作业:
编写一个脚本,输入一行加密文本和一个距离值,并使用凯撒密码输出明文。
我的问题:
正在打印 hello^3 world^2
我不太清楚为什么。我将不胜感激任何修复或理解我的代码中的错误的帮助。 (这不是为成绩分配的,即没有截止日期为了我自己的利益而做)
它希望我能够使用其各自的输入:
利普斯${svph%
4
Jhss'tl'Pzothls5
7
Zu&hk2&ux&tuz&zu&hk
2
khoor#zruog$
3
我的代码:
code = input("enter coded text: ")
distance = int(input("enter value: "))
plainText = ""
for ch in code:
ordvalue = ord(ch)
ciphervalue = ordvalue - distance
if ciphervalue < ord('a'):
ciphervalue = ord('z') - \
(distance - (ord('a')-ordvalue - 1))
plainText += chr(ciphervalue)
print(plainText)
最佳答案
首先,您应该将输入/输出与处理分离,因为您的代码将更容易测试:
def cipher(code, distance):
plainText = ""
for ch in code:
ordvalue = ord(ch)
ciphervalue = ordvalue - distance
if ciphervalue < ord('a'):
ciphervalue = ord('z') - (distance - (ord('a')-ordvalue - 1))
plainText += chr(ciphervalue)
return plainText
code = input("enter coded text: ")
distance = int(input("enter value: "))
print(cipher(code, distance))
现在,您拥有:
>>> cipher("Lipps${svph%", 4)
'\x8aello²world±'
我猜您期待的是“Hello world!”之类的内容。如果删除这两行,就会得到以下结果:
if ciphervalue < ord('a'):
ciphervalue = ord('z') - (distance - (ord('a')-ordvalue - 1))
看:
def cipher2(code, distance):
plainText = ""
for ch in code:
ordvalue = ord(ch)
ciphervalue = ordvalue - distance
plainText += chr(ciphervalue)
return plainText
>>> cipher2("Lipps${svph%", 4)
'Hello world!'
>>> cipher2("Jhss'tl'Pzothls5", 7)
'Call me Ishmael.'
>>> cipher2("Zu&hk2&ux&tuz&zu&hk", 6) # was 2, but I guess it's 6
'To be, or not to be'
>>> cipher2("khoor#zruog$", 3)
'hello world!'
这是这个问题有趣的部分。但我认为您有正确的直觉:当生成的ordvalue 值不在预期范围内(即负值或过大)时会发生什么?
>>> cipher2("hello", 103)
Traceback (most recent call last):
...
ValueError: chr() arg not in range(0x110000)
函数 ord
生成一个 0 到 1114111 之间的 unicode 代码点,但我认为对于练习,您可以将范围限制为 0 - 127(ASCII 字符):
def cipher3(code, distance):
assert abs(distance) < 128
plainText = ""
for ch in code:
ordvalue = ord(ch)
ciphervalue = ordvalue - distance
if ciphervalue < 0:
ciphervalue += 128
elif ciphervalue >= 128: # don't forget distance can be negative
ciphervalue -= 128
plainText += chr(ciphervalue)
return plainText
>>> cipher3("hello", 103)
'\\x01~\\x05\\x05\\x08'
>>> cipher3('\x01~\x05\x05\x08', -103)
'hello'
请注意:
if ciphervalue < 0:
ciphervalue += 128
elif ciphervalue >= 128: # don't forget distance can be negative
ciphervalue -= 128
相当于:
ciphervalue = ciphervalue % 128
如果您只想使用可打印字符,您可以使用string
模块:
import string
# string.printable is '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~ \t\n\r\x0b\x0c'
def cipher4(code, distance):
assert abs(distance) < len(string.printable)
plainText = ""
for ch in code:
ordvalue = string.printable.index(ch) # this is clearly suboptimal
ciphervalue = (ordvalue - distance) % len(string.printable)
plainText += string.printable[ciphervalue]
return plainText
>>> cipher4("hello", 80)
'ByFFI'
>>> cipher4('ByFFI', -80)
'hello'
关于Python:解密凯撒密码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57103130/