怎样用常规密码学解加密python脚本
本篇文章为大家展示了怎样用常规密码学解加密python脚本，内容简明扼要并且容易理解，绝对能使你眼前一亮，通过这篇文章的详细介绍希望你能有所收获。
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一、凯撒加解密（python3）
def change(c,i):
c = c.lower()
num = ord(c)
if num >= 97 and num <= 122:
num = 97 + ((num - 97) + i) % 26
return chr(num)
def kaisa_jiami(string,i):
string_new = ''
for s in string:
string_new += change(s,i)
print(string_new)
return string_new
def kaisa_jiemi(string):
for i in range(25):
print('\n', i, '\n')
i += 1
kaisa_jiami(string,i)
def main():
print('请选择需要的操作：')
print('1：凯撒加密')
print('2：凯撒解密')
choice = input()
if choice == '1':
string = input('请输入需要加密的字符串：')
num = int(input('请输入需要偏移的位数：'))
kaisa_jiami(string,num)
elif choice == '2':
string = input('请输入需要解密的字符串：')
kaisa_jiemi(string)
else:
print('输入错误，请重试！')
main()
if __name__ == '__main__':
main()二、base64和base32解密（python2）
import base64
readfile = open('base.txt','r')
writefile = open('flag.txt','w')
txt = readfile.readlines()[0]
while True:
try:
txt = base64.b32decode(txt)
except:
txt = base64.b64decode(txt)
finally:
print(txt)
writefile.write(txt)
writefile.write('\n')
writefile.close()
readfile.close()三、RC-4加解密（python3）
def get_text():
text=input('输入明文：')
return text
def get_key():
key=''
active=True
while active:
key=input('输入密钥：')
if key == '':
print('密钥不能为空，请重新输入！')
else:
active=False
return key
def init(key):
s=list(range(256))
j=0
for i in range(256):
j=(j+s[i]+ord(key[i%len(key)]))%256
s[i],s[j] =s[j],s[i]
print('s初始置换数组为：')
print(s)
return s
def trans_stream(message,s_box):
result=[]
i=j=0
for s in message:
i=(i+1)%256
j=(j+s_box[i])%256
s_box[i],s_box[j]=s_box[j],s_box[i]
t=(s_box[i]+s_box[j])%256
k=s_box[t]
result.append(chr(ord(s)^k))
print('密文为：')
print(''.join(result))
def start():
print('欢迎使用RC-4加密！')text=get_text()
key=get_key()
print('明文为：'+text)
print('密钥为：'+key)
s=init(key)
trans_stream(text,s)
temp=input('回车结束程序。')
start()四、差分曼彻斯特解密（python2）
msg1 = 0x9a9a9a6a9aa9656699a699a566995956996a996aa6a965aa9a6aa596a699665a9aa699655a696569655a9a9a9a595a6965569a59665566955a6965a9596a99aa9a9566a699aa9a969969669aa6969a9559596669
s = bin(msg1)[2:]
print s
r = ""
tmp = 0
for i in xrange(len(s) / 2):
c = s[i * 2]
if c == s[i * 2 - 1]:
r += '1'
else:
r += '0'
print hex(int(r, 2))[2:-1].decode('hex')五、莫斯电码解密（python3）
from __future__ import print_function
while 1:
a = input("input the string:")
s = a.split(" ")
dict = {'01': 'A',
'1000': 'B',
'1010': 'C',
'100':'D',
'0':'E',
'0010':'F',
'110': 'G',
'0000': 'H',
'00': 'I',
'0111':'J',
'101': 'K',
'0100': 'L',
'11': 'M',
'10': 'N',
'111': 'O',
'0110': 'P',
'1101': 'Q',
'010': 'R',
'000': 'S',
