Question

不使用对数，如何在BASIC中将2的幂序列转换为线性整数？

Input:  0, 1, 2, 4, 8, 16, 32, 64, 128
Output: 0, 1, 2, 3, 4,  5,  6,  7,   8

Answer 1

是的，您将数字转换为二进制。例如，二进制值64是：1000000。由于1位于第七个位置，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，值会附加到字符串，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长度获得2的幂。例如，字符串“1001010”的长度为7，这意味着该数字介于64和127之间。

Answer 2

这是c＃算法的Visual Basic转换。它是一个整数Log2函数。它使用先前初始化的数组。这可以在这里找到（以及许多其他有点狡猾的黑客攻击）：http://graphics.stanford.edu/~seander/bithacks.html

这种算法的优点及其快速的原因是因为它只用几个简单的算术运算和数组查找就可以执行基数为2的对数。

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。它是我不久前制作的更大的BitHelper类的子集。

/// <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]);
        }
    }
}

将2的幂转换为没有对数的线性序列

2 个答案: