pada pertemuan 7 membahas tentang array,,dan pada pertemuan itu dosen menyuruh untuk membuat algoritma tentang array,,dan dibuat bersama-sama kelompok tp,dan hasilnya cukup bagus dari kelompok saya,,hanya sedikit lagi menuju sempurna............
Jumat, 20 April 2012
refleksi minggu ke 6
pada pertemuan ke 6 materinya adalah perulangan secara iteraktif dan rekursif,,dalam pertemuan dikelas bapak dosen menjelaskan dengan baik,,tapi memang materinya sangat sulit,dan saya belum paham,,,,harus banyak belajar lagi............
Minggu, 15 April 2012
Algorithm& Flowchart Determining Number 100 divisible by 3&5
Algorithm & Flowchart Determining Numbers 100 divisible
by 3 & 5
v Flowchart:
v Description
of:
1. Advance
from the start.
2. Then we
give control of the control input with "i ß 0".
3. Our input
recurrence (Looping).
4. After that
we specify the first conditioning: "i> 100".
5. After the
conditioning of both: "i% 3 = 1 and i% 5 = 1".
6. but if the conditioning value "no"
then the output will be in print as "i" and will be entered in the
processing of "i ß i +15"
If the conditioning value "yes" then it will be entered in the processing
of "i ß i
+15".
7. The process
will continue until the "i> 100" trsbt met.
8. Completed.
v Syntax C +
+:
|
# Include <iostream>
# Include <string> using namespace std; int main () { int i; i = 0; while (! (i> 100)) { if (i% 3 == 1 && i% 5 == 1) { } else { cout << i << endl;} i = i +15; } system ("PAUSE"); return 0; } |
v Syntax C +
+ with Class structure:
|
# Include <cstdlib>
# Include <iostream> using namespace std; class number { friend ostream & operator << (ostream &, & number); friend istream & operator >> (istream &, & number); public: number () {}; void process () { i = 0; while (! (i> 100)) {If (i% 3 == 1 && i% 5 == 1) { } else { cout << i << endl;} i = i +15; } } private: int i; }; istream & operator >> (istream & in, & enter the numbers) { cout << "Numbers divisible by 3 & 5 =" << endl; in >> masukkan.i; return in; } ostream & operator << (ostream & out, numbers & output) { cout << "Bilangannya is" << keluaran.i << endl; return out; } int main (int argc, char * argv []) { number bil; cin >> bil; bil.proses (); cout << bil; cout << endl; system ("PAUSE"); return 0; } |
Algorithm & Flowchart Determining Build space [square / rectangle]
v Flowchart:
v Description of:
1. Advance
from the start.
2. We have
input the length of the first.
3. After
that we have input width.
4. And we set
pengkondisiaannya with input "p = l".
5. If the
conditioning value "yes" then it will go in processing the
"PUT" Wake Such is the Square "and in outputkan.
6. But if the
conditioning value "no" then the output will be processed sbgmana
"PUT" Wake Such is the Rectangle ".
7. Completed.
v Syntax C + +:
|
# Include <iostream> # Include <string> using namespace std; int main () { raptor_prompt_variable_zzyz string; int p; int l; raptor_prompt_variable_zzyz = "Enter the length:"; cout << endl << raptor_prompt_variable_zzyz; cin >> p; raptor_prompt_variable_zzyz = "Put L:"; cout << endl << raptor_prompt_variable_zzyz; cin >> l; if (p == l) { cout << "Build Such is the square" << endl;} else { cout << "Build Such is the Rectangle" << endl;} system ("PAUSE"); return 0; } |
v Syntax C + + with Class structure:
|
# Include <cstdlib>
# Include <iostream> using namespace std; bngn_ruang class { friend ostream & operator << (ostream &, bngn_ruang &); friend istream & operator >> (istream &, bngn_ruang &); public: bngn_ruang () {}; void process () { if (p == l) { result = "Wake Such is the Square";} else result = "Wake Such is the Rectangle"; } private: int p; int l; char * result; }; istream & operator >> (istream & in, bngn_ruang & insert) { cout << "Enter the length:" << endl; in >> masukkan.p; cout << "Enter the width:" << endl; in >> masukkan.l; return in; } ostream & operator << (ostream & out, bngn_ruang & output) { cout << "The output is as follows:" << endl << keluaran.hasil; return out; } int main (int argc, char * argv []) { bngn_ruang space; cin >> space; ruang.proses (); cout << space; cout << endl; system ("PAUSE"); return 0; } |
Sabtu, 14 April 2012
program membalik kata yang diinputkan(Rekursif)
program C++ !!!!mas brooo
#include <iostream.h>
#include <conio.h>
class Balik{
friend istream& operator>>(istream&, Balik&);
friend ostream& operator<<(ostream&, Balik&);
public:
void balik_kata();
void balik(int s);
private:
char kata[50];
int n;
};
void Balik::balik_kata(){
cout<<"Masukkan Kata Yang Akan Dibalik : ";
cin>>kata;n=strlen(kata);
cout<<"Kata Setelah Dibalik : ";
balik(n);
}
void Balik::balik(int s){
if (s != -1){
cout<<kata[s];
balik(s-1);
}
}
int main()
{
Balik X;
X.balik_kata();
system ("Pause");
return EXIT_SUCCESS;
}
Jumat, 06 April 2012
Menghitung nilai x^y
Menghitung nilai x^y dengan x bilangan real dan y bilangan bulat secara rekursif
Senin, 02 April 2012
THE OBSERVATION GROUP
At the meeting alpro week 4 we analyze the interation of the clean floor in the mosque ablution places UAD 111 campus,and we will describe a tittle what are the things that need to be done:
1.The first thing to do is set up tools to clean the clear.
2.Next go to the place to be in the clear.
3.After that process is to determine:
-Spray cleaners crust on the floor and wait for sereval minutes.
-Then do the cleaning by brushing the floor they will be evenly and with repetitive motion.
-After that,rinse remaining dirt with a water hose to the intermediary.
-Once the process is equaly clean,clear water that is still remembered on the surface of the floor with the moop(made of rubber)and do repeatedly.
4.After that return to the place cleaner tool storage.
5.Completed.
A few observations from our group results my be usefull and sorry if there is an error writing.
Refleksi Minggu Ke 5
Di pertemuan ke 5 alpro,membahas tentang sub program,ternyata memang sangat sulit,tapi dengan terus mencoba dan terus belajar insya allah hasilnya akan bagus.pesan saya jangan takut salah!!!!!!terus mencoba ,,,,
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