'1': 'T',
'001': 'U',
'0001': 'V',
'011': 'W',
'1001': 'X',
'1011': 'Y',
'1100': 'Z',
'01111': '1',
'00111': '2',
'00011': '3',
'00001': '4',
'00000': '5',
'10000': '6',
'11000': '7',
'11100': '8',
'11110': '9',
'11111': '0',
'001100': '?',
'10010': '/',
'101101': '()',
'100001': '-',
'010101': '.',
'110011':',',
'011010':'@',
'111000':':',
'101010':':',
'10001':'=',
'011110':"'",
'101011':'!',
'001101':'_',
'010010':'"',
'10110':'(',
'1111011':'{',
'1111101':'}'
};
for item in s:
print (dict[item],end='')
print("\n")六、栅栏密码解密（python3）
def zhalan(e):
elen = len(e)
field = []
for i in range(2, elen):
if (elen % i == 0):
field.append(i)
for f in field:
b = elen // f
result = {x: '' for x in range(b)}
for i in range(elen):
a = i % b;
result.update({a: result[a] + e[i]})
d = ''
for i in range(b):
d = d + result[i]
print(d)
d.lower()
if __name__ == '__main__':
e = 'qddpqwnp-cplen%prqwn_{_zz*d@gq}'
zhalan(e)七、转轮机密码解密（python3）
# 五、栅栏密码解密
```c
def zhalan(e):
elen = len(e)
field = []
for i in range(2, elen):
if (elen % i == 0):
field.append(i)
for f in field:
b = elen // f
result = {x: '' for x in range(b)}
for i in range(elen):
a = i % b;
result.update({a: result[a] + e[i]})
d = ''
for i in range(b):
d = d + result[i]
print(d)
d.lower()
if __name__ == '__main__':
e = 'qddpqwnp-cplen%prqwn_{_zz*d@gq}'
zhalan(e)九、递推算法（python3）
n=['33DB76A7C594BFC3','CD36C2E32A371480',
'8CEE9FF3933365BC','57373FE3C783A78F',
'59B322834BB73B59','423719DD973C6AD3',
'C858FBEABF480DA3','3CC8C789BA7B8135']
#func1
a=1;b=2
f1={}
for i in range(3,100):
t=a+b
f1[hex(t)[-16:]]=str(str(i))
a,b=b,t
#func2
a=1;b=2;c=3
f2={}
for i in range(4,100):
t=a+b+c
f2[hex(t)[-16:]]=str(str(i))
a,b,c=b,c,t
#func3
a=1;b=2;c=3;d=4
f3={}
for i in range(5,100):
t=a+b+c+d
f3[hex(t)[-16:]]=str(str(i))
a,b,c,d=b,c,d,t
#func-4
a=1;b=2;c=3;d=4;e=5
f4={}
for i in range(6,100):
t=a+b+c+d+e
f4[hex(t)[-16:]]=str(str(i))
a,b,c,d,e=b,c,d,e,t
flag='flag{'
for i in range(0,8):
n[i]=n[i].lower()
if n[i] in f1.keys():
flag=flag+f1[n[i]]+"_"
elif n[i] in f2.keys():
flag=flag+f2[n[i]]+"_"
elif n[i] in f3.keys():
flag=flag+f3[n[i]]+"_"
elif n[i] in f4.keys():
flag=flag+f4[n[i]]+"_"
flag=flag[:-1]+'}'
print(flag)十、二进制幂数（python2）
#! /usr/bin/env python
#coding=utf-8
a="8842101220480224404014224202480122"
a=a.split("0")
flag=''
for i in range(0,len(a)):
str = a[i]
list=[]
sum=0
for j in str:
list.append(j)
length = len(list)
for k in range(0,length):
sum+=int(list[k])
flag+=chr(sum+64)
print flag上述内容就是怎样用常规密码学解加密python脚本，你们学到知识或技能了吗？如果还想学到更多技能或者丰富自己的知识储备，欢迎关注创新互联行业资讯频道。
本文标题：怎样用常规密码学解加密python脚本
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