python简单的字符串凯撒加密
本文发布于1423天前,最后更新于 1423 天前,其中的信息可能已经有所发展或是发生改变。
def StrEncrypt(
# 待加密字符串
ordstr,
# 偏移量
npos):
    productstr = ""
    strlen = len(ordstr)
    for i in range(strlen):
        asc = ord(ordstr[i])
        # 如果在字母表范围内就进行偏移
        if(asc >= ord('a') and asc <= ord('z')):
            productstr += chr((ord(ordstr[i]) - ord('a') +
                                npos) % 26 + ord('a'))
        elif(asc >= ord('A') and asc <= ord('Z')):
            productstr += chr((ord(ordstr[i]) - ord('A') +
                                npos) % 26 + ord('A'))
        # 如果不在就直接连接字符串
        else:
            productstr += ordstr[i]
    return productstr


def StrDecrypt(
# 待解密字符串
ordstr,
# 偏移量
npos):
    productstr = ""
    strlen = len(ordstr)
    for i in range(strlen):
        asc = ord(ordstr[i])
        # 如果在字母表范围内就进行偏移
        if(asc >= ord('a') and asc <= ord('z')):
            productstr += chr((ord(ordstr[i]) - ord('a') -
                               npos) % 26 + ord('a'))
        elif(asc >= ord('A') and asc <= ord('Z')):
            productstr += chr((ord(ordstr[i]) - ord('A') -
                               npos) % 26 + ord('A'))
        # 如果不在就直接连接字符串
        else:
            productstr += ordstr[i]
    return productstr


def StrKeyEncrypt(
    # 待加密字符串
        ordstr,
    # 偏移量
        key):
    productstr = ""
    strlen = len(ordstr)
    keylen = len(key)
    for i in range(strlen):
        asc = ord(ordstr[i])
        # 如果在字母表范围内就进行偏移
        npos = int(key[i % keylen]) # 取key
        if(asc >= ord('a') and asc <= ord('z')):
            productstr += chr((ord(ordstr[i]) - ord('a') +
                               npos) % 26 + ord('a'))
        elif(asc >= ord('A') and asc <= ord('Z')):
            productstr += chr((ord(ordstr[i]) - ord('A') +
                               npos) % 26 + ord('A'))
        # 如果不在就直接连接字符串
        else:
            productstr += ordstr[i]
    return productstr


def StrKeyDecrypt(
    # 待解密字符串
        ordstr,
    # 偏移量
        key):
    productstr = ""
    strlen = len(ordstr)
    keylen = len(key)
    for i in range(strlen):
        asc = ord(ordstr[i])
        # 如果在字母表范围内就进行偏移
        npos = int(key[i % keylen])  # 取key
        if(asc >= ord('a') and asc <= ord('z')):
            productstr += chr((ord(ordstr[i]) - ord('a') -
                               npos) % 26 + ord('a'))
        elif(asc >= ord('A') and asc <= ord('Z')):
            productstr += chr((ord(ordstr[i]) - ord('A') -
                               npos) % 26 + ord('A'))
        # 如果不在就直接连接字符串
        else:
            productstr += ordstr[i]
    return productstr

借用老师的电脑发布的博文

暂无评论

发送评论 编辑评论


				
|´・ω・)ノ
ヾ(≧∇≦*)ゝ
(☆ω☆)
(╯‵□′)╯︵┴─┴
 ̄﹃ ̄
(/ω\)
∠( ᐛ 」∠)_
(๑•̀ㅁ•́ฅ)
→_→
୧(๑•̀⌄•́๑)૭
٩(ˊᗜˋ*)و
(ノ°ο°)ノ
(´இ皿இ`)
⌇●﹏●⌇
(ฅ´ω`ฅ)
(╯°A°)╯︵○○○
φ( ̄∇ ̄o)
ヾ(´・ ・`。)ノ"
( ง ᵒ̌皿ᵒ̌)ง⁼³₌₃
(ó﹏ò。)
Σ(っ °Д °;)っ
( ,,´・ω・)ノ"(´っω・`。)
╮(╯▽╰)╭
o(*////▽////*)q
>﹏<
( ๑´•ω•) "(ㆆᴗㆆ)
😂
😀
😅
😊
🙂
🙃
😌
😍
😘
😜
😝
😏
😒
🙄
😳
😡
😔
😫
😱
😭
💩
👻
🙌
🖕
👍
👫
👬
👭
🌚
🌝
🙈
💊
😶
🙏
🍦
🍉
😣
Source: github.com/k4yt3x/flowerhd
Source: https://github.com/zhaoolee/ChineseBQB
Source: https://github.com/zhaoolee/ChineseBQB
Source: https://github.com/zhaoolee/ChineseBQB
颜文字
Emoji
小恐龙
花!
滑稽大佬
演奏
程序员专属
上一篇
下一篇