Module Module1
Sub MoveLeft()
Console.WriteLine("You Moved Left")
End Sub
Sub MoveRight()
Dim Number As String
Console.WriteLine("You Moved Right")
Randomize()
Number = Int(Rnd() * 10) + 1
If Number = 1 To 2 Then
Console.WriteLine(" An Enemy Appeared")
ElseIf
Console.WriteLine(" The Way Is Clear")
End If
End Sub
Sub Shoot()
Console.WriteLine("Pew")
End Sub
Sub Main()
Dim quit As Boolean = False
Dim key As String = ""
Do While quit = False
If Console.KeyAvailable Then
key = Console.ReadKey(0).Key
If key = 81 Then
quit = True
End If
Select Case key
Case 37
MoveLeft()
Case 39
MoveRight()
Case 32
Shoot()
End Select
End If
Loop
End Sub
End Module
如何获取随机数滚动并使某些值使敌人出现而其他值不会出现。
答案 0 :(得分:1)
而不是
@Api(
name = "myApi",
version = "v1",
namespace = @ApiNamespace(
ownerDomain = "awesome.me.com",
ownerName = "awesome.me.com",
packagePath = ""
)
)
public class MyEndpoint{
@ApiMethod(name = "getPeriod")
public Period getPeriod(@Named("days")long days) {
return Handler.getPeriod(days);
}
//other api methods
}
尝试
Randomize()
Number = Int(Rnd() * 10) + 1
If Number = 1 To 2 Then
Console.WriteLine(" An Enemy Appeared")
ElseIf
End If
此代码创建一个名为Dim rnd As New Random
Select Case rnd.Next(5)
Case 1
Console.WriteLine(" An Enemy Appeared")
Case Else
Console.WriteLine(" The Way Is Clear")
End Select
的新随机随机数生成器,生成rnd
和1
之间的随机数,如果随机数为5
则会出现敌人。如果随机数是其他任何东西,方法很清楚。
您只需要一系列1
并检查结果是否为5
,因为它与1
的范围相同,并检查10
的结果1}}或1
虽然计算速度在您的程序中并不重要,但我认为使用2
语句可能比原始代码更快 - 并且以后可以更容易维护
答案 1 :(得分:0)
' Initialize the random-number generator.
Randomize()
' Generate random value between 1 and 10.
Dim number As Integer = CInt(Int((10 * Rnd()) + 1))
' Show enemy if the integer number is smaller than 3 (= 1, 2)
If number < 3 Then
'your method here to have enemy appear
Else
Console.WriteLine(" The Way Is Clear")
End If