数字处理程序.doc

8. 验证下列命题：设n是一个s1是n的各位数字之和，s2是2n的各位数字之和。若s1=s2,则n是9的倍数。根据图2和程序填写对象的相关属性，X代表该对象无属性。对象名称（Name）标题（Caption）正文（Text）窗体Frm1(验证命题)X标签LblInput(输入一个两位数)X文本框TxtInputX空判断按钮(CmdVerify)验证X退出按钮CmdExit退出XPrivate Sub CmdVerify_Click()Dim n As Integer,s1 As Integer,sa As Integern=Val_（TxtInput_)s1=Sm(n)s2=Sm(2*n)If _（s1=s2 and n Mod 9=0）_ThenMsgBox 该命题成立ElseMsgBox 该命题不成立End If TxtInput.Text=TxtInput.SetFocusEnd SubPrivate Sub CmdExit_Click()Unload MeEnd SubPrivate Function Sm(_（ByVal n As Integer）_) As IntegerDim sm1 as integersm1=0Do While n0sm1=sm1+n Mod 10_（n=n10）_LoopSm=sm1End Function8. 下面程序的功能是验证回文数猜想。任取一自然数，将其倒过来，并将这两个数相加，然后把这个和数倒过来，与原来的和数相加重复这个过程有限次，必将得到一个回文数。所谓回文数指左右数字完全对称的自然数，例如，121，12321，898，111均是回文数。根据猜想，假如取一自然数6868+86=154154+451=605605+506=1111，经过三步运算就得到回文数1111。Option ExplicitPrivate Sub Form_Click()Dim S As Long,D As Long,Sum As LongDim Flg As BooleanS=InputBox(输入一个正整数)Do D=Fun(S)Sum =S+DPrint Sum;=;S;+;DCall Comp(Sum,Flg)S=_(Sum)_Loop Until FlgPrint Sum;是回文数End SubPrivate Function Fun(N As Long)As LongDim Char As String,J As Long Dim I As Integer,Slen As IntegerChar=Ltrim(Str(N)_(Slen=Len(Char)_J=1For I=1 To SlenFun=Fun+Val(Mid(Char,I,1)*J_(J=J*10或J=10i)_Next IEnd FunctionPrivate Sub Comp(N As Long,F As Boolean)Dim Ch As String,S As Integer,I As IntegerS=Len(Ltrim(Str(N)F=FalseCh=Ltrim(Str(N)For I=1 To_(S)_If Mid(Ch,I,1)Mid(CH,S+1-I,1)Then Exit SubNext I_(F=True)_End Sub9.下面程序的功能是从键盘上输入一个不超过4位的十进制正整数，先把该十进数的每一位数字转换位一个4位二进制数，再把得到的二进制数依次合并成为一个16位的二进制数。例如输入十进制数359，会得到二进制数0000001101011001。Private Sub Form_Click()Dim num As Integer, s As String, M As IntegerDim i As Integer, j As Integer, n As Integernum = InputBox(输入一个长度不超过4位的十进制正整数：, , 1)_M=num_Do While num0j=j + 1n=num Mod 10s=cover(n) & s_num=num10_LoopFor i=1 To 4-js=0000 & sNext iText1.Text=十进制数&CStr(M)&的转换结果是： & sEnd SubPrivate Function cover(ByVal n As Integer)As StringDim i As Integer, s As String, k As IntegerDo Until n1_k=n Mod 2_s=CStr(k) & sn=n2LoopIf Len(s) 4 Thens=0000 & Scover=_right(s,4)_Elsecover=sEnd If End Function7下面程序的功能是将以“12；3；15”形式输入的字符串分解为数字12、3、15，分别赋给3个数组元素，并对其进行简单运算（图2为本程序执行画面）。Private Sub Command1_Click（）Dim a（3） As Integer, i As Integer, n As IntegerDim s As Strings = Text1.TextFor i = 1 To 2n = InStr(s, ;)a（i） = Left（s, n - 1）s = Mid（s, n + 1, Len(s) - n）Next iA（3）= Val( s )Text2.Text = a（1） * a（2） + a（3）End Sub图28本程序是找零巧数。所谓零巧数是具有下述特性的四位正整数：其百位数为0，如果去掉0，得到一个三位正整数，而该正整数乘以9，等于原数。例如20252259，所以2025是零巧数（如图3所示）。