根据凯撒密码\Playfair密码的原理写代码，对输入的符号串能够根据设置的密钥分别正确实现加密和解密功能

非常感谢您的提问，我可以为您提供以下代码实现凯撒密码和Playfair密码的加密和解密功能：

凯撒密码：

def caesar_cipher(plain_text, key):
    cipher_text = ""
    for ch in plain_text:
        if ch.isalpha():
            new_ch = chr((ord(ch) - ord('a') + key) % 26 + ord('a'))
            cipher_text += new_ch
        else:
            cipher_text += ch
    return cipher_text

def caesar_decipher(cipher_text, key):
    return caesar_cipher(cipher_text, 26 - key)

Playfair密码：

def gen_playfair_table(key):
    key = key.lower().replace(" ", "")
    key_set = set(key)
    alphabet = "".join([chr(ord('a') + i) for i in range(26)])
    table_list = []
    for ch in key:
        if ch in alphabet:
            alphabet = alphabet.replace(ch, "")
            table_list.append(ch)
    for ch in alphabet:
        table_list.append(ch)
    table_str = "".join(table_list)
    table = []
    for i in range(5):
        row = table_str[i * 5 : (i + 1) * 5]
        table.append(row)
    return table

def playfair_cipher(plain_text, key):
    table = gen_playfair_table(key)
    cipher_text = ""
    plain_text = plain_text.lower().replace(" ", "")
    if len(plain_text) % 2 == 1:
        plain_text += "x"
    for i in range(0, len(plain_text), 2):
        ch1, ch2 = plain_text[i], plain_text[i + 1]
        row1, col1 = 0, 0
        row2, col2 = 0, 0
        for j in range(5):
            if ch1 in table[j]:
                row1, col1 = j, table[j].find(ch1)
            if ch2 in table[j]:
                row2, col2 = j, table[j].find(ch2)
        if row1 == row2:
            cipher_ch1 = table[row1][(col1 + 1) % 5]
            cipher_ch2 = table[row2][(col2 + 1) % 5]
        elif col1 == col2:
            cipher_ch1 = table[(row1 + 1) % 5][col1]
            cipher_ch2 = table[(row2 + 1) % 5][col2]
        else:
            cipher_ch1 = table[row1][col2]
            cipher_ch2 = table[row2][col1]
        cipher_text += cipher_ch1 + cipher_ch2
    return cipher_text

def playfair_decipher(cipher_text, key):
    table = gen_playfair_table(key)
    plain_text = ""
    cipher_text = cipher_text.lower().replace(" ", "")
    for i in range(0, len(cipher_text), 2):
        ch1, ch2 = cipher_text[i], cipher_text[i + 1]
        row1, col1 = 0, 0
        row2, col2 = 0, 0
        for j in range(5):
            if ch1 in table[j]:
                row1, col1 = j, table[j].find(ch1)
            if ch2 in table[j]:
                row2, col2 = j, table[j].find(ch2)
        if row1 == row2:
            plain_ch1 = table[row1][(col1 - 1) % 5]
            plain_ch2 = table[row2][(col2 - 1) % 5]
        elif col1 == col2:
            plain_ch1 = table[(row1 - 1) % 5][col1]
            plain_ch2 = table[(row2 - 1) % 5][col2]
        else:
            plain_ch1 = table[row1][col2]
            plain_ch2 = table[row2][col1]
        plain_text += plain_ch1 + plain_ch2
    return plain_text

请注意，这里的加密和解密函数使用了称之为“求模算术”的技巧，可以防止加密结果大于字母表中最后一个字母的 ASCII 码或小于字母表中第一个字母的 ASCII 码。同时，playfair_cipher 和 playfair_decipher 函数还处理了明文长度为奇数的情况，添加了一个"X"来填充。