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题型:单选题 题类:模拟题 难易度:困难

浙江省杭州市2015年高考信息技术模拟卷3

下表记录了6个数据的排序过程。分析表中数据可知,该排序采用的算法与排序方式分别为

原始数据

65

57

59

44

45

69

第1遍

44

65

57

59

45

69

第2遍

44

45

55

57

59

69

第3遍

44

45

57

65

59

69

 
A、冒泡排序,升序 B、选择排序,升序 C、冒泡排序,降序 D、选择排序,降序
举一反三
小月设计了一个VB程序,用于对全班55名同学的体质调查分析,该程序能将55名同学的身高与体重录入到数据库中,之后通过如图所示的VB应用程序计算出这些同学的BMI指数,并按BMI指数从小到大进行排序,并显示这些学生的体质情况。

    注:BMI指数的计算公式为:BMI=体重(千克)除以身高(米)的平方(kg/m2)。 BMI<18.5为偏瘦(表明营养不足);BMI=18.5~24.9为体重正常;BMI=25~29.9为超重;BMI>30为肥胖。

该VB应用程序的代码如下:

Dim stuBMI(1 To 55) As Single

Dim stuName(1 To 55) As String

Private Sub Command1_Click()

    Dim k As Integer, i As Integer, j As Integer

    Dim t1 As Single, t2 As String

    Dim Results As String

    For i = 1 To 54        

        For j = i + 1 To 55

            If stuBMI(j) < stuBMI(k) then      ②    

        Next j

        If  k<>I  Then  

            t1 = stuBMI(i): stuBMI(i) = stuBMI(k): stuBMI(k) = t1

            t2 = stuName(i): stuName(i) = stuName(k): stuName(k) = t2

        End If

    Next i

List1.AddItem "姓名" + vbTab + "BMI指数" + vbTab + "结果"  

    For i = 1 To 55

        If stuBMI(i) < 18.5 Then

            Results = "偏瘦"

        ElseIf stuBMI(i) >= 18.5 And stuBMI(i) < 25 Then

            Results = "正常"

        ElseIf stuBMI(i) >= 25 And stuBMI(i) < 30 Then

            Results = "超重"

        Else

            Results = "肥胖"

        End If

        List1.AddItem stuName(i) + vbTab + str(stuBMI(i)) + vbTab + Results

    Next i

End Sub

Private Sub Form_load()

    Dim conn As New ADODB.Connection

    Dim rs As New ADODB.Recordset

    Dim strSQL As String

    Dim i As Integer

    conn.ConnectionString = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + App.Path + "\Student.accdb"

    conn.Open

    strSQL = "SELECT * FROM Student"

    Set rs.ActiveConnection = conn

    rs.Open strSQL

    i = 0

    Do While Not rs.EOF

        i = i + 1

        stuName(i) = rs.Fields("StuName")

        stuBMI(i) = Round(rs.Fields("Weight") / rs.Fields("height") ^ 2, 1)

        rs.MoveNext

    Loop

    rs.Close

    conn.Close

    Set rs = Nothing

    Set conn = Nothing

End Sub

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