数字和字符处理题型分析.doc

数字和字符处理题型分析(2)执行下面的程序，单击窗体后在窗体上显示的第一行结果是1，第三行结果是_2_。Private Sub Command1_Click()Dim Mystr, MyStr1, Mystr2 As StringMyStr1 = BFor i = 1 To 3Mystr2 = LCase(MyStr1)MyStr1 = MyStr1 & Mystr2Mystr = Mystr & MyStr1Print MystrMyStr1 = Chr(Asc(MyStr1)+ i)Next iEnd SubiMyStr1MyStr2MyStr打印B1BbbBbBbC2CccEBbCcBbCc3EeeBbCcEeBbCcEe答案：(1)Bb(2)BbCcEe分析：Asc()函数只取第一个字符的数值。7.下面是一个加密/解密程序，加密的算法是：将原文中每个字符的ASCII代码加上该字符在原文中的位置数(设字符在原文中是第n个字符，在其位置是n被10除的余数，当余数为0时，其位置数为10。例如原文字符串为“meet me at sunset”，第一个字符“m”，其位置数为1，而“s”为第12个字符，其位置数是被10除的余数2，其余依次类推)作为密文字符的代码，再将密码字符逆序排列(例如ABCD，逆序排列为DCBA)即为密文。解密算法是加密算法的逆运算。Option ExplicitPrivate Sub Command1_Click() 加密 Dim enc As String, dec As String, i As Integer Dim s As String * 1, n As Integer dec = Text1 For i = 1 To Len(dec) s = Mid(dec, i, 1) (19) If n = 0 Then n = 10 enc = Chr(Asc(s) + n) & enc Next i Text2 = encEnd SubPrivate Sub Command2_Click() 解密 Dim enc As String, dec As String, i As Integer Dim n As Integer, s As String * 1, p As Integer enc = Text2 (20) For i = 1 To n s = Mid(enc, i, 1) p = n Mod 10 If p = 0 Then p = 10 dec = (21) n = n - 1 Next i Text1 = decEnd Sub答案：n = i Mod 10n = Len(enc)Chr(Asc(s) - p) & dec(4)运行下面的程序，单击窗体后在窗体上显示的第一行结果是1，第三行结果是2。Private Sub Form_Click()Dim mst As String, mst1 As String, mst2 As StringDim i As Integermst1 = CeBbAaFor i = Len(mst1) To 1 Step -2mst2 = Mid(mst1, i - 1, 2)mst = mst & mst2Print mstNext iEnd Submst1mst2msti打印CeBbAaAaAa6AaBbAaBb4AaBbCcAaBbc2AaBbc答案：(8)Aa(9)AaBbCe分析：Mid()函数的操作方式是从后向前取字符。(2)执行下面程序，第二行输出结果是1，第三行输出结果是2。Private Sub Form_Click()Dim ch As String, i As Integerch = ABCFor i = 1 To 3ch = Mid(ch, 2 * i - 1, 1)+ Left(ch, Len(ch)Print chNext iEnd Subchi打印ABC1Mid(ABC,1,1)+Left(ABC,3)A”+”ABCAABCAABC2Mid(AABC,3,1)+Left(AABC,4)B+AABCBAABCBAABC3Mid(BAABC,5,1)+Left(BAABC,5)C+BAABCCBAABCCBAABC答案：(1)BAABC(2)CBAABC35. 执行下面的程序，单击窗体后在窗体上显示的结果是_35_。Private Sub Form_Click() Dim Str1 As String, Str2 As String Dim Str3 As String, I As Integer Str1=e For I=1 To 2 Str2=UCase(Str1) Str1=Str2 & Str1 Str3=Str3 & Str1 Str1=Chr(Asc(Str1)+I) Next I Print Str3End SubIStr1Str2Str3打印1eEEeEe2FFEeFFFFHEeFF(A)EeFF(B)eEFF(C)EEFF(D)eeFF答案：A分析：35. 在窗体Form1上，有一个列表框控件List1，在窗体的Click事件中有如下代码：Private Sub Form_Click()Dim k As IntegerDim entry As String, item As Stringentry=EDCBAFor k=Len(entry)To 1 Step -1item=LCase(Mid(entry,k,1) & kList1.AddItem itemNext kEnd Sub运行此程序，单击窗体后在窗体的列表框中显示的第四个列表项内容是_35_。