明文：ziboshaokao k1=5,k0=3 n=26

首先，我们需要将明文转换成数字，使用A对应0，B对应1，以此类推，Z对应25的方式进行转换。

z i b o s h a o k a o 25 8 1 14 18 7 0 14 10 0 14

根据仿射密码的映射函数，我们可以依次将每个数字进行加密：

f(25) = (525+3) mod 26 = 0 -> A f(8) = (58+3) mod 26 = 15 -> P f(1) = (51+3) mod 26 = 8 -> I f(14) = (514+3) mod 26 = 21 -> V f(18) = (518+3) mod 26 = 1 -> B f(7) = (57+3) mod 26 = 10 -> K f(0) = (50+3) mod 26 = 3 -> D f(14) = (514+3) mod 26 = 21 -> V f(10) = (510+3) mod 26 = 13 -> N f(0) = (50+3) mod 26 = 3 -> D f(14) = (5*14+3) mod 26 = 21 -> V

因此，加密后的密文为：APIVBKDNDV

使用Python语言实现：

plain_text = "ziboshaokao" k1 = 5 k0 = 3 n = 26

将明文转换成数字

num_list = [] for c in plain_text: num = ord(c) - ord('a') num_list.append(num)

加密

cipher_text = "" for num in num_list: new_num = (k1*num + k0) % n cipher_text += chr(new_num + ord('a'))

print("明文：", plain_text) print("密文：", cipher_text