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昌吉市住房和城乡建设局网站沈阳网站排名优化

news 2025/9/17 18:30:33

昌吉市住房和城乡建设局网站,沈阳网站排名优化,佛山专业建站公司,校园网络设计方案ensp✨费纳姆密码：加解密都需要密钥，以二进制形式表示的密码。（密钥多是一次性的，称位一次性密码本） 加密过程： char_num {A: 1000001, B: 1000010, C: 1000011, D: 1000100,E: 1000101, F: 1000110, G: 100…

✨费纳姆密码：加解密都需要密钥，以二进制形式表示的密码。（密钥多是一次性的，称位一次性密码本）

加密过程：

char_num = {'A': '1000001', 'B': '1000010', 'C': '1000011', 'D': '1000100','E': '1000101', 'F': '1000110', 'G': '1000111', 'H': '1001000','I': '1001001', 'J': '1001010', 'K': '1001011', 'L': '1001100','M': '1001101', 'N': '1001110', 'O': '1001111', 'P': '1010000','Q': '1010001', 'R': '1010010', 'S': '1010011', 'T': '1010100','U': '1010101', 'V': '1010110', 'W': '1010111', 'X': '1011000','Y': '1011001', 'Z': '1011010'}num_char = {'1000001': 'A',  '1000010': 'B', '1000011': 'C', '1000100': 'D','1000101': 'E', '1000110': 'F', '1000111': 'G', '1001000': 'H','1001001': 'I', '1001010': 'J', '1001011': 'K', '1001100': 'L','1001101': 'M', '1001110': 'N', '1001111': 'O', '1010000': 'P','1010001': 'Q', '1010010': 'R', '1010011': 'S', '1010100': 'T','1010101': 'U', '1010110': 'V', '1010111': 'W', '1011000': 'X','1011001': 'Y', '1011010': 'Z', }# 转换
def txt_num(txt):txt = txt.upper()result = ""for i in txt:result = result + char_num[i]return  result# 最终数字转换为字母
def num_list(txt):num = 0s = []while True:s.append(txt[num:num + 7])num += 7if num > len(txt) - 7:breakreturn s# 加密 文本内容
def encipher(txt, k):result = ''if txt.isalpha() and k.isalpha():txt = txt_num(txt)k = txt_num(k)j = len(k)for i in range(0, len(txt)):if txt[i] == k[i % j]:result += '0'else:result += '1'else:result = "明文与密钥应为纯字母"return result# 解密 文本为数字格式
def decipher(txt, k):result = ''if k.isalpha():k = txt_num(k)j = len(k)for i in range(0, len(txt)):if txt[i] == k[i % j]:result += '0'else:result += '1'result = num_list(result)result_char = ""for i in result:try:result_char = result_char + num_char[i]except KeyError:result_char = result_char + '?'result = result_charelse:result = "密钥应为纯字母"return  resultif __name__ == '__main__':# 加解密,密文text,明文text_2text = ''# 读文本文件key = "polar"text_2 = "polarCTF"print(encipher(text_2, key))# print(decipher(text, key))

解密过程：

char_num = {'A': '1000001', 'B': '1000010', 'C': '1000011', 'D': '1000100','E': '1000101', 'F': '1000110', 'G': '1000111', 'H': '1001000','I': '1001001', 'J': '1001010', 'K': '1001011', 'L': '1001100','M': '1001101', 'N': '1001110', 'O': '1001111', 'P': '1010000','Q': '1010001', 'R': '1010010', 'S': '1010011', 'T': '1010100','U': '1010101', 'V': '1010110', 'W': '1010111', 'X': '1011000','Y': '1011001', 'Z': '1011010'}num_char = {'1000001': 'A',  '1000010': 'B', '1000011': 'C', '1000100': 'D','1000101': 'E', '1000110': 'F', '1000111': 'G', '1001000': 'H','1001001': 'I', '1001010': 'J', '1001011': 'K', '1001100': 'L','1001101': 'M', '1001110': 'N', '1001111': 'O', '1010000': 'P','1010001': 'Q', '1010010': 'R', '1010011': 'S', '1010100': 'T','1010101': 'U', '1010110': 'V', '1010111': 'W', '1011000': 'X','1011001': 'Y', '1011010': 'Z', }# 转换
def txt_num(txt):txt = txt.upper()result = ""for i in txt:result = result + char_num[i]return  result# 最终数字转换为字母
def num_list(txt):num = 0s = []while True:s.append(txt[num:num + 7])num += 7if num > len(txt) - 7:breakreturn s# 加密 文本内容
def encipher(txt, k):result = ''if txt.isalpha() and k.isalpha():txt = txt_num(txt)k = txt_num(k)j = len(k)for i in range(0, len(txt)):if txt[i] == k[i % j]:result += '0'else:result += '1'else:result = "明文与密钥应为纯字母"return result# 解密 文本为数字格式
def decipher(txt, k):result = ''if k.isalpha():k = txt_num(k)j = len(k)for i in range(0, len(txt)):if txt[i] == k[i % j]:result += '0'else:result += '1'result = num_list(result)result_char = ""for i in result:try:result_char = result_char + num_char[i]except KeyError:result_char = result_char + '?'result = result_charelse:result = "密钥应为纯字母"return  resultif __name__ == '__main__':# 加解密,密文text,明文text_2text = '00000000000000000000000000000000000001001100110110001010'# 读文本文件key = "polar"text_2 = "polarCTF"# print(encipher(text_2, key))print(decipher(text, key))

✨杰弗逊转轮加密：需要密钥并用很多行乱序英文字母表示密文的密码。首先要将密文按密钥顺序进行调整，再根据密文顺序整每行字符串即可得到明文。

方法1：

除了每行第一个字母组成的，依次往后向下寻找有意义的字符串，直到找到为止！！！

方法2：

import re
PasswordList = '''
1: < ZWAXJGDLUBVIQHKYPNTCRMOSFE <
2: < KPBELNACZDTRXMJQOYHGVSFUWI <
3: < BDMAIZVRNSJUWFHTEQGYXPLOCK <
4: < RPLNDVHGFCUKTEBSXQYIZMJWAO <
5: < IHFRLABEUOTSGJVDKCPMNZQWXY <
6: < AMKGHIWPNYCJBFZDRUSLOQXVET <
7: < GWTHSPYBXIZULVKMRAFDCEONJQ <
8: < NOZUTWDCVRJLXKISEFAPMYGHBQ <
9: < XPLTDSRFHENYVUBMCQWAOIKZGJ <
10: < UDNAJFBOWTGVRSCZQKELMXYIHP <
11: < MNBVCXZQWERTPOIUYALSKDJFHG <
12: < LVNCMXZPQOWEIURYTASBKJDFHG <
13: < JZQAWSXCDERFVBGTYHNUMKILOP <
'''
ciphertext = "UOFJLBLSXJPJY"  # 存放密文
content = re.findall(r'<(.*?)<', PasswordList, re.S)  # 匹配所有的密码表字符串
key = [2, 3, 7, 5, 13, 12, 9, 1, 8, 10, 4, 11, 6]   # 密钥
list = []   # 定义空列表list，和空串ans
ans = ""
for i in range(13):index = content[key[i]-1].index(ciphertext[i])list.append((index))
for i in range(0, 26):flag = ""for j in range(13):flag += content[key[j]-1][(list[j]+i) % 26]print(flag)