10 2021 档案

仿射加密
摘要：str1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' str2 = str1.lower() a = input("请输入原文") b = int(input("请输入密钥一")) c = int(input("请输入密钥二")) for i in a : if i in str2 阅读全文

posted @ 2021-10-31 10:06 碍你 阅读(30) 评论(0) 推荐(0)

凯撒的解密和加密
摘要：str1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' str2 = str1.lower() x = int(input("请选择加密还是解密")) a = input("请输入密文或是明文") if x == 0: for i in a : if i in str2: b = s 阅读全文

posted @ 2021-10-30 21:35 碍你 阅读(62) 评论(0) 推荐(0)

摩斯解密
摘要：a = input("input the string:") s = a.split(" ") dict = {'.-': 'A', '-...': 'B', '-.-.': 'C', '-..':'D', '.':'E', '..-.':'F', '--.': 'G', '....': 'H', 阅读全文

posted @ 2021-10-30 17:12 碍你 阅读(69) 评论(0) 推荐(0)

凯撒的加密和解密
摘要：str1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' str2 = str1.lower() a = input() for i in a : if i in str2: b = str2.find(i) c = (b + 3) % 26 print(str2[c],end='') 阅读全文

posted @ 2021-10-30 17:01 碍你 阅读(278) 评论(0) 推荐(0)

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