內容 ：
有 1 個 bit，可以表示 0 與 1。
有 2 個 bit，可以表示 00，01，10，11。
有 n 個 bit，請產生所有 n 個 bit 所能表示的 2 進位數字。
有 2 個 bit，可以表示 00，01，10，11。
有 n 個 bit，請產生所有 n 個 bit 所能表示的 2 進位數字。
輸入說明 ：
每行一個數字 n ( 0 < n < 15 )
代表 n 個bit
代表 n 個bit
輸出說明 ：
請參考範例輸出
範例輸入 ：
1
2
範例輸出 ：
0
1
00
01
10
11
提示 ：
字串是你的好朋友
Dim aa(1000000) As String
回覆刪除Dim stri As String
Private Sub Form_Load()
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
Do Until EOF(1)
Input #1, x
aa(0) = String(x, "0"): aa(1) = String(x - 1, "0") & 1
m = 2: n = 2
For i = 1 To x - 1
stri = String(x - m, "0") & "1"
k = n - 1
For j = 0 To k
aa(n) = stri & Right(aa(j), x - Len(stri))
n = n + 1
Next j
m = m + 1
Next i
For i = 0 To 2 ^ x - 1
Print #2, aa(i)
Next i
Loop
Close #2
Close #1
End Sub
BY 阿揚
Dim StrA As String
回覆刪除Public Sub Form_Load()
Me.Hide
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
While Not EOF(1)
Input #1, N
ReDim A(N) As Integer
StrA = ""
For i = 1 To N
A(i) = 0
StrA = StrA & A(i)
Next i
Print #2, StrA
For i = 1 To 2 ^ N - 1
StrA = ""
A(1) = A(1) + 1
For k = 1 To N - 1
If A(k) > 1 Then A(k) = A(k) - 2: A(k + 1) = A(k + 1) + 1
Next k
For j = N To 1 Step -1
StrA = StrA & A(j)
Next j
Print #2, StrA
Next i
Wend
Close #2
Close #1
End
End Sub
BY 小白
Dim aa(1000000) As String
回覆刪除Dim stri As String
Public Sub Form_Load()
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
Do Until EOF(1)
Input #1, x
Call check(x, 2)
Loop
Close #2
Close #1
End Sub
Public Function check(x, num) As String
For i = 0 To num ^ x - 1
check = ""
k = i
Do Until k = 0
step = k Mod num
k = k \ num
check = step & check
Loop
If Len(check) <> x Then check = String(x - Len(check), "0") & check
Print #2, check
Next i
End Function
BY 阿揚
Dim aa(1000000) As String
回覆刪除Dim stri As String
Public Sub Form_Load()
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
Me.Hide
Do Until EOF(1)
Input #1, x
Call check(x, 2, 2)
Loop
Close #2
Close #1
End
End Sub
Public Function check(x, num, stp) As String
For i = 0 To num - 1
aa(i) = String(x - 1, "0") & i
Print #2, aa(i)
Next i
n = i
k = 10
Do Until n = num ^ x
For i = 0 To stp - 1
aa(n) = k + Val(aa(i))
aa(n) = String(x - Len(aa(n)), "0") & aa(n)
Print #2, aa(n)
n = n + 1
Next i
stp = n
k = k * 10
Loop
End Function
阿揚
兩個的function意義錯誤。
回覆刪除function是會傳回值的。
sub是一段副程式，不用傳回東西的。
這題照我的想法是用 function寫沒有錯，但是你們的程式，
是拿function來做sub的事。
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
回覆刪除ListBox1.Items.Clear()
Dim count As Integer = TextBox1.Text '把輸入的bit個數放入count
Dim final As String '存放解的變數
length = count '把bit個數存入length 相當為長度
Call com(count, final)
End Sub
Sub com(ByVal a, ByRef final)
For i As Integer = 0 To 1
If (a >= 0) Then
Call com(a - 1, final & i.ToString) '這是一個遞迴式，因為每一次都是在做同樣的事情(產生0跟1做排序)，我用變數a來控制我呼叫的次數，然後把目前串連的值傳到下一層
End If
'判斷長度是否為當初輸入的 是則為解 加入到List後把當下的解清空
If (Len(final) = length) Then
ListBox1.Items.Add(final)
final = ""
End If
Next
End Sub