我的程序中的加密方法未正确加密。我想我已经弄明白了为什么要使用调试模式,因为它读取单词之间的空格作为必须加密的内容。所以我试着键入一条没有空格的消息,但仍然没有正确显示出来。
我想问题是if语句和键。我试过注释行、更改语句、将if语句更改为for循环,但仍然不正确。
def main():
vig_square = create_vig_square()
message = input("Enter a multi-word message with punctuation: ")
input_key = input("Enter a single word key with no punctuation: ")
msg = message.lower()
key = input_key.lower()
coded_msg = encrypt(msg, key, vig_square)
print("The encoded message is: ",coded_msg)
print("The decoded message is: ", msg)
def encrypt(msg,key,vig_square):
coded_msg = ""
key_inc = 0
for i in range(len(msg)):
msg_char = msg[i]
if key_inc == len(key)-1:
key_inc = 0
key_char = key[key_inc]
if msg_char.isalpha() and key_char.isalpha():
row_index = get_row_index(key_char,vig_square)
col_index = get_col_index(msg_char,vig_square)
coded_msg = coded_msg+vig_square[row_index][col_index]
else:
coded_msg = coded_msg + " "
key_inc = key_inc+1
return coded_msg
def get_col_index(msg_char, vig_square):
column_index = ord(msg_char) - 97
return column_index
def get_row_index(key_char, vig_square):
row_index = ord(key_char) - 97
return row_index
def create_vig_square():
vig_square = list()
for row in range(26):
next_row = list()
chr_code = ord('a') + row
for col in range(26):
letter = chr(chr_code)
next_row.append(letter)
chr_code = chr_code + 1
if chr_code > 122:
chr_code = ord('a')
vig_square.append(next_row)
return vig_square
main()
我们举了一个例子:
Enter a multi-word message with punctuation: The eagle has landed.
Enter a single word key with no punctuation: LINKED
The encoded message is: epr oejwm ukw olvqoh.
The decoded message is: the eagle has landed.
但我的编码信息是:
epr iloyo sif plvqoh
