4.3 运用选择结构描述问题求解过程 知识点题库

Private Function bw(x As Integer, y As String) As String

'函数bw功能：若二进制位数不足，通过在前面添"0"使得位数为4的倍数

  Dim r As Integer, i As Integer

  r = x Mod 4

  If r <> 0 Then

    For i = 1 To 4 - r

      y = "0" + y

    Next i

    x = Len(y)

  End If

  bw =

End Function

Private Sub Command1_Click()

  Dim m As String, n As Integer, i As Integer, a As Integer

  Dim s As Integer, s1 As String, j As Integer

  m = Text1.Text

  n = Len(m)

  Text2.Text = ""

  m =       '调整m的位数为4的倍数

  For i = 1 To n - 3 Step 4

    s = 0

    For j = i To i + 3

      a = Val(Mid(m, j, 1))

      s =

    Next j

    If s >= 10 Then

      s1 = Chr(Asc("A") + s - 10)

    Else

      s1 = CStr(s)      '函数CStr类似于Str，但无前导空格

    End If

    Text2.Text = Text2.Text + s1

  Next i

End Sub

①给定24小时内的租赁起始时间和结束时间：输入租赁开始的时间和结束的时间。注：时间的输入格式为hh：mm：ss，如10：23：11。

②计算租车时间time：计算租车起始时间和结束时间的时间差time。

③计算租车费用pay：若time小于等于1，则pay为0；若，则pay为time；若time大于10， 则pay为10。

Stime=input('租车开始时间：')

etime=input('租车结束时间：')

#stime.split("：")[0]切出的是租车开始时间的时

#stimespli("：")[1]切出的是租车开始时间的分

#stime.split("：")[2]切出的是租车开始时间的秒

if int(etime. split("：")[0])>=int(stime.split("：")[0])：

time=int(etime. split(":")[0])-int(stime.spilt(":")[0])

if int(etime.split("：")[1]>int (stime.split(":"))[1)：

    time=time+1

elif int (etime.split(":")[1])= int(stime.split(":")[1]) and int(etime.split(":")[2]>= int(stime.split(":")[2])：

    time= time+1

if time<=1：

    pay=0

elif

    pay=time

else：

print(pay)

Private Sub Command1_Click()

    Dim a As Integer, b As Integer, qz As Integer, sum As Integer

    Dim tx As String, c As String, tmp As String

        tx = Text1.Text: tmp=""

        n = Len(tx):sum = 0

        For i = 1 To n

            c = Mid(tx, i, 1)

            If c >= "0" And c <= "9" Then

            ElseIf c = "*" Then

                a = Val(tmp): tmp = ""

            Else

                b = Val(tmp): tmp = ""

            End If

            If  Then

            qz = 1

            Do While a > 0

                x = a Mod 2

                qz = qz * 2

                a = a \ 2

            Loop

            s = s & Str(sum)

            a = 0: b = 0: sum = 0

        End If

    Next i

    Text2.Text = s

End Sub

Const maxn= 1005

Dima(1 To 3 * maxn) As Integer

Dim ticket(1 To 3 * maxn) As Integer

Dim i，n，k，x，y，cost As Integer

Dim tail，head As Integer

Dim s As String

Dim found As Boolean

Private Sub Command1_Click()

    s= Text1.Text

    k= 1

    For i=1 To L en(s)

        c= Mid(s，i，1)

        If c="，" Then k=k+1 Else

    Next i

    '将各条出行数据输出在列表框List1中，代码略

End Sub

Private Sub Command2_ Click()

    head=1：tail= 1

    For i=1 To k\ 3

        If a(3 * i-2)=0 Then

            cost= cost+a(3 * i-1)

            ticket(3 * tail)=a(3 * i)+45

            tail= tail+1

        Else

            Do While head < tail And ticket(3 * head) < < a(3 * i)

                head= head+1

            Loop

            If found= False Then cost=cost+a(3 * i-1)

        End If

    Next i

    Label2.Caption= Str( cost)

End Sub

Function pd(x As Integer, y As Integer) As Boolean

    pd= False

    For j=x Toy

        If ticket(3 * j-1) >=a(3 * i-1) And ticket(3 * j-2)=0 Then

            pd=True

ticket(3 * j-2)=1

            Exit Function

        End If

    Next j

End Function

Dim a(1 To 50) As Integer

Private Sub Command1_Click()