(A)b4(B)b2(C)d4(D)d2答案：D分析：Kentryitem添加5EDCBAa5a54b4b43c3c32d2d21e1e11. 执行下面的程序，在窗体上显示的输出结果的第一行是_this is a book._，第二行结果是_This Is A Book._。Option ExplicitPrivate Sub Form_Click()Dim s As String, i As Integer, flag As Booleans = THIS IS A BOOK.s = LCase(s)Print sflag = TrueFor i = 1 To Len(s)If Mid(s, i, 1)= Thenflag = TrueElseIf flag Thens = Left(s, i - 1) & UCase(Mid(s, i, 1) & Right(s, Len(s)- i)flag = FalseEnd IfNext iPrint sEnd Subsiflag打印THIS IS A BOOK.1Truethis is a book.this is a book.This is a book.False5TrueThis Is a book.6False8TrueThis Is A book.9False10TrueThis Is A Book.11FalseThis Is A Book.分析：用flag来控制进程的走向，4. 执行下面的程序，图片框第一行内容是 CBA ，图片框第二行内容是 ABC 。Option ExplicitPrivate Sub Command1_Click()Dim S As StringS = ABCBack SPicture1.PrintPicture1.Print SEnd SubPrivate Sub Back(St As String)If Len(St) 1 ThenBack(Right(St, Len(St)- 1)End IfPicture1.Print Left(St, 1);End SubSt层次打印ABC一BC二C三CBC二CBABC一CBA0ABC分析：这是递归过程。3执行下面的程序，单击窗体上的按钮，则在窗体上显示的第一行是(4)，第二行是(5)，第三行是(6)。Private Sub Command1_Click()Dim s As String, t As StringDim k As Integer, m As Integers = BASICYk = 1: m = kFor k = 1 To Len(s) Step m + 1t = t & Chr(Asc(Mid(s, m, 1)+ k)k = k + 1If Mid(s, k, 1)= Y Then Exit Form = m + kPrint tNext kPrint mEnd Substkm打印BASICYC11C21+2=3CW43+5=8CW78答案：(4)C(5)CW(6)8分析：要记住Chr(65)A，Chr(97)a。Chr(x)返回以x为ASCII代码值的字符, Asc(x)为求单个字符ASCII代码函数，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。注意：循环变量k在循环体参与了运算，实际步长为3。变量m的赋值过程为1、3、8。2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 St0 As StringDim St1 As String, St2 As StringSt0 = ASFRSDCFRSKXCall Change_String(St0, 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)112，实际上只循环了4次，读到“R”就退出了 tem = Mid(S, i, 1)读取一个字符, A、S、F、R、S、If tem = S ThenSt2 = St2 & atElseIf tem = R ThenSt1 = St1 & TExit ForElseSt1 = St1 & temSt2 = St2 & temEnd IfNext iEnd Sub答案：iStemSt1St21ASFRSDCFRSKXAAA2SA & atAat3FA & FAFAat & FAatF4RAF & TAFT(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。3执行下面程序，单击命令按钮Command1，则在窗体上显示的第一行是(4)，第二行是(5)，第三行是(6)。