今天上班路上，看到新一期电脑报出来了，就随手买了一份来看
也就看到了下面的题目，无事，便写了一段
题目如下：
对如下矩阵，要求经过四次顺时针旋转，最后又得到最初的矩阵，并显示过程。
123 741 987 369 123
456 > 852 > 654 > 258 > 456
789 963 321 147 789
我的解题思路很简易，直接抽取对像来对比一下就能看出规律来：
第一个矩阵我用二维数组a(2,2)表示，第二个矩阵用二维数组b(2,2)表示，对比结果如下：
b(0,0)=a(2,0)
b(0,1)=a(1,0)
b(0,2)=a(0,0)
b(1,0)=a(2,1)
b(1,1)=a(1,1)
b(1,2)=a(0,1)
b(2,0)=a(2,2)
b(2,1)=a(1,2)
b(2,2)=a(0,2)
很明显的规律了，看下标的变化哈，然后就有了下面的程序了，vb.net的
删去了窗体设计代码，一个ＦＯＲＭ１，一个Ｔｅｘｔｂｏｘ１，一个ＢＵＴＴＯＮ１，一个ＴＩＭＥＲ１。
Public
Class Form1
Inherits System.Windows.Forms.Form
Public a, b As Integer(,) '定义两个二维数组
#Region " Windows 窗体设计器生成的代码 "
Public Sub New()
MyBase.New()
'该调用是 Windows 窗体设计器所必需的。
InitializeComponent()
'在 InitializeComponent() 调用之后添加任何初始化
End Sub
'窗体重写 dispose 以清理组件列表。
Protected Overloads Overrides Sub Dispose(ByVal disposing As Boolean)
If disposing Then
If Not (components Is Nothing) Then
components.Dispose()
End If
End If
MyBase.Dispose(disposing)
End Sub
'Windows 窗体设计器所必需的
Private components As System.ComponentModel.IContainer
'注意: 以下过程是 Windows 窗体设计器所必需的
'可以使用 Windows 窗体设计器修改此过程。
'不要使用代码编辑器修改它。
Friend WithEvents Button1 As System.Windows.Forms.Button
Friend WithEvents TextBox1 As System.Windows.Forms.TextBox
Friend WithEvents Timer1 As System.Windows.Forms.Timer
<System.Diagnostics.DebuggerStepThrough()> Private Sub InitializeComponent()
Me.components = New System.ComponentModel.Container
Me.Button1 = New System.Windows.Forms.Button
Me.TextBox1 = New System.Windows.Forms.TextBox
Me.Timer1 = New System.Windows.Forms.Timer(Me.components)
Me.SuspendLayout()
'
'Button1
'
Me.Button1.Location = New System.Drawing.Point(24, 112)
Me.Button1.Name = "Button1"
Me.Button1.Size = New System.Drawing.Size(80, 32)
Me.Button1.TabIndex = 0
Me.Button1.Text = "开始演示"
'
'TextBox1
'
Me.TextBox1.Location = New System.Drawing.Point(24, 32)
Me.TextBox1.Multiline = True
Me.TextBox1.Name = "TextBox1"
Me.TextBox1.Size = New System.Drawing.Size(80, 72)
Me.TextBox1.TabIndex = 1
Me.TextBox1.Text = ""
Me.TextBox1.TextAlign = System.Windows.Forms.HorizontalAlignment.Center
'
'Timer1
'
Me.Timer1.Interval = 1000
'
'Form1
'
Me.AutoScaleBaseSize = New System.Drawing.Size(6, 14)
Me.ClientSize = New System.Drawing.Size(128, 165)
Me.Controls.Add(Me.TextBox1)
Me.Controls.Add(Me.Button1)
Me.Name = "Form1"
Me.Text = "Form1"
Me.ResumeLayout(False)
End Sub
#End Region
Public Sub print() '在文本框里显示结果的过程
TextBox1.Text = Nothing
Dim i, j As Integer
For i = 0 To 2
For j = 0 To 2
TextBox1.Text += CStr(a(i, j))
Next
TextBox1.Text += vbCrLf
Next
End Sub
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
a = New Integer(2, 2) {} '重新定义a下标值
b = New Integer(2, 2) {} '重新定义b下标值
Dim i, j, g As Integer
For i = 0 To 2 '循环方式初始化数组a的值
For j = 0 To 2
g += 1
a(i, j) = g
Next
Next
Timer1.Enabled = True '启动时间控件
End Sub
Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick
print() '调用显示过程
Dim i, j As Integer
For i = 0 To 2 '利用循环给数组了赋值
For j = 0 To 2
b(i, j) = a(2 - j, i)
Next
Next
For i = 0 To 2 '因为直接a=b的只是句柄相等了，所以用循环重新给数组a赋值
For j = 0 To 2
a(i, j) = b(i, j)
Next
Next
