Jumat, 13 Mei 2016

Diktat Bab 9



Assalamu'alaikum wr.wb

Pada kesempatan kali ini, saya akan membahas kasus-kasus pada diktat bab 9. Langsung saja.

Workshop

Kasus 1

Algoritma

deklarasi
A[maks][maks],baris,kolom::integer
deskripsi
read (baris,kolom)
for i <-- 1 to n -1 do
for j <-- 1 to n-1 do
baris,kolom := A[i][j]
for i <-- 1 to n-1 do
for j <-- 1 to n-1 do
for z <-- 1 to n-1 do
hasil := A[i][j]*2[m]
endfor
endfor
endfor

C++

#include <iostream>

#define maks 10
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char** argv) {
int A[maks][maks], baris, kolom;
cout<<"input baris :"; cin>>baris;
cout<<"input kolom :"; cin>>kolom;
for(int i=0; i<baris; i++){
for(int j=0; j<kolom; j++){
cout<<"Masukan data ke ["<<i<<"] ["<<j<<"] : "; cin>>A[i][j];
}
}
cout<<"Isi Matriks A :\n";
for(int i=0; i<baris; i++){
for(int j=0; j<kolom; j++){
cout<<A[i][j]<<" ";
}
cout<<endl;
}
int m[baris],n;
cout<<"penghitungan vektor \n";
cout<<"berapa variabel : ";
cin>>m[kolom];
for(int z=0; z<kolom; z++){
cout<<"Masukan data ke ["<<z<<"] "" : "; cin>>m[z];
}
cout<<endl;

cout<<" vektor b : \n";
for(int z=0; z<kolom; z++){
cout<<m[z]<<"\n";
}
cout<<"perkalian matrik dan vektor \n";
for(int i=0; i<baris; i++){
for(int j=0; j<kolom; j++){
for(int z=0; z<kolom; z++){
n=A[i][j]*m[z];
cout<<n<<" ";
}
}
}
return 0;
}



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Kasus 2




C++




#include<iostream>




using namespace std;

main(){ int i,j,n;

cout<<"masukkan ordo matriks : "; cin>>n;

for(i=1;i<=n;i++){

for(j=1;j<=n;j++){

if(i==j) cout<<"1 ";

else cout<<"0 ";

}

cout<<endl;

}

}




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Flowchart




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Kasus 3




C++




#include <iostream>

#include <conio.h>

#include <cstdlib>




using namespace std;

main(){ int i,j,m,n,o;

cout<<"Program transpose matriks C(m,n)\nMasukkan nilai m : ";

cin>>m; cout<<"masukkan nilai n : "; cin>>n;

int mat[m][n];

for(i=0;i<m;i++){

for(j=0;j<n;j++){

cout<<"nilai C("<<i+1<<","<<j+1<<")=";

cin>>mat[i][j];

}

cout<<endl;

}

system("cls");

cout<<"matriks diatas diilustrasikan seperti gambaran dibawah ini\n\n";

for(i=0;i<m;i++){

for(j=0;j<n;j++){

cout<<mat[i][j]<<" ";

}

cout<<endl;

}

o=m; m=n; n=o;

cout<<"\nTranspose matriksnya adalah seperti gambar dibawah ini\n\n";

for(j=0;j<m;j++){

for(i=0;i<n;i++){

cout<<mat[i][j]<<" ";

}

cout<<endl;

}

return 0;

}




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Flowchart


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Kasus 4




C++




#include <iostream>

#include <stdio.h>

#include <stdlib.h>

using namespace std;

void Masukan(int *q, int brs, int klm) {

int i, j;

for(i = 0; i < brs; i++) {

for(j = 0; j < klm; j++) {

cout << " Elemen [" << (i + 1) << "][" << (j + 1) << "]? ";

scanf("%d", (q + i * klm + j));

}

}

}




void Tampilkan(int *q, int brs, int klm) {

int i, j;

for(i = 0; i < brs; i++) {

cout << endl;

for(j = 0; j < klm; j++) {

cout <<" ";

cout << *(q + i * klm + j) << "\t";

}

}

cout << endl;

}




void Kali(int *p, int *q, int *z, int bA, int kA, int bB, int kB) {

int i, j, k;

for(i = 0; i < bA; i++) {

for(j = 0; j < kB; j++) {

*(z + i * bB + j) = 0;

for(k = 0; k < kA; k++) {

*(z + i * bB + j) += *(p + i * kA + k) * (*(q + k * kB + j));

