Olá estou com uma duvida , estou criando um app que calcula matrizes , fiz um layout simples com
-- [xmlpublic class MainActivity extends Activity implements OnClickListener{
EditText Text1,Text2;
Button Btcalcula;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Text1=(EditText) findViewById(R.id.Text1);
Text2=(EditText) findViewById(R.id.Text2);
Btcalcula=(Button) findViewById(R.id.Btcalcula);
Btcalcula.setOnClickListener(new View.OnClickListener(){
@Override
public void onClick(View arg0) {
double num1 = Double.parseDouble(
Text1.getText().toString());
[/xml]
e esse é o codigo em java da matriz que seria uma decomposição em LU
[JAVA]/**
* LU Decomposition, a structure to access L, U and piv.
*
* @param A
* Rectangular matrix
*/
public LUDecomposition(Matrix A) {
// Use a "left-looking", dot-product, Crout/Doolittle algorithm.
LU = A.getArrayCopy();
m = A.getRowDimension();
n = A.getColumnDimension();
piv = new int[m];
for (int i = 0; i < m; i++) {
piv[i] = i;
}
pivsign = 1;
double[] LUrowi;
double[] LUcolj = new double[m];
// Outer loop.
for (int j = 0; j < n; j++) {
// Make a copy of the j-th column to localize references.
for (int i = 0; i < m; i++) {
LUcolj[i] = LU[i][j];
}
// Apply previous transformations.
for (int i = 0; i < m; i++) {
LUrowi = LU[i];
// Most of the time is spent in the following dot product.
int kmax = Math.min(i, j);
double s = 0.0;
for (int k = 0; k < kmax; k++) {
s += LUrowi[k] * LUcolj[k];
}
LUrowi[j] = LUcolj[i] -= s;
}
// Find pivot and exchange if necessary.
int p = j;
for (int i = j + 1; i < m; i++) {
if (Math.abs(LUcolj[i]) > Math.abs(LUcolj[p])) {
p = i;
}
}
if (p != j) {
for (int k = 0; k < n; k++) {
double t = LU[p][k];
LU[p][k] = LU[j][k];
LU[j][k] = t;
}
int k = piv[p];
piv[p] = piv[j];
piv[j] = k;
pivsign = -pivsign;
}
// Compute multipliers.
if (j < m & LU[j][j] != 0.0) {
for (int i = j + 1; i < m; i++) {
LU[i][j] /= LU[j][j];
}
}
}
}
[/JAVA]
gostaria de inserir esta formula bara o button calcular.
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