End Sub
End Class
也就看到了下面的题目，无事，便写了一段
题目如下：
对如下矩阵，要求经过四次顺时针旋转，最后又得到最初的矩阵，并显示过程。
123 741 987 369 123
456 > 852 > 654 > 258 > 456
789 963 321 147 789
我的解题思路很简易，直接抽取对像来对比一下就能看出规律来：
第一个矩阵我用二维数组a(2,2)表示，第二个矩阵用二维数组b(2,2)表示，对比结果如下：
b(0,0)=a(2,0)
b(0,1)=a(1,0)
b(0,2)=a(0,0)
b(1,0)=a(2,1)
b(1,1)=a(1,1)
b(1,2)=a(0,1)
b(2,0)=a(2,2)
b(2,1)=a(1,2)
b(2,2)=a(0,2)
很明显的规律了，看下标的变化哈，然后就有了下面的程序了，vb.net的
删去了窗体设计代码，一个ＦＯＲＭ１，一个Ｔｅｘｔｂｏｘ１，一个ＢＵＴＴＯＮ１，一个ＴＩＭＥＲ１。
Public
Class Form1 Inherits System.Windows.Forms.Form Public a, b As Integer(,) '定义两个二维数组#Region " Windows 窗体设计器生成的代码 " Public Sub New() MyBase.New() '该调用是 Windows 窗体设计器所必需的。 InitializeComponent() '在 InitializeComponent() 调用之后添加任何初始化 End Sub '窗体重写 dispose 以清理组件列表。 Protected Overloads Overrides Sub Dispose(ByVal disposing As Boolean) If disposing Then If Not (components Is Nothing) Then components.Dispose() End If End If MyBase.Dispose(disposing) End Sub 'Windows 窗体设计器所必需的 Private components As System.ComponentModel.IContainer '注意: 以下过程是 Windows 窗体设计器所必需的 '可以使用 Windows 窗体设计器修改此过程。 '不要使用代码编辑器修改它。 Friend WithEvents Button1 As System.Windows.Forms.Button Friend WithEvents TextBox1 As System.Windows.Forms.TextBox Friend WithEvents Timer1 As System.Windows.Forms.Timer <System.Diagnostics.DebuggerStepThrough()> Private Sub InitializeComponent() Me.components = New System.ComponentModel.Container Me.Button1 = New System.Windows.Forms.Button Me.TextBox1 = New System.Windows.Forms.TextBox Me.Timer1 = New System.Windows.Forms.Timer(Me.components) Me.SuspendLayout() ' 'Button1 ' Me.Button1.Location = New System.Drawing.Point(24, 112) Me.Button1.Name = "Button1" Me.Button1.Size = New System.Drawing.Size(80, 32) Me.Button1.TabIndex = 0 Me.Button1.Text = "开始演示" ' 'TextBox1 ' Me.TextBox1.Location = New System.Drawing.Point(24, 32) Me.TextBox1.Multiline = True Me.TextBox1.Name = "TextBox1" Me.TextBox1.Size = New System.Drawing.Size(80, 72) Me.TextBox1.TabIndex = 1 Me.TextBox1.Text = "" Me.TextBox1.TextAlign = System.Windows.Forms.HorizontalAlignment.Center ' 'Timer1 ' Me.Timer1.Interval = 1000 ' 'Form1 ' Me.AutoScaleBaseSize = New System.Drawing.Size(6, 14) Me.ClientSize = New System.Drawing.Size(128, 165) Me.Controls.Add(Me.TextBox1) Me.Controls.Add(Me.Button1) Me.Name = "Form1" Me.Text = "Form1" Me.ResumeLayout(False) End Sub#End Region Public Sub print() '在文本框里显示结果的过程 TextBox1.Text = Nothing Dim i, j As Integer For i = 0 To 2 For j = 0 To 2 TextBox1.Text += CStr(a(i, j)) Next TextBox1.Text += vbCrLf Next End Sub Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click a = New Integer(2, 2) {} '重新定义a下标值 b = New Integer(2, 2) {} '重新定义b下标值 Dim i, j, g As Integer For i = 0 To 2 '循环方式初始化数组a的值 For j = 0 To 2 g += 1 a(i, j) = g Next Next Timer1.Enabled = True '启动时间控件 End Sub Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick print() '调用显示过程 Dim i, j As Integer For i = 0 To 2 '利用循环给数组了赋值 For j = 0 To 2 b(i, j) = a(2 - j, i) Next Next For i = 0 To 2 '因为直接a=b的只是句柄相等了，所以用循环重新给数组a赋值 For j = 0 To 2 a(i, j) = b(i, j) Next Next End SubEnd Class
