第五章 控制结构编程练习.docx

1、第五章 控制结构编程练习第5章 控制结构程序设计No . 1绿色文字，用上引号隔开的，是注释，不需要写，起到说明作用。No . 2设置text1的maxlength设置为3方法1程序代码：方法2程序代码：Private Sub Command1_Click() Dim a As Integer 如果思考成熟的话，当然可以不定义那么多变量 a = Val(Text1.Text) 但往往我们办不到。 Text4.Text = a 100 求百位数 Text3.Text = (a Mod 100) 10 求十位数 Text2.Text = a Mod 10 求个位数End Sub方法3程序代码：Pr

2、ivate Sub Command1_Click() 往往我们做不到先知先觉，所以定义一些变量也无妨。 Dim a As Integer a = Val(Text1.Text) b = a 100 求百位数 c= a Mod 10 求个位数d= (a-c-100*b) 10 求十位数Text2.Text = bText3.Text = dText4.Text = cEnd SubNo . 3No . 4方法1程序代码: 一般来说，题目都提供了判断的规则。需要完成的，是如何接近这个规则。取字符mid()函数，绝对是一个很方便的手段。方法2程序代码:Private Sub Command1_Cli

3、ck() Dim a As Integer, b As Integer, c As Integer, d As Integer a = Val(Text1.Text) b = a 100 用数学方法取出个位，十位，百位上的数 c = (a Mod 100) 10 d = a Mod 10 If a = b 3 + c 3 + d 3 Then Label1.Caption = 是水仙花数 Else Label1.Caption = 不是水仙花数 End IfEnd SubNo . 5方法1程序代码Private Sub Command1_Click() Dim a As Integer, i

4、As Integer, s As Long, sum As Long a = Val(Text1.Text) For i = 0 To 5 s = s + a * 10 i sum = sum + s Next i Text2.Text = sumEnd Sub方法2程序代码Private Sub Command2_Click() Dim a As Integer, a = Val(Text1.Text) Text2.Text = a*(1+11+111+1111+11111+111111)End Sub选择结构编程No . 6Private Sub Command1_Click()Dim x

5、 As Double, y# 熟练掌握选择的几种结构，单分支，双分支，多分支，以及select casex = Val(Text1.Text)If x 10 Then y = x - 1ElseIf x 20 Then y = 2 * x - 1Else y = 3 * x - 1End IfText2.Text = yEnd SubNo . 7Private Sub Command1_Click()Dim a As Single, b As Single, c As Singlea = Val(Text1.Text) 因为等下要进行数学大小比较，所以转换类型。b = Val(Text2.Te

6、xt)c = Val(Text3.Text)If a = 0 Or b = 0 Or c c And b + c a And a + c b Then Text4.Text = 能Else Text4.Text = 否End IfEnd SubNo . 8方法1程序代码： 策略：总体来说，先比较两个。比较大的，令他暂时为max；比较小的，令他暂时为min。然后再把第三个c跟目前的max比较，跟目前的min比较。Private Sub Command1_Click()Dim max As Integer, min As IntegerA=Val(Text2.Text)B=Val(Text2.Te

7、xt)C=Val(Text2.Text)max = amin = maxIf b max Then max = bIf b max Then max = cIf c min Then min = cText4.Text = maxText5.Text=minEnd Sub方法2程序代码：Private Sub Command1_Click()Dim a%, b%, c As Integer, max%, min%a = Val(Text1.Text)b = Val(Text2.Text)c = Val(Text3.Text)If a b max = a min = bEnd IfIf c ma

8、x Then max = cIf c b Then max = aElse ba max = bEnd IfIf c max Then max = cIf a b Then min = bElse min = aEnd IfIf min c Then min = cText4.Text = maxText5.Text = minEnd Sub方法4程序代码: 最复杂的一种Private Sub Command1_Click()Dim a%, b%, c As Integer, max%, min%a = Val(Text1.Text)b = Val(Text2.Text)c = Val(Tex

