION SORTING

program Selection;
uses crt;
var input : array [1..10] of integer;
banding,tukar,jumlah,i,j,k,l,temp:integer;
begin
clrscr;
writeln('PROGRAM TEKNIK SELECTION');
write('Masukkan jumlah data , maksimal 10 : ');readln(jumlah);
for i := 1 to jumlah do
    begin
        write('Masukkan data ke -',i,' : ');readln(input[i]);
    end;
for j:= 1 to jumlah-1 do
    begin
    k := j;
    for l := j+1 to jumlah do
        begin
        if input[k] > input[l] then
           begin
           k := l;
           end;
        banding:=banding+1;
        end;
    temp:=input[k];
    input[k] := input[j];
    input[j] := temp;
    tukar:=tukar+1;
    end;
for i:= 1 to jumlah do
begin
write('Data ke -',i,' yang sudah di sortir : ');writeln(input[i]);
end;
write('Jumlah perbandingan : ');writeln(banding);
write('Jumlah pertukaran : ');writeln(tukar);
readln;
end.

PROGRAM PASCAL TEKNIK EXCHANGE SORTING

program Exchange;
uses crt;
var input : array [1..10] of integer;
banding,tukar,jumlah,i,j,k,temp:integer;
begin
clrscr;
writeln('PROGRAM TEKNIK EXCHANGE');
write('Masukkan jumlah data , maksimal 10 : ');readln(jumlah);
for i := 1 to jumlah do
    begin
        write('Masukkan data ke -',i,' : ');readln(input[i]);
    end;
for j:= 1 to jumlah do
    begin
    for k:= j+1 to jumlah do
        begin
        banding:=banding+1;
        if input[j]>input[k] then
            begin
            temp:=input[k];
            input[k]:=input[j];
            input[j]:=temp;
            tukar:=tukar+1;
            end;
        end;
    end;
for i:= 1 to jumlah do
begin
write('Data ke -',i,' yang sudah di sortir : ');writeln(input[i]);
end;
write('Jumlah perbandingan : ');writeln(banding);
write('Jumlah pertukaran : ');writeln(tukar);
readln;
end.

Dragon Nest T4 lv 50 Crusader PvE/PvP Hybrid Build

Today i will post about my once favored game, Dragon Nest. My first char was a Cleric and advance into Paladin, and later on 50 cap become a Crusader. This build is my own preference for casual PvP and PvE players. This is not the best nor the worst, because all builds have their weaknesses and strong point, hence i believed the best build is the build that you build by your own preference.

Crusader is a Cleric's 3rd job and advanced from a Paladin. All Crusader skills are focused more as DPS and losing its path through defence. If you want more tanky and stunny(justice crash) you can choose Guardian as your 3rd job. Here is a quick link of my build, you can see damage, cooldown and all other info in this skill calculator :
http://dnarmory.com/skill-calc?i=4-20-42-50#AnGEX.kNjXgWp_obN84H

You can build your very own Crusader build, following this link:
http://dnarmory.com/skill-calc?i=4-20-42-60

Here is my build picture :

Here's my quick explanation on my build :
1st - Cleric's tree

• Lv 1 Righteous Bolt : a requirement for Lightning Zap(Paladin's Tree).
• Lv 1 Paralyze Bolt : i think you already know how useful this skill is in PvP.
• Lv 1 Dive Kick : to catch opponet that are far from you, and continue your combos.
• Lv 6 Sliding Knee Kick : lv 6 of this will give you extra kick when activated, and for me its very useful for bouncing your opponet again for continuation of combos.
• Lv 4 Block : Super useful in PvE and PvP Comp ON (not softban).
• Lv 8 Heal : Okay. I think you guys agreed to max it :)
• Lv 4 Toughness, Lv 4 Aerial Evasion, Lv 5 Sliding Step, Lv 1 Mental Fortitude, Lv 3 Attituned Mind : passives are useful in pvp, and for me i like regeneration more than maxing the mp pool.

