Without using logarithms, how can I convert a sequence of powers of 2 into linear integers in BASIC?
Input: 0, 1, 2, 4, 8, 16, 32, 64, 128
Output: 0, 1, 2, 3, 4, 5, 6, 7, 8
是的,您将数字转换为二进制。例如,64 的二进制值是:1000000。由于 1 在第七位,因此您知道所需的值是 7。这是一个 Visual Basic 程序来执行此操作:
Public Function DecimalToBinary(DecimalNum As Long) As String
Dim tmp As String
Dim n As Long=
n = DecimalNum
tmp = Trim(Str(n Mod 2))
n = n \ 2
Do While n <> 0
tmp = Trim(Str(n Mod 2)) & tmp
n = n \ 2
Loop
DecimalToBinary = tmp
End Function
在此算法中,值被附加到字符串中,但您也可以将它们存储在 1 和 0 的数组中。另请注意,您始终可以通过上述算法生成的字符串的长度来获得 2 的幂。例如,字符串“1001010”的长度为 7,表示该数字介于 64 和 127 之间。
这是 C# 算法进一步向下的 Visual Basic 转换。这是一个整数 Log2 函数。它使用先前初始化的数组。这个(和许多其他的小技巧)可以在这里找到:http: //graphics.stanford.edu/~seander/bithacks.html
这个算法的优点,以及为什么它很快,是因为它执行以 2 为底的对数,只需要几个简单的算术运算和数组查找。
Public Class BitHelper
' Methods
Shared Sub New()
BitHelper.logTable256(0) = BitHelper.logTable256(1) = 0
Dim i As Integer
For i = 2 To 256 - 1
BitHelper.logTable256(i) = (1 + BitHelper.logTable256((i / 2)))
Next i
End Sub
Public Shared Function Log2(ByVal number As Integer) As Byte
If (number <= 65535) Then
If (number > 255) Then
Return CByte((8 + BitHelper.logTable256((number >> 8))))
End If
Return CByte(BitHelper.logTable256(number))
End If
If (number <= 16777215) Then
Return CByte((16 + BitHelper.logTable256((number >> 16))))
End If
Return CByte((24 + BitHelper.logTable256((number >> 24))))
End Function
Private Shared ReadOnly logTable256 As Integer() = New Integer(256 - 1) {}
End Class
这是原始的 c# 代码。它是我前段时间制作的一个更大的 BitHelper 类的一个子集。
/// <summary>
/// Helper methods for bit twiddling. Much of the ideas used come
/// from http://graphics.stanford.edu/~seander/bithacks.html
/// </summary>
public static class BitHelper
{
private static readonly int[] logTable256 = new int[256];
/// <summary>
/// Initialize BitHelper class.
/// </summary>
static BitHelper()
{
logTable256[0] = logTable256[1] = 0;
for (int i = 2; i < 256; i++) {
logTable256[i] = 1 + logTable256[i / 2];
}
}
/// <summary>
/// Determines the integer logarithm base 2 (Floor(Log2(number))) of the specified number.
/// </summary>
/// <param name="number">The number for which the base 2 log is desired.</param>
/// <returns>The base 2 log of the number.</returns>
public static byte Log2(int number) {
if (number <= 0xffff) {
if (number > 0xff) {
return (byte) (8 + logTable256[number >> 8]);
} else {
return (byte) logTable256[number];
}
} else if (number <= 0xffffff) {
return (byte) (16 + logTable256[number >> 16]);
} else {
return (byte) (24 + logTable256[number >> 24]);
}
}
}