}

}

}

}




int main() {

cout<<"\t\t\tPROGRAM : PERKALIAN MATRIKS DENGAN SKALAR\n";

awal:

int *a, *b, *c, brsA, klmA, brsB, klmB;

cout<<"\n SYARAT...!!!\n";

cout<<" Jumlah Kolom A HARUS SAMA dengan Baris B\n\n";

cout << " Jumlah Baris Matriks A? ";

cin >> brsA;

cout << " Jumlah Kolom Matriks A? ";

cin >> klmA;

cout << " Jumlah Baris Matriks B? ";

cin >> brsB;

cout << " Jumlah Kolom Matriks B? ";

cin >> klmB;

if(klmA != brsB) {

cout << " Perkalian TIDAK Dapat Dilakukan." << endl;

cout << " Jumlah Kolom A TIDAK Sama Dengan Jumlah Baris B." << endl;

} else {

a = (int *) malloc(brsA * klmA * 4);

b = (int *) malloc(brsB * klmB * 4);

c = (int *) malloc(brsA * klmB * 4);




cout << "\n Kordinat Inputan [y][x]" ;

cout << "\n Contoh: ordo 2x2" ;

cout << "\n [1][1]\t[1][2]\t" ;

cout << "\n [2][1]\t[2][2]\t" ;

cout << "\n\n Input Matriks A:\n" ;

cout << " ----------------\n";

Masukan(a, brsA, klmA);

cout << "\n Input Matriks B:\n" ;

cout << " ----------------\n";

Masukan(b, brsB, klmB);




cout << "\n Matriks A:" << endl;

Tampilkan(a, brsA, klmA);

cout << "\n Matriks B:" << endl;

Tampilkan(b, brsB, klmB);

Kali(a, b, c, brsA, klmA, brsB, klmB);

cout << "\n Matriks A x B:" << endl;

Tampilkan(c, brsA, klmB);

}

cout<<"\n\n Apakah Anda Ingin Mengulang [Y/N] ? " ;

char m ;

cin>>m;

if (m == 'y' || m== 'Y')

goto awal;

else if (m == 'n' || m== 'N')

{

goto akhir;

}

akhir:

cout<<"\n\n\t\t\t \"TERMA KASIH\"";

return 0;

}







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Kasus 5




C++




#include <iostream>




using namespace std;




main(){ int i,j; float det,temp;

float mat[2][2], c[2][2], ci[2][2], ch[2][2];

for(i=0;i<2;i++){

for(j=0;j<2;j++){

cout<<"nilai C("<<i+1<<","<<j+1<<")=";

cin>>mat[i][j];

c[i][j]=mat[i][j];

}

cout<<endl;

}

cout<<"matriks diatas diilustrasikan seperti gambaran dibawah ini\n\n";

for(i=0;i<2;i++){

for(j=0;j<2;j++){

cout<<mat[i][j]<<" ";

}

cout<<endl;

}

mat[0][1]=-mat[0][1]; mat[1][0]=-mat[1][0];

temp=mat[0][0]; mat[0][0]=mat[1][1]; mat[1][1]=temp;

cout<<"\nadjoin dari matriks diatas adalah\n";

for(i=0;i<2;i++){

for(j=0;j<2;j++){

cout<<mat[i][j]<<" ";

}

cout<<endl;

}

det=mat[0][0]*mat[1][1]-mat[1][0]*mat[0][1];

det=1/(float)det;

cout<<"\nSedangkan untuk inversnya adalah sebagai berikut\n";

for(i=0;i<2;i++){

for(j=0;j<2;j++){

ci[i][j]=det*mat[i][j];

cout<<ci[i][j]<<" ";

}

cout<<endl;

}

for(i=0;i<2;i++){

for(j=0;j<2;j++){

for(int k=0;k<2;k++){

temp=c[i][k]*ci[k][j];

ch[i][j]=ch[i][j]+temp;

}

}

}

cout<<"\nhasil perkalian matriks asal dengan\nmatriks yang telah di invers adalah\n";

for(i=0;i<2;i++){

for(j=0;j<2;j++){

cout<<ch[i][j]<<" ";

}

cout<<endl;

}

}







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Flowchart




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Alhamdulillah pembahasan kasus pada diktat bab 9 terselesaikan. Sekian pembahasan dari saya, kurang lebihnya saya mohon maaf.

Kunjungi link blog temen kelompok saya:-)
Greentimeuad.blogspot.com
hadie97.blogspot.com


Wassalamu"alaikum wr.wb

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