For i = 1 To 8
a(i) = (Int(Rnd * 10) + 1) * (i Mod 2 + 1)
Next i
k = 1
For i = 1 To 3
For j = 1 To 8 - 2 * i
If k * a(j) < k * a(j + 2) Then
t = a(j): a(j) = a(j + 2): a(j + 2) = t
End If
k = -k
Next j
执行该程序后,a(1)~a(8)可能的值为( )