9、t3.Text)If a b Then If b c If a c Then max = b min = c Else a b If a c If b c Then max = a min = c Else max = a min = b End If End IfEnd IfText4.Text = maxText5.Text = minEnd SubNo . 9Private Sub Command1_Click()Dim a As Integera = Val(Text1.Text)If a Mod 4 = 0 And a Mod 100 0 Or a Mod 400 = 0 Then

10、根据题意书写表达式Label2.Caption = 闰年ElseLabel2.Caption = 非闰年End IfEnd SubNo . 10Private Sub Command1_Click() 不得删除本行注释 Dim n As Integer n = Val(Text1.Text) If n Mod 4 = 0 And n Mod 100 0 Or n Mod 400 = 0 Then Text2.Text = 366 Else Text2.Text = 365 End IfEnd Sub循环结构编程No . 12Private Sub Command1_Click()Dim a A

11、s Integera = Val(Text1.Text)For i = 1 To aSum = Sum + i 2 通常是sum=sum+i，但是，并不是说第i项就一定是i，本题第i项就是i2NextText2.Text = SumEnd SubNo . 13 参考第六章-单选按钮由于是两个单选框,所以互相排斥,所以option.value=true的时候,肯定option.value=false.因此可以建立双分支结构.方法2程序代码：Private Sub Command1_Click() 不得删改本行注释 Dim n As Integer, i As Integer, sum As Lo

12、ng n = Val(Text1.Text) sum = 0 If Option1.Value = True Then For i = 1 To n Step 2 sum = sum + i Next i Else For i = 2 To n Step 2 sum = sum + i Next i End If Text2.Text = sumEnd Sub方法3程序代码：Private Sub Command1_Click() 不得删改本行注释 Dim n As Integer, i As Integer, sum As Long n = Val(Text1.Text) sum = 0 I

13、f Option1.Value = True Then A=1 Else A=2End IfFor i=a to n step 2Sum=sum+iNext i Text2.Text = sumEnd SubNo . 14方法1：Private Sub Command1_Click()Dim i As IntegerFor i = 1 To 200 If i Mod 3 = 0 And i Mod 7 = 0 Then Text1.Text = Text1.Text & & i 第二个Text1.Text代表过去的积累，第一个代表新的总的。 End IfNextEnd Sub方法2Privat

14、e Sub Command1_Click()Dim i As IntegerFor i = 1 To 200 If i Mod 3 = 0 And i Mod 7 = 0 Then Text1.Text = Text1.Text & space(2) & i space(2)代表输出2个空格。 End IfNextEnd SubNo . 15逐一判断， 因为要逐一，所以要循环；因为要判断，所以用选择。Private Sub Command1_Click()Dim i As Integer ,n as integer,sum as integerN=val(text1.text)For i =

15、1 To n If i Mod 3 = 0 Then sum = sum+i End IfNextText2.text=sumEnd SubNo . 16Private Sub Command1_Click()Dim i As IntegerFor i = 1 To 100 If i Mod 9= 0 Then Text1.Text = Text1.Text & & i2 End IfNextEnd SubNo . 17Private Sub Command1_Click()Dim i As Integer ,n as integer,sum as integerN=val(text1.tex

16、t)Sum=1 累乘的sum，需要事先设置为1，否则sum默认为0，每次累乘的结果都为0。For i = 1 To n If i Mod 7 = 0 Then sum = sum*i End IfNextText2.text=sumEnd SubNo . 18策略：先生成一个，令他为max，再陆续生成9个。每生成一个a，就和之前的max和min比较。比max大，把max里面的值改为当前这个a；比min小，把min里面的值改为当前这个a。Private Sub Command1_Click()Dim i As Integer, min As Integer, max As Integer, a

17、As Integermax = Int(Rnd * 101) min = maxFor i = 1 To 9 a = Int(Rnd * 101) If a max Then max = a If a min Then min = aNextText1.Text = maxText2.Text = minEnd SubNo . 21Private Sub Command1_Click()Dim n, i As Integern = int(Text1.Text)for i=2 to n-1 if n mod I =0 then exit fornextif i=n-1 then 或者本语句写成