图3Private Function Change（s As String） As StringDim p As String, i As IntegerFor i = 1 To Len（s）If i 2 Then 是否为左边第二位数p = p + Mid（s, i, 1）将每位字符串连接End IfNext iChange = pEnd FunctionPrivate Sub Command1_Click（）Dim i As Integer, s As StringFor i = 1000 To 9999在四位数之间查找s = CStr（i）将数值转变成字符串If Mid（s, 2, 1） = 0 Then取百位数，并判断是否为0If Val（Change（s） * 9 = i Then 去掉百位数后的数乘9后是否等于原来的数List1.AddItem CStr（i）+ = +Change（s）+ *9 连接成字符串后，添加到列表框内End IfEnd IfNext iEnd Sub9本程序是把给定的二进制整数转换为八进制整数。Private Sub Command1_Click（）Dim a As String, b As String, c As Stringa = InputBox（请输入一个二进制数, 输入框）L = 3 -（ Len（a）Mod 3a = String（L, 0） & an = Len（a）/ 3 For m = 1 To nb = Mid（a, 3 * m - 2, 3）c = c + zh（b）Next mText1.Text = cEnd SubPrivate Function zh（s As String） As StringDim i As Integer, n As Integer, p As Integerp = 1For i = 2 To 0 Step -1n = n * 2 + Val（Mid（s, p, 1）p = p + 1Next izh = Str（n）End Function10找出由两个不同数字组成的平方数，并将结果按图4中的格式显示在列表框List1中。Option ExplicitPrivate Sub Command1_Click（）Dim I As Long, N As LongFor I = 11 To 300N = I * I If Verify（N） ThenList1.AddItem Str（I*I）+ =+ Str（I）+ * + Str（I）End IfNext IEnd SubPrivate Function Verify（N As Long） As BooleanDim A（0 To 9） As Integer, I As Integer, J As IntegcrDo While N 0A（N Mod 10）= 1N = N 10LoopFor I = 0 To 9Js = Js + A（I）Next IIf Js = 2 Then Verify = TrueEnd Function图42运行下面的程序，当单击窗体时，文本框Text1中的内容是（2），文本框Text2中的内容是（3）。Private Sub Form_Click()Dim St As StringDim St1 As String, St2 As StringSt = ASFRSDCFRSKXCall Change_String(St, St1, St2)Text1.Text = St1Text2.Text = St2End SubPrivate Sub Change_String(S As String, St1 As String, St2 As String)Dim tem As String, i As IntegerFor i = 1 To Len(S)tem = Mid(S, i, 1)If tem = S ThenSt2 = St2 & atElseIf tem = R ThenSt1 = St1 & TExit ForElseSt1 = St1 & temSt2 = St2 & temEnd IfNext iEnd Sub（2）AFT（3）AatF分析：Mid（x，n1，n2）从x字符串左边第n1个位置开始向右起取n2个字符。过程Change_String的功能是，将字符串St中的“S” 置换为“at”，“R”置换为“T”，并退出For循环。6运行下面程序，单击命令按钮Command1，则在窗体上显示的第一行内容是（13），第二行内容是（14），文本框Text1中显示的内容是（15）。Private Sub Command1_Click()Dim S As StringS = LAUSIVText1.Text = Display(S, Len(S)End SubPrivate Function Display(S As String, L As Integer) As StringIf L 1 Then Display = Display(S, L - 1)S = Left(S, L - 1) & Right(S, 1) & Mid(S, L, Len(S) - L)Display = SPrint DisplayEnd FunctionSL打印层数LAUSIV6一LAUSIV5二LAUSIV4三LAUSIV3四LAUSIV2五LAUSIV1Left(S, 0) & Right(S, 1) & Mid(S, 1, 5)VLAUSI六VLAUSI2Left(S, 1) & Right(S, 1) & Mid(S, 1, 4)VILAUS五VILAUS3Left(S, 2) & Right(S, 1) & Mid(S, 1, 3)VISLAU四VISLAU4Left(S, 3) & Right(S, 1) & Mid(S, 1, 2)VISULA三