加密与解密算法 知识点题库

Private Sub Command1_Click()

   Dim s As String, ch As String, mw As String

   Dim i As Integer, j As Integer

   s = Text1.Text

   i = 1  

   j = 1

   mw = “” 

   Do While i <= Len(s)

       ch =      ①       

       j = j + 1

       mw =      ②     

       i = i + j

   Loop

   Text2.Text =    ③     

End Sub

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Private Sub Command1_Click()

  Dim i As Integer, j As Integer, k As Integer Dim m As Integer, n As Integer, t As Integer Dim count As Integer

  Dim a(1 To 12) As Integer

  Dim mingw As String, miw As String

  mingw = Text1.Text

  For i = 1 To Len(mingw)

    n = Asc(Mid(mingw, i, 1))

    j = 8

    Do While

      a(j) = (n + 1) Mod 2 count = count + a(j)

      n = n \ 2

      j = j - 1 Loop

      For k = 1 To 4

        m = 8 + k - 1

        Do While m >= 2 ^ (k - 1)

          a(m + 1) = a(m)

          m = m –1

        Loop

      Next k

      For t = 1 To 3

        miw =

      Next t

      Text2.Text = miw

  Next i

End Sub

Function rev_btoh(b(i) As Integer, start As Integer, ending As Integer) As String
  Dim i, sum As Integer

  Const code = "0123456789ABCDEF"

  sum = b(ending)

  For i = start To ending - 1

    sum = sum * 2 + b(i)

  Next i

  rev_btoh = Mid(code, sum + 1, 1)

End Function

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def change(code,key):

   #change函数功能：实现要进行加密的字符code由大写字母转小写字母

    #判断转换后的字符code是否为小写字母，若是则进行加密

    m=ord(code)

return chr(m)

code=input(“请输入要加密的明文”)

key=int(input(“请输入密钥”))

code_new=“”

for i in code：

    code_new+=change(i,key)

print(code_new)

Const n = 3

Dim a(1 To n * n) As Integer

Private Sub Command1_Click()

    Dim mw(1 To 2*n*n) As Integer

    '存储密文数据，奇数位存储加密值，偶数位存储该数据在原文中的位置

    Dim yw(1 To n*n) As Integer         '存储解密数据

    Dim k As Integer

    '生成规模为n*n 的密钥矩阵存入数组a中，代码略

    For i = 1 To Len(Text1.Text)

        c = Mid(Text1．Text, i, 1)

        If c <> "," Then

            t = t * 10 + Val(c)

        Else

            mw(k) = Val(t): t = 0

        End If

    Next i

    For i = 1 To k\2      '对密文解密，结果存数组yw

        yw(mw(2 * i)) =

    Next i

    s = ""

    For i = 1 To k\2

        s = s + Chr((yw(i)) + 64)

    Next i

    Text2.Text = s

End Sub

Function getmm(x As Integer, n As Integer)

    col = (x - 1) Mod n + 1

    row = (x - 1) \ n + 1

    getmm = a((col - 1) * n + row)

    If row Mod 2 = 0 Then getmm =

End Function

def dtb(num):     #十进制转8位二进制，高低4位分别处理

    i=0;result=""

    while i<8:

        if i<4:

            y=1-num % 2

        else:

            y=num % 2

        num=num//2

        result=str(y)+result

        i+=1

    return result

mw=input("请输入待加密的字符：")

jm=""

mm={"Q":"A","W":"B","E":"C","R":"D","T":"E","Y":"F","U":"G","I":"H","O":"I","P":"J","A":"K","S":"L","D":"M","F":"N","G":"O","H":"P","J":"Q","K":"R","L":"S","Z":"T","X":"U","C":"V","V":"W","B":"X","N":"Y","M":"Z"}

for i in range(len(mw)):

    if "A"<=mw[i]<="Z":

    else:

        jm+=mw[i]

jm1=""

for i in range(len(jm)):

zf="0123456789ABCDEF";result1="";sum=0

for i in :

    sum=int(jm1[i])*8+int(jm1[i+1])*4+int(jm1[i+2])*2+int(jm1[i+3])

    result1+=zf[sum]

    sum=0

print("密文为：",result1)

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