07秋理科VB考前辅导_3

数组定义
dim <数组名>(<下标说明>) as <数据类型>
例如：
dim a(1 to 10) as integer
dim b(10) as integer
对于第二种声明方式，若希望下标从1或0（默认）开始可以用option base {1|0} （必须写在代码模块最前面）
数组元素的使用
dim a(1 to 10) as integer
a(1) = 1 : a(2) := 2
for k = 3 to 10
a(k) = k
next k
print a(k) …此语句执行的结果是什么？
例：
private sub Form_Click() ’（教材P80程序阅读题4） dim a(5) as byte,i as byte
a(0)=1
for i =1 to 5
a(i) =a(i-1)+i : print a(i);
next i
end sub
应用举例
数组元素应用举例：（兔子问题）
查找数组最大元素的算法讲解
试题讲解：调试21、调试22、调试31
数组的排序算法（选择排序、优化选择排序、冒泡排序）
（调试25）
二维数组的定义
dim a(1 to 3, 1 to 3) as integer
二维数组的遍历
for k = 1 to 3
for j = 1 to 3
a(k,j) = k + j
next j
next k
…显示二维数组
for k = 1 to 3
for j = 1 to 3
print a(k,j);
next j
next k
试题讲解：调试28、调试29
二维数组的应用举例
例1
有一5*5二维数组，首先在每一行的前四个元素赋[10，99]之间的随机整数，然后将每一行前四个元素的平均值写入到第5个元素中，最后将此二维数组显示在窗体上
例2：
矩阵转置：
转置前的矩阵转置后的矩阵
1 2 3 4 1 1 1 1
1 2 3 4 2 2 2 2
1 2 3 4 3 3 3 3
1 2 3 4 4 4 4 4
试题讲解：调试27
动态数组
dim n as integer
n = val(inputbox(“n =”))
dim a(1 to n) as integer
…上述代码是否可实现数组元素个数的动态改变？
动态数组的定义
dim a() as integer
动态数组元素个数的重定义
n = val(inputbox(“n =”))
redim a(1 to n)
…
动态数组考题举例
private sub command1_click()
dim s as string, c as string
dim L as integer, k as integer, j as integer dim a() as string
s = inputbox(“s=”)
L = len(s)
______
for k = 1 to L
a(k) = mid(s,k,1)
next k
for k = 1 to L – 1
for j = k + 1 to L
c = a(k) : a(k)=a(j) :a(j)=c
next j
next k
for k =1 to L
print a(k);
next k
end sub
选项
A: dim a(1 to L) as string
B: redim a() as string
C: dim a(1 to L)
D: redim a(L)
试题讲解：调试24
算法补充
判断是否为回文（调试11）
随机不相等数组元素的生成
二进制到十进制转换（调试40）
程序阅读题练习
程序1：
Option Base 1
Private Sub Form_Click()
Dim a(5, 5) As Integer
Dim i As Integer, j As Integer
For i = 1 To 5
For j = 1 To 5
If j <= i Then
a(i, j) = 1
End If
Print a(i, j); " ";
Next j
print
Next i
End sub
程序2：（教材P80程序阅读题4）
Private sub Form_Click()
dim a(5,5) as byte, i as byte, j as byte for i=1 to 5
for j=1 to 5
a(i,j)=i * j
next j
next i
for i=1 to 5
print a(i,i);
next i
end sub
程序3：（教材P81程序阅读题6）
private sub Form_Click()
dim a(1 to 2, 1 to 3) as integer
dim i as integer, j as ineger
for i=1 to 2
for j =1 to 3
a(i,j) = i+j : print tab(j*5+2);a(i,j), next j
print
next i
end sub。