求两个矩阵的乘积,Java转成Delphi的

[代斯Max]

public class MultiMatrix {
double[]multiplyMatrix;
public static void main(String args[]){
double[] a={12,44,188};
double[][] b={{3,5,9},{6,2,7},{1,4,8}};
MultiMatrix mm=new MultiMatrix();
mm.mMatrix(a,b);
mm.display();
}
public void mMatrix(double[] a,double[][] b){
multiplyMatrix=new double[a.length];
for (int i = 0; i<a.length; i++) {//rows of a
for (int j = 0; j<b[0].length; j++) {//columns of b
//columns of a = rows of b
multiplyMatrix[i]+=a[j]*b[j][i];
}
}
}
public void display(){
for (int i = 0; i<multiplyMatrix.length; i++) {
System.out.print(multiplyMatrix[i]+" ");
}
}


}

[代斯Max]

或者写一个求两个数组的矩阵乘积的函数

[s11ss]

uses Types;

{函数}
function m(a:TDoubleDynArray; b: array of TDoubleDynArray): TDoubleDynArray;
var
  I, J: Integer;
begin
  SetLength(Result, Length(a));
  for I:=0 to High(a) do
  begin
    Result[I] := 0;
    for J:=0 to High(b) do
      Result[I] := Result[I] + a[J]*b[J][I]
  end;
end;

{调用}
procedure TForm1.Button1Click(Sender: TObject);
var
  a:TDoubleDynArray;
  b: array of TDoubleDynArray;
  r: TDoubleDynArray;
  I: Integer;
begin
  SetLength(a, 3);
  a[0]:=12;a[1]:=44;a[2]:=188;

  SetLength(b, 3);
  for I:=0 to High(b) do
    SetLength(b[I], 3);
  b[0][0]:=3;b[0][1]:=5;b[0][2]:=9;
  b[1][0]:=6;b[1][1]:=2;b[1][2]:=7;
  b[2][0]:=1;b[2][1]:=4;b[2][2]:=8;

  r := m(a, b);
  showmessagefmt('%f,%f,%f', [r[0],r[1],r[2]])
end;