2nd - Paladin's Tree

• Lv 12 Holy Relic : i think its one of a main DPS for crusaders. short cooldown, and Holy Relic EX will give you more wide area and 50% damage. And also you can change your Relic-EX placement during casting, not like the non-EX version. 
• Lv 6 Lightning Zap : i just take it lv 6 because it gives much damage on lv 6 as the skill damage amplifies. I personally dont think to max it because warriors can easily use "Relieve" and your lightning zap would be useless on pvp.
• Lv 7 Electric Smite : a must-maxed skill. main DPS skill for all paladins. and a super armor breaker.
• Lv 6 Armor Break : same as Smite, this one has a short cooldown and a medium super armor breaker.
• Lv 4 Auto BLock : helps you from being combo'ed by the other player , by luck :)
• Lv 3 Elemental Aura : i think its useful both in pve and pvp(againts elemental lords).
• Lv 2 Conviction Aura : very useful for a solo player and 1vs1.
• The others lv 1 i think not worth spending SP on those skills. Shield Charge is just for mobility and flinching. Just sacred hammering who is worth SP-spending on the next 60 cap, because of Sacred Hammering EX. However sacred hammering has a long cooldown on pvp, so i found it not so useful.

3rd - Crusader's Tree

• Lv 1 Judgement Hammer : medium-armor  breaker, low damage but have a short cooldown and long ranged, its good to combo this with your holy relic ex.
• Lv 1 Crusader's Zeal : only useful in PvE, i think. it will not give much bonus in pvp.
• Holy RELIC EX : ITS A MUST :) I LOVE THOSE GIANT RELICS.

For the heraldy skill plates, i recommend :

• Half Turn Kick : Action speed
• Electric Smite : Damage
• Armor Break : Damage
• Holy Relic : Damage
• Holy Kick : Action speed

Ok thats all i hope you guys get something useful from this post, thank you :D

PROGRAM PASCAL TEKNIK INSERTION

Berikut ini adalah program implementasi teknik Insertion Sorting menggunakan program Turbo Pascal :

program Insertion;
uses crt;
var input : array [1..10] of integer;
banding,tukar,jumlah,i,j,k,l,m,temp:integer;
begin
clrscr;
writeln('PROGRAM TEKNIK INSERTION');
write('Masukkan jumlah data , maksimal 10 : ');readln(jumlah);
for i := 1 to jumlah do
    begin
        write('Masukkan data ke -',i,' : ');readln(input[i]);
    end;
for j:= 2 to jumlah do
    begin
    l := j;
    m := j-1;
    for k:= j-1 downto 1 do
        begin
        if input[l] < input[m] then
            begin
            temp := input[l];
            input[l] := input[m];
            input[m] := temp;
            tukar := tukar+1;
            end;
        if m > 1 then
            begin
            l := l-1;
            m := m-1;
            banding := banding+1;
            end;
        end;
    end;
for i:= 1 to jumlah do
begin
write('Data ke -',i,' yang sudah di sortir : ');writeln(input[i]);
end;
write('Jumlah perbandingan : ');writeln(banding);
write('Jumlah pertukaran : ');writeln(tukar);
readln;
end.

Apa itu Big Oh? sebuah pertanyaan besar

Big Oh

Big Oh adalah fungsi yang berkaitan dengan kelajuan proses, dan biasanya digunakan dalam analisis suatu algoritma untuk menentukan efisiensi dan kompleksitasnya.

-T(n) = O(f(n))

 Bila ada suatu konstanta C, maka :

-T(n) ≤ C.(f(n))

Contoh Lain :

·        Jika f(n) = 5n² maka T(n) = O(n²)

·        Jika f(n) = 7n² + 3n + 5 maka T(n) = O(n²)

·        Jika f(n) = n³ – 2n + 8 maka T(n) = O(n³)

Klasifikasi Algoritma berdasarkan notasi Big Oh :

1.      T(n) = c

Disebut sebagai algoritma konstan, dimana programnya hanya dieksekusi dengan suatu nilai yang konstan. Bersifat konstan dan tidak dipengaruhi parameter atau banyak ada yang dieksekusi. Merupakan algoritma yang paling ideal.

Contoh Program Pascal :

Begin

A:=7;

End.

