密碼分析（cryptanalysis）是指把某個人寫的密文加以分解。這個程序通常會對密文訊息做統計分析。你的任務就是寫一個程式來對密文作簡單的分析。
輸入規範
輸入檔案的第一列有一個正整數n，代表以下有多少列需要分析的密文。接下來的n列，每列含有0或多個字元（可能包含空白字元）
輸出規範
每列包含一個大寫字元（A~Z）和一個正整數。這個正整數代表該字元在輸入檔案中出現的次數。輸入中大小寫（例如：A及a）視為相同的字元。輸出時請按照字元出現的次數由大到小排列，如果有2個以上的字元出現次數相同的話，則按照字元的大小（例如：A在H之前）由小到大排列。
請注意：如果某一字元未出現在輸入檔案中，那它也不應出現在輸出檔案中。
輸入範例（test5.txt）
3
This is a test.
Count me 1 2 3 4 5.
Wow!!!! Is this question easy?
輸出範例（result5.txt）
S 7
T 6
I 5
E 4
O 3
A 2
H 2
N 2
U 2
W 2
C 1
M 1
Q 1
Y 1
Dim word(26) As Integer
回覆刪除Private Sub Form_Load()
Me.Hide
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
Input #1, mycount
For i = 1 To mycount
Input #1, temp
strr = strr & temp
Next
For ii = 65 To 90
ori = Len(strr)
strr = Replace(strr, Chr(ii), "")
strr = Replace(strr, Chr(ii + 32), "")
word(ii - 64) = ori - Len(strr)
If word(ii - 64) <> 0 Then List1.AddItem word(ii - 64) - 100000
Next
For iii = 0 To List1.ListCount
For iiii = 1 To 26
If Val(List1.List(iii)) + 100000 = word(iiii) Then Print #2, Chr(iiii + 64) & " " & word(iiii): List1.RemoveItem iii
Next
Next
Close #2
Close #1
End
End Sub
Dim a(26), k, c() As Integer
回覆刪除Dim b() As String
Private Sub Form_Load()
Me.Hide
Open App.Path & "\in.txt" For Input As #1
Open App.Path & "\out.txt" For Output As #2
Input #1, n
List1.Clear
For w = 1 To Val(n)
Line Input #1, x
x = UCase(x)
For i = 1 To Len(x)
If Asc(Mid(x, i, 1)) >= 65 And Asc(Mid(x, i, 1)) <= 90 Then a(Asc(Mid(x, i, 1)) - 64) = a(Asc(Mid(x, i, 1)) - 64) + 1
Next i
Next w
For i = 1 To 26
If a(i) = "" Then a(i) = 0
List1.AddItem a(i) & Chr(i + 64)
Next i
k = 1
For i = List1.ListCount - 1 To 0 Step -1
If Left(List1.List(i), 1) <> 0 Then
ReDim Preserve b(k), c(k)
c(k) = Left(List1.List(i), Len(List1.List(i)) - 1): b(k) = Right(List1.List(i), 1)
k = k + 1
End If
Next i
For i = 1 To UBound(b)
For j = i + 1 To UBound(b)
If c(i) < c(j) Then g = b(i): b(i) = b(j): b(j) = g: h = c(i): c(i) = c(j): c(j) = h
If c(i) = c(j) And b(i) > b(j) Then g = b(i): b(i) = b(j): b(j) = g: h = c(i): c(i) = c(j): c(j) = h
Next j
Next i
For i = 1 To UBound(b)
Print #2, b(i) & " " & c(i)
Next i
Close #2
Close #1
End
End Sub