Private Sub Command1_Click()Dim s_data As Integer, d_data As String, k As Integer, p As Strings_data = 29Do Until s_data = 0 And st = 9 Then p = p & st Else If p Then List1.AddItem p p = End If Next i If p Then List1.AddItem p ReDim num(List1.ListCount) For i = 1 To UBound(num) p = List1.List(i - 1) num(i) = Val(Left(p, Len(p) - 1) n = Val(Right(p, 1) List2.AddItem change(num(i), n) Next iEnd SubPrivate Function change(x As Integer, y As Integer) As Integer Dim p As Integer, k As Integer Do p = x Mod 10 change = change + p * y k x = x 10 k = k + 1 Loop Until x 0j = j + 1j记录有几组4位的十进制n=num Mod 10用n取num的个位数s=cover(n) & s用cover函数求n的二进制数，用s连接起来_ _去掉num的个位数num=num 10LoopFor i=1 To 4-j如果j小于4就在前面加0000s=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_ _取n除2的余数，这是十进制转变成二进制的方法k=n Mod 2s= CStr(k) & s用s将二进制数连接起来n=n 2LoopIf Len(s)4 Then如果二进制数不够4位，在左边加0000s=0000 & Scover=_ _只取右边的4位，截去多余的0。right(s,4)Elsecover=sEnd IfEnd 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)。图2Private Sub Command1_Click()Dim a(3)As Integer, i As Integer, n As IntegerDim s As Strings = Text1.TextFor i = 1 To 2n = (16) n记录“;”的位置InStr(s, ;)或Mid(s, n + 1, Len(s)-n)a(i)= Left(s, n - 1)将“；”左边的字符记录给a(1)和a(2)s = (17) 将“；”右边的字符记录给s。Mid(s, n + 1, Len(s)-n)Next i(18) 将最右边的字符（数）赋值给a(3)。A(3)= Val( s )Text2.Text = a(1)* a(2)+ a(3)End Sub答案：(16)InStr(s, ;)(17)Mid(s, n + 1, n + 1)或Mid(s, n + 1, Len(s)-n)(18) A(3)= Val( s )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，Left(x，n)从x字符串左边起取n个字符。本题的难点是如何利用字符串函数从字符串中分离出数字的字符串。8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 Thenp = (19) 用Mid函数逐个读取s中的字符，i=2除外。p & Mid(s, i, 1)End IfNext iChange = pEnd FunctionPrivate Sub Command1_Click()Dim i As Integer, s As StringFor i = 1000 To 9999(20) 将i转变成字符串s，s = CStr(i)If Mid(s, 2, 1)= 0 Then判别s的百位是否为“0”。If Val(Change(s)* 9 = i Then将去掉百位数为“0”的数乘9是否等于i(21)按图例中的格式添加到列表框中List1.AddItem CStr(i)+ = +Change(s)+ *9End IfEnd IfNext iEnd Sub答案：(19)p + Mid(s, i, 1)(20)s = CStr(i)(21)List1.AddItem CStr(i)+ = +Change(s)+ *9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，CStr(n)将一数值n转换成第一位非空字符的字符串型。过程Change的功能是将字符串s左边第二个“0”字符剔除掉，重新组成新字符串p返回。选择(21)要根据图3的内容来确定。8下面的程序可从一个由字母与数字相混的字符串中选出数字串，并把数字串构成的数写入一个名为List1的列表框中(图1是程序执行的画面)。图1 运行界面Option ExplicitPrivate Sub Cmd1_Click()Dim s As String, k As Integer, c()As StringDim p As String, i As Integers = Text1.Text(19) k记录数据段的个数k = 1For i = 1 To Len(s)If Mid(s, i, 1)= 0 And Mid(s, i, 1)= 0 And Mid(s, i + 1, 1)= 9 And i 1 Then 下一个字符是否为字母If p Then如果p不为空的话，即其中还有内容(21) 增加一个数组ReDim Preserve c(k)，下列句子有提示c(k)= p数值记录到数组k = k + 1(22) 清空p的内容p = End IfEnd IfNext iReDim Preserve c(k)c(k)= pFor i = 1 To k(23) 将数组的内容逐个添加到列表框内List1.AddItem c(i)Next iEnd Sub答案：(19)k = 1(20)p & Mid(s, i, 1)(21)ReDim Preserve c(k)(22)p = (23)List1.AddItem c(i)分析：由于在Dim s()时没有定义数组的大小，所以在过程中一定要用ReDim重新定义数组。