Maka T(n) = O(1), karena program hanya dieksekusi satu kali, dan konstan.

2.      T(n) = O(n)

Disebut algoritma linear, dimana waktu eksekusinya sebanding dengan jumlah data, dan merupakan kondisi optimal dalam membuat algoritma.

Contoh Program Pascal :

Begin

For I := 1 to n do

            Begin

            A:=A+1;

            End;

End.

Maka T(n) = O(n) karena program dieksekusi sejumlah n kali.

3.      T(n) = O(n²)

Disebut algoritma kuadratik, karena waktu eksekusi program akan sebanding dengan jumlah kuadrat jumlah data. Biasanya timbul karena adanya nested loop atau loop bersarang pada suatu program.

Contoh Program Pascal :

Begin

For I := 1 to n do

            Begin

            For J:=1 to n do

                        Begin

A:=A+1;

                        End;

            End;

End.

Loop dalam di eksekusi n kali. Loop luar di eksekusi n kali. Oleh karena itu T(n) = O(n²) dan program di eksekusi n x n atau n².

4.      T(n) = O(log n)

Disebut algoritma logaritmik, karena waktu eksekusi sebanding dengan logaritma dari jumlah data. Biasanya algoritma ini digunakan untuk memecahkan masalah besar menjadi masalah yang lebih kecil.

Contoh : algoritma Binary Search dan algoritma fungsi rekursif.

5.      T(n) = O(n log n)

Disebut algoritma linearitmik, karena merupakan gabungan dari linear dan logaritmik, biasanya digunakan untuk memecahkan masalah yang besar menjadi masalah yang lebih kecil dan diselesaikan secara terpisah, lalu kemudian hasilnya digabungkan.

Contoh : algoritma Quick Sort.

6.      T(n) = O(n³)

Merupakan algoritma yang buruk dan sebisa mungkin dihindari, biasanya algoritma ini dihasilkan dari 3 buah nested loop atau 3 buah loop bersarang.

Contoh program Pascal :

Begin

For I := 1 to n do

            Begin

            For J:=1 to n do

                        Begin

For K:= 1 to n do

            Begin

            A:=A+1;

            End;

                        End;

            End;

End.

Loop terdalam di eksekusi n kali. Loop tengah di eksekusi n kali. Loop luar di eksekusi n kali. Oleh karena itu T(n) = O(n³) dan program di eksekusi n x n x n atau n³.

Program Pascal Binary Search

program binary_search;
uses crt;
var input:Array [1..30,1..3] of integer;
    jumlah,i,j,cari,langkah:integer;
    isi:boolean;

procedure mencari;

function cektengah(var z:integer):integer;
var     x:integer;
    ada: boolean;
begin
    x := 1;
    ada := false;
    while not(ada) do
    begin
        if (z = input[x,2]) then
        begin
            cektengah := x;
            ada := true;
        end
        else
            x := x+1;
    end;
end;

function cekkirkan(var z:integer): integer;
var     x:integer;
    ada: boolean;
begin
    x := 1;
    ada := false;
    while not(ada) do
    begin
        if (z = input[x,1]) or (z =input[x,3]) then
        begin
            cekkirkan := x;
            ada := true;
        end
        else
            x := x+1;
    end;
end;

begin
j := cektengah(cari); langkah := 1;
        while (j > 1) do
        begin
            i := input[j,2]; j := cekkirkan(i);
            if (input[j,1] = i) or (input[j,3] = i) then
            begin
                langkah := langkah + 1;
            end;
        end;
writeln('Angka ditemukan dalam ',langkah,' langkah');
end;


procedure masuk;

function cektengah(var z:integer):integer;
var     x:integer;
    ada: boolean;
begin
    x := 1;
    ada := false;
    while not(ada) do
    begin
        if (z = input[x,2]) then
        begin
            cektengah := x;
            ada := true;
        end
        else
            x := x+1;
    end;
end;

