Algoritma & Flowchart Menentukan Bilangan 100 habis dibagi 3 & 5

Indonesian version :

• Flowchart :
  
  
  

• Pendeskripsiannya :

1. Terlebih dahulu mulai dari start.
2. Kemudian kita berikan penguasaan dgn inputan penguasaan “i  0”.
3. Kita masukan perulangannya (Looping).
   
4. Setelah itu kita tentukan pengkondisian pertamanya : “i>100”.
5. Setelah itu pengkondisian keduanya : “i%3=1 and i%5=1”.
6. Jika pengkondisian bernilai “yes” maka akan di masuk di pemrosesan “i  i+15” tetapi jika pengkondisian bernilai “no” maka output akan di cetak sbg “i” dan akan di masuk di pemrosesan “i  i+15”.
7. Proses tersebut akan terus dilakukan sampai nilai “i>100” trsbt terpenuhi.
8. Selesai.

• 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; }

• Syntax C++ dengan structure Class :

#include <cstdlib> #include <iostream> using namespace std; class bilangan{ friend ostream& operator<<(ostream&, bilangan&); friend istream& operator>>(istream&, bilangan&); public: bilangan(){}; void proses(){ 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, bilangan& masukkan){ cout<<"Bilangan Yang Habis Dibagi 3 & 5 = "<<endl; in >>masukkan.i; return in; } ostream& operator<<(ostream& out, bilangan& keluaran){ cout<<"Bilangannya adalah "<<keluaran.i<<endl; return out; } int main(int argc, char *argv[]) { bilangan bil; cin >> bil; bil.proses(); cout<< bil; cout<< endl; system("PAUSE"); return 0; }

English Version :

Ø  Flowchart:

 

Ø  Description of:

1.       Advance from the start.

2.        Then we give control of the control input with "i

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.       If the conditioning value "yes" then it will be entered in the processing  i +15" but if the conditioning value "no" then the output willßof "i   i +15".be in print as "i" and will be entered in the processing of "i+15".

7.    The process will continue until the "i> 100" the met.

8.       Completed.

Ø  C + + syntax:

Ø  # 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; }

Ø  C + + syntax 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; }

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