变量k为数组的大小数，变量p为单个数字的字符串，数组c被赋予各数字的字符串。AddItem的功能是向列表框添入一项数据。8.以下是一个加密程序，其中r1与r2是两个密钥表，分别对应字母az及空格。加密方法是从原文的字符串中取一个字符，再根据该字符在原文中的排位次序是奇数还是偶数，分别从r1或r2密钥表中查出相应的替代字符。如原文为“meet me at sunset”，其首字符为m，排位为1，则使用密钥表r1，由于在字母表中m为第13个字符，而r1中对应字符为t，故将m替换成t。其余类推。原文中如包含大写字母，则应先全部转换为小写字母。Const r1 As String * 27 = qzawsxedcrfvtgbyhnujm ikolp Const r2 As String * 27 = poi uytrewqasdfghjklmnbvcxzPrivate Sub Command1_Click()Dim I As Integer, code As String, s As String * 1Dim n As Integer, decode As Stringcode =【23】 读取Text1中的内容Text1.textFor i = 1 To Len( code )s = Mid( code , i , 1 )读取一个字符赋值给sIf s then字符是否为空格，不是空格【24】n记录字符在字母表的序号，例如“b”为2。n=Asc(s)- 96If i Mod 20 then判断奇偶数Decode = decode & Mid( r1,n,1 )Else【25】参照上一句decode = decode &Mid(r2,n,1)End IfElse是空格If i Mod 20 then奇数【26】参照下一句，第27个字符为空格Else偶数Decode = decode & Mid( r2, 27, 1 )第27个字符为空格End IfEnd IfNext iText2 = decodeEnd Sub答案：【23】Text1.text答案：【24】n=Asc(s)- 96答案：【25】decode = decode &Mid(r2,n,1)答案：【26】decode = decode & Mid(r1,27,1)7.本程序是一个采用矩阵变换对西文进行加密的程序。取大于或等于原文长度的最小平方数n2，构造一个nn的矩阵，将原文中的字符逐个按行写入该矩阵，多余的矩阵元素则写入空格字符，再按列读出此矩阵，即为密文。程序界面参见图1。(2004.9)图1Option ExplicitPrivate Sub Command1_Click()Dim msg As Stringmsg = Text1Text2 = encode(msg)返回字符串，显示在Text2中End SubPrivate Function sr(n As Integer) As IntegerDim k As Integerk = nk为矩阵大小kn为字符串长度DoIf (18) Thenk是否能整数开方 sqr(k)=Int(sqr(k) sr = Sqr(k): Exit DoElse(19)k加1 kk+1End IfLoopEnd FunctionPrivate Function encode(ore As String) As StringDim n1 As Integer, n2 As Integer, m() As String * 1Dim i As Integer, j As Integer, k As Integern1 = Len(ore)n2 = sr(n1)n2为矩阵的大小k = 1(20)重新定义数组的大小ReDim m(n2,n2)For i = 1 To n2For j = 1 To n2If k 1 And Len(p) Len(sta) Thensta = _ sta是上一次的最长的字符串。pEnd If_下一组的开头一个字符。p = Mid(st, i + 1, 1)End IfNext iIf Len(p) 1 And Len(p) Len(sta)Then_p为最长的字符串max_st pElsemax_st = sta否则上一组字符串为最长End IfEnd Function答案：p & Mid(st, i + 1, 1) pp = Mid(st, i + 1, 1)max_st = p8下面程序的功能是找出由两个不同的数字组成的回文平方数。程序界面参见图 2 。图2Option ExplicitPrivate Sub Command1_Click()Dim a(0 To 9) As Integer, i As Long, flg As BooleanDim L As Long, j As Integer, sum As IntegerFor i = 10 To 1000L = I * I_清除数组a的内容，Erase aCall _调用函数sun1,参照形参的内容，L，a，flg。 sub1(L), a, flg)If flg Thenflg是实参返回的值For j = 0 To 9sum = sum + a(j)统计数组中有几个1Next jIf sum = 2 Then如果有2个1，就满足条件，否则不满足条件List1.AddItem CStr(i) & 2= & S