Java 打印矩阵乘法步骤
我需要打印两个矩阵如何相乘的分步解释。例如，如果有两个矩阵：Java 打印矩阵乘法步骤,java,swing,for-loop,matrix,Java,Swing,For Loop,Matrix,我需要打印两个矩阵如何相乘的分步解释。例如，如果有两个矩阵： ┌ ┐ │ 1 2 │ A = │ 4 3 │ └ ┘ ┌ ┐ │ 6 9 │ B = │ -8 -5 │ └ ┘ …我需要打印： **Explanation:** C11 = 1•6 + 2•(-8) = -10 C12 = 1•9 + 2•(-5) = -1 C21 = 4•6 + 3•(-8) = 0 C22 = 4
┌ ┐
│ 1 2 │
A = │ 4 3 │
└ ┘
┌ ┐
│ 6 9 │
B = │ -8 -5 │
└ ┘
**Explanation:**
C11 = 1•6 + 2•(-8) = -10
C12 = 1•9 + 2•(-5) = -1
C21 = 4•6 + 3•(-8) = 0
C22 = 4•9 + 3•(-5) = 21
**Result**
┌ ┐
│ -10 -1 │
A•B = │ 0 21 │
└ ┘
int frstMtxLen = frstMtx.length;
int scndMtxLen = secMtx.length;
for(int i = 0; i < frstMtxLen; i++)
{
for(int j = 0; j < frstMtx[i].length; j++)
{
resltMtxP[i][j] = "";
for (int k = 0; k < scndMtxLen; k++)
{
resltMtxP[i][j] = "\t"+resltMtxP[i][j]+" + "+ frstMtx[i][k] +" X "+secMtx[k][j]+" ;
}
try
{
doc.insertString(doc.getLength(), resltMtxP[i][j]+" ", headings); //I'm writing to a JTextPane
}
catch (Exception ex)
{
ex.printStackTrace();
}
}
try
{
doc.insertString(doc.getLength(), "\n", keyWord);
}
catch (BadLocationException e1)
{
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
告诉我哪里出错。这个程序解释了两个给定矩阵的乘法。输出被传递到
消费者import java.util.Locale;
interface Consumer<T>
{
void accept(T t);
}
public class MatMulExplainer
{
public static void main(String[] args)
{
test2x2_2x2();
test3x2_2x4();
}
private static void test2x2_2x2()
{
double A[][] = {
{ 1, 2 },
{ 4, 3 },
};
double B[][] = {
{ 6, 9 },
{ -8, -5 },
};
double C[][] = explain(A, B, console());
}
private static void test3x2_2x4()
{
double A[][] = {
{ 1, 2 },
{ 4, 3 },
{ 5, 6 },
};
double B[][] = {
{ 6, 9, -3, 8},
{ -8, -5, -1, 7},
};
double C[][] = explain(A, B, console());
}
private static Consumer<String> console()
{
return new Consumer<String>()
{
@Override
public void accept(String t)
{
System.out.println(t);
}
};
}
private static double[][] explain(
double A[][], double B[][], Consumer<String> consumer)
{
consumer.accept("Multiply\n"+toString(A));
consumer.accept("and\n"+toString(B));
int rA = A.length;
int cA = A[0].length;
int cB = B[0].length;
double C[][] = new double[rA][cB];
for (int r=0; r<rA; r++)
{
for (int c=0; c<cB; c++)
{
float sum = 0;
StringBuilder sb = new StringBuilder();
sb.append("C"+r+","+c+" = ");
for (int n=0; n<cA; n++)
{
sb.append(toString(A[r][n])+" * "+toString(B[n][c]));
sum += A[r][n] * B[n][c];
if (n < cA - 1)
{
sb.append(" + ");
}
}
sb.append(" = "+toString(sum));
consumer.accept(sb.toString());
C[r][c] = sum;
}
}
consumer.accept("Result:\n"+toString(C));
return C;
}
private static String toString(double A[][])
{
StringBuilder sb = new StringBuilder();
for (int r=0; r<A.length; r++)
{
for (int c=0; c<A[r].length; c++)
{
sb.append(toString(A[r][c])+" ");
}
sb.append("\n");
}
return sb.toString();
}
private static String toString(double d)
{
final String format = "%6.2f";
return String.format(Locale.ENGLISH, format, d);
}
}
您当前的代码在哪里？您有什么具体问题？如果您是来这里获取代码的，那么很抱歉！可能没有人会免费给你代码，而你自己必须为此而努力！是否希望完整的GUI解决方案是Swing？@admdrew我已编辑了问题以添加代码和问题。请看一下，让我知道问题出在哪里is@shekharsuman别担心！提供的答案对我有用。。谢谢你的关心。正是我需要的：）非常感谢：）
import java.util.Locale;
interface Consumer<T>
{
void accept(T t);
}
public class MatMulExplainer
{
public static void main(String[] args)
{
test2x2_2x2();
test3x2_2x4();
}
private static void test2x2_2x2()
{
double A[][] = {
{ 1, 2 },
{ 4, 3 },
};
double B[][] = {
{ 6, 9 },
{ -8, -5 },
};
double C[][] = explain(A, B, console());
}
private static void test3x2_2x4()
{
double A[][] = {
{ 1, 2 },
{ 4, 3 },
{ 5, 6 },
};
double B[][] = {
{ 6, 9, -3, 8},
{ -8, -5, -1, 7},
};
double C[][] = explain(A, B, console());
}
private static Consumer<String> console()
{
return new Consumer<String>()
{
@Override
public void accept(String t)
{
System.out.println(t);
}
};
}
private static double[][] explain(
double A[][], double B[][], Consumer<String> consumer)
{
consumer.accept("Multiply\n"+toString(A));
consumer.accept("and\n"+toString(B));
int rA = A.length;
int cA = A[0].length;
int cB = B[0].length;
double C[][] = new double[rA][cB];
for (int r=0; r<rA; r++)
{
for (int c=0; c<cB; c++)
{
float sum = 0;
StringBuilder sb = new StringBuilder();
sb.append("C"+r+","+c+" = ");
for (int n=0; n<cA; n++)
{
sb.append(toString(A[r][n])+" * "+toString(B[n][c]));
sum += A[r][n] * B[n][c];
if (n < cA - 1)
{
sb.append(" + ");
}
}
sb.append(" = "+toString(sum));
consumer.accept(sb.toString());
C[r][c] = sum;
}
}
consumer.accept("Result:\n"+toString(C));
return C;
}
private static String toString(double A[][])
{
StringBuilder sb = new StringBuilder();
for (int r=0; r<A.length; r++)
{
for (int c=0; c<A[r].length; c++)
{
sb.append(toString(A[r][c])+" ");
}
sb.append("\n");
}
return sb.toString();
}
private static String toString(double d)
{
final String format = "%6.2f";
return String.format(Locale.ENGLISH, format, d);
}
}
Multiply
1.00 2.00
4.00 3.00
and
6.00 9.00
-8.00 -5.00
C0,0 = 1.00 * 6.00 + 2.00 * -8.00 = -10.00
C0,1 = 1.00 * 9.00 + 2.00 * -5.00 = -1.00
C1,0 = 4.00 * 6.00 + 3.00 * -8.00 = 0.00
C1,1 = 4.00 * 9.00 + 3.00 * -5.00 = 21.00
Result:
-10.00 -1.00
0.00 21.00