begin
j:=1;isi:=false;
while not(isi) do
      begin
           if (input[i,2]>input[j,2])then
           begin
                if (input[j,3]=0)then
                begin
                input[j,3]:=input[i,2];
                isi:=true;
                end
                else j:=cektengah(input[j,3]);
           end
           else
           begin
                if (input[j,1]=0) then
                begin
                input[j,1]:=input[i,2];
                isi:=true;
                end
                else j:=cektengah(input[j,1]);
           end;
      end;
end;

begin
clrscr;
writeln('***************************');
writeln('  Program  Binary  Search  ');
writeln('***************************');
writeln('*****PROGRAM DIMULAI*******');
write('Masukan jumlah input : ');readln(jumlah);
for i:= 1 to jumlah do
    begin
    write('Masukkan elemen ke -',i,' : ');readln(input[i,2]);
    input[i,1]:=0;input[i,3]:=0;
    end;
for i:= 2 to jumlah do
    begin
    masuk;
    end;
writeln('****************************');
write('Masukan angka yang dicari : ');readln(cari);
mencari;
writeln('***********************************');
writeln('Terima kasih atas partisipasi anda!');
writeln('*********PROGRAM SELESAI***********');
readln;
end.

Tugas Pascal Moving Text

Program berikut ini akan membuat tulisan berjalan ke samping lalu ke bawah :

uses crt;
var
a,b:string;
i,j:integer;
x,y:byte;
begin
j:=45;
clrscr;
a:='HELLO WORLD';
for i:=length(a) downto 1 do
    begin
    j:=j-1;x:=1;y:=1;
    b:=copy(a,i,1);
    repeat
        delay(20);
        gotoxy(x,y);write(b);
        gotoxy(x-1,y);write(' ');
        x:=x+1;
        until x>j;
    end;
x:=44;
for i:=length(a) downto 1 do
    begin
    for y:=1 to 20 do
        begin
        gotoxy(x,y);write(A[i]);
        gotoxy(x,y-1);write(' ');
        delay(20);
        end;
    x:=x-1;
    delay(50);
    end;
readln;
end.

Program Pascal STACK

program infix_postfix;
uses crt;
type s100 = string[100];
stacks = record
stack : s100;
top : 0..100
end;
var infix, postfix : s100;
operator : set of char ;
x,y:byte;
i:integer;

procedure push (var s : stacks; elements : char);
begin
s.top := s.top + 1;
s.stack[s.top] := elements
end;

function pop (var s : stacks) : char;
begin
pop := s.stack[s.top];
s.top := s.top - 1;
end;

function priority (tanda : char) : integer;
begin
case tanda of
'^' : priority := 3;
'*', '/' : priority := 2;
'+', '-' : priority := 1;
'(' : priority := 0
end
end;

procedure convert (infix : s100);
var i : integer;
test : boolean;
temp, kar : char;
s : stacks;
begin
postfix:='';
for i := 1 to length(infix) do
    begin
    kar := infix[i];
    if kar = '(' then push(s, kar)
    else if kar = ')' then
        begin
        while s.stack[s.top] <> '(' do
        postfix:=postfix + pop(s);
        temp := pop(s)
        end
    else if kar in operator then
        begin
        while (s.top <> 0) and (priority(kar) <= priority(s.stack[s.top])) do
        postfix:=postfix + pop(s);
        push(s, kar)
        end
    else if kar <> ' ' then postfix:=postfix + kar;
    end;
    if s.top <> 0 then
        repeat
        postfix:=postfix + pop(s)
        until s.top = 0;
end;

procedure animate(postfix:string);
begin
x:=13;
    for i:=1 to length(postfix) do
        begin
        for y:=7 to 14 do
            begin
            gotoxy(x,y);write(postfix[i]);
            gotoxy(x,y-1);write(' ');
            delay(20);
            end;
    x:=x+1;
    delay(50);
    end;
end;

begin
clrscr;
operator:= ['^','*','/','+','-'];
writeln('===================================');
writeln('|  mengubah infix menjadi postfix  |');
writeln('===================================');
gotoxy (5,7);
write('infix = ');
readln(infix);
gotoxy (5,14) ;
write('hasil = ');
convert (infix);
animate(postfix);
gotoxy(13,7); write(infix);
gotoxy (5,16);
writeln('terimakasih atas partisipasi anda');
readln;
end.