Matrix chain multiplication: Difference between revisions
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1773740 (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))
</pre>
=={{header|FreeBASIC}}==
{{trans|VBA}}
This is a translation of the Python iterative solution.
<syntaxhighlight lang="vbnet">Dim Shared As Integer U(), V()
Sub Aux(i As Integer, j As Integer)
Dim As Integer k = U(i, j)
If k < 0 Then
Print Str(i);
Else
Print "(";
Aux(i, k)
Print "*";
Aux(i + k, j - k)
Print ")";
End If
End Sub
Sub Optimize(a() As Integer)
Dim As Integer i, j, k, c
Dim As Integer n = Ubound(a) - 1
Redim U(n, n), V(n, n)
For i = 1 To n
U(i, 1) = -1
V(i, 1) = 0
Next i
For j = 2 To n
For i = 1 To n - j + 1
V(i, j) = &H7FFFFFFF
For k = 1 To j - 1
c = V(i, k) + V(i + k, j - k) + a(i) * a(i + k) * a(i + j)
If c < V(i, j) Then
U(i, j) = k
V(i, j) = c
End If
Next k
Next i
Next j
Print V(1, n); " ";
Aux(1, n)
Print
Erase U, V
End Sub
Dim As Integer A1(1 To 4) = {5, 6, 3, 1}
Optimize(A1())
Dim As Integer A2(1 To 13) = {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}
Optimize(A2())
Dim As Integer A3(1 To 12) = {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}
Optimize(A3())
Sleep</syntaxhighlight>
{{out}}
<pre> 48(1*(2*3))
38120((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))
1773740(1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))</pre>
=={{header|Go}}==
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