18、if i=n then Label2.Caption = 非质数 相应的，这边是“质数”Else Label2.Caption = 质数 相应的，这边是“非质数”End if End SubPrivate Sub Command1_Click()Dim n, i As Integern = Val(Text1.Text)For i = 2 To n - 1 If n Mod i = 0 Then Label2.Caption = 非质数 Else Label2.Caption = 质数 End IfNext iEnd Sub错误的编程方式！因为大部分n/(n-1)不能整除。那么，n14时，在循

19、环中一直到i=13。最终14 mod 13 ,反而是质数No .22Private Sub Command1_Click() 不得删除本行注释，参考课本P96 Dim m As Integer, n As Integer, r% m = Val(Text1.Text) n = Val(Text2.Text) r = m Mod n Do While r 0 m = n n = r r = m Mod n Loop Label3.Caption = nEnd SubNo . 23方法1程序代码：Private Sub Command1_Click()不得删改本行注释 Dim i As Integ

20、er, j As Integer, m As Integer, s As Long For i = Len(Text1.Text) To 1 Step -1 m = val(Mid(Text1.Text, i, 1) s = s + m * 2 j 二进制转换为十进制，规定用位权展开式。寻找多项式的规律， j = j + 1 Next i Text2.Text = sEnd Sub方法2程序代码：Private Sub Command1_Click()不得删改本行注释 Dim i As Integer, m As Integer, s As Long a = Len(Text1.Text) 两

21、处要用到长度，所以就定义一个变量a来接收 For i = 1 To a Step 1 m = Val(Mid(Text1.Text, i, 1) s = s + m * 2 (a - i) 二进制转换为十进制，规定用位权展开式。寻找多项式的规律， Next iText2.Text = sEnd SubNo . 25方法1：Private Sub Command1_Click()For i = 1 To Len(Text1.Text) step 1 a = Mid(Text1.Text, i, 1) Text2.Text = a + Text2.TextNextEnd Sub方法2：Privat

22、e Sub Command1_Click()For i = Len(Text1.Text) To 1 step -1 a = Mid(Text1.Text, i, 1) Text2.Text = Text2.Text +a NextEnd Sub方法3：Private Sub Command1_Click()I=1while i =100000 Then 排除两种情况：输入不够6位；输 For i = 2 To 6 入的8542ab这种样式的。因为 val(8542ab)的结果是8542。Val函数的特点 n = val(Mid(Text1.Text, i, 1) sum = sum + n

23、Next i If val(Mid(Text1.Text, 1, 1) = sum Mod 10 Then 题目意思是sum可能是两位数。求个位数 Text2.Text = 合法会员 那么632146也是合法的，6等于16的个位数 Else Text2.Text = 非法会员 End IfElse Msgbox “请你输入6位会员号”End ifEnd SubNo . 28策略： 取出第一个字符max，假定它是最大的字符。 利用循环，从第二个字符开始逐一取字符A。在循环内嵌套选择语句， A是否比max大。是，就把a的值赋给max。方法1程序代码：Dim m As String, i As In

24、teger, s As String max= Mid(Text1.Text,1, 1) 字符比较，默认按照ASCII比较。 For i = 2 To Len(Text1.Text) s = Mid(Text1.Text, i, 1) If smax Then max = s Next i Label2.Caption = maxEnd Sub方法2程序代码：Dim m As String, i As Integer, s As String max=Mid(Text1.Text,1, 1) 字符比较，默认按照ASCII比较。 For i = 2 To Len(Text1.Text) s =

25、Mid(Text1.Text, i, 1) If asc(s)asc(max) Then max = s Next i Label2.Caption = maxEnd SubNo . 28可参考课本P160-Function过程Option ExplicitDim m As Integer, n As IntegerPrivate Sub Command1_Click() 不得删除本行注释n = Val(Text1.Text)m = Val(Text2.Text)Text3.Text = Fact(m) / (Fact(n) * Fact(m - n)End Sub#Function Fact(k As Integer) As Double Dim i As Integer, F As Double F = 1 For i = 1 To k F = F * i Next i Fact = FEnd Function#