二進位數相乘
Solomon
posted @ 2010年9月08日 10:08
in Uncategorized
, 1705 阅读
Public Class Form1 Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load Dim fileContents As String fileContents = My.Computer.FileSystem.ReadAllText(Application.StartupPath & "\Test1.txt") Dim d() = Split(fileContents, " x ") Dim n1 = d(0), n2 = d(1) Dim Len = n1.Length * 2 Dim str1 = Space(Len) & n1 & vbNewLine & Space(Len) & n2 & vbNewLine & "--------------------" & vbNewLine For i As Integer = n1.Length To 1 Step -1 If Mid(n2, i, 1) = "0" Then str1 = str1 & Space(len) & StrDup(n1.Length, "0") & vbNewLine Else str1 = str1 & Space(len) & n1 & vbNewLine End If len = len - 1 Next Dim x = c10to2(c2to10(n1) * c2to10(n2)) str1 = str1 & "--------------------" & vbNewLine & Space(len) & x My.Computer.FileSystem.WriteAllText(Application.StartupPath & "\result1.txt", str1, False) End End Sub Function c2to10(ByVal x As String) Dim c = 0 Dim n = x.ToString For i As Integer = n.Length To 1 Step -1 If Mid(n, i, 1) = "1" Then c = c + (2 ^ (n.Length - i)) Next Return c End Function Function c10to2(ByVal y As String) Dim x = Val(y) Dim str1 = "" Do Until x = 1 Or x = 0 str1 = IIf(x Mod 2 = 0, "0" & str1, "1" & str1) x = x \ 2 Loop str1 = IIf(x = 0, "0" & str1, "1" & str1) Return str1 End Function End Class
上課半夜
2022年12月21日 22:21
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2023年1月01日 15:36
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2023年1月16日 13:39
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