    Dim i As Integer, j As Integer, zb As Integer

    Dim cha(1 To 10) As Integer      ‘记录一条线路中各个岔路口坐标

    Dim lu(1 To 10) As Integer        ‘记录路线经过的坐标

    zb = 1: j = 0:  lu(1) = 1: i = 1

    Do While i <= 9      ‘第1个和最后1个固定，中间有9个需要路过的坐标

        If zb Mod 6 <> 0 And a(zb + 1) = 1 Then     ‘向右畅通

            If zb + 6 <= 30 and a(zb + 6) = 1 Then    ‘记录岔路口

                j = j + 1

            End If

            zb = zb + 1

            i = i + 1

            lu(i) = zb

        ElseIf zb + 6 <= 30 And a(zb + 6) = 1 Then    ‘向下畅通

            zb = zb + 6

            i = i + 1

            lu(i) = zb

        Else      ‘坐标退回到上一个岔路口

            If j = 0 Then Exit Do

                Do While lu(i) <> cha(j)

                Loop

                a(cha(j) + 1) = 0    ‘标记此路口右侧不可用

                j = j - 1

            End If

        End If

    Loop

    If j = 0 Then

        Text1.Text = "没有畅通道路"

    Else

        For i = 1 To 10

            Text1.Text = Text1.Text + Str(lu(i))

        Next i

    End If

End Sub

Private Sub Form_Load()

End If ‘生成图形矩阵的数组a各个元素，并在列表框显示，代码略

End Sub

① ② ③ 

Dim a(1 To 20) As Integer     ‘依次存储5张骨牌上的数字

Dim d(1 To 5) As Integer       ‘依次存储5张骨牌，上下两行的差值

Const n = 5

Private Sub Form_Load()

'从文本框Text1读取5张骨牌上的数字，依次存入数组a

'a(1)～a(4)依次存储第一张骨牌从左到右从上到下四个数字

'a(5)到a(20)依次类推

'代码略

End Sub

Private Sub Command1_Click()

    Dim i As Integer, up As Integer, down As Integer, t As Integer, k As Integer

    Dim id As Integer, times As Integer, x As Integer, ans As Integer

    For i = 1 To n

        up = a(4 * i - 3) + a(4 * i - 2)

        down = a(4 * i - 1) + a(4 * i)

        d(i) = up - down

        t = t + d(i)

    Next i

    id = 0 : times = 0

    ans =

    If ans <> 0 Then

        k = 1

        Do While k <= n

            For x = 1 To 3

                temp =

                If temp < ans Then

                    ans = temp

                    id = k

                End If

                If temp = 0 Then Exit Do

            Next x

            k = k + 1

        Loop

    End If

    Text2.Text = CStr(id)

    Text3.Text = CStr(times)

    Text4.Text = CStr(ans)

End Sub

Function turn(ByVal k As Integer, ByVal x As Integer) As Integer

    ‘求将第k张骨牌，顺时针旋转x次后，上下两行的差值，代码略

End Function

Private Sub Command1_Click()

    Dim s As String, ans As String

    Dim ch As String, n As Integer

    Dim i As Integer

    s= Text1.Text

    ans =““: n=0

    For i=1 To Len(s)

        If ch>= "A" And ch<= "Z" Or ch>>="a" And ch<=" z" Then

            n=n+1

        ElseIf n > 0 Then

            ans=ans+Str(n) + ","

            List1.AddItem

            n=0

        End If

    Next i

    List1.AddItem"连续字母的数量依次为:"+ans

End Sub

1)a医生值班儿比C医生晚一天

2)D医生值班儿比e医生晚两天

3)B医生值班儿笔记医生早三天

4)F医生值班儿日在BC医生值班日之间，且在星期四。打印出每个医生的值班表。

某医院内科有a,b,c,d,e,f,g七位医生，他们在一星期内每天值一次班，根据排班要求打印出值班表。

# 用列表结构列出Monday~Sunday

days=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday",]

for a in range(7):

  for b in range(7):

    if a!=b:

      for c in range(7):

        if(c!=a)and(c!=b):

          for d in range(7):

            if(d!=a)and(d!=b)and(d!=c):

              for e in range(7):

                if(e!=a)and(e!=b)and(e!=c)and(e!=d):

                  for f in range(7):

                    if(f!=a)and(f!=b)and(f!=c)and(f!=d)and(f!=e):

                      for g in range(7):

                        if(g!=a)and(g!=b)and(g!=c)and(g!=d)and(g!=e)and(g!=f):

                        #用条件表达式表示出符合要求的排班

                        if(a==c+1)and(d==)and(b==g-3)and(f>b)and(f<c)and():

                        #打印输出每个医生的排班结果

                      print("a医生:",,'\n')

                  print("b医生:",days[b],'\n')

              print("c医生:",days[c],'\n')

          print("d医生:",days[d],'\n')

      print("e医生:",days[e],'\n')

    print("f医生:",days[f],'\n')

print("g医生:",days[g],'\n')

编写完成后原名保存并关闭应用软件。