ctfshow刷题记录-cry方向-1
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题目来源：ctfshow 菜狗杯 crypto方向 base47
题目描述：
神必字符： E9CVT+HT5#X36RF4@LAU703+F$E-0N$@68LMXCVDRJJD5@MP#7MUZDTE?WWLG1S#L@+66H@59KTWYK8TW0RV
神必字典：
0123456789ABCDEFGHJKLMNPQRSTUVWXYZ?!@#$%^&*-+
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第一次做这种base换表的题目，在网上查了查相关wp，感觉自己对base家族还不太熟悉，于是自己先用py写了个base64的加解密的脚本，代码如下：
string1 = "E9CV^T+HT5#X36RF4@LAU703+F$E-0N$@68LMXCVDRJJD5@MP#7MUZDTE?WWLG1S#L@+^66H@59KTWYK8TW0RV"
test_string = "abc123"
dict1 = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ?!@#$%^&*-+"
def base64_encode(test_string):
dict2 = ("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/")# base64表
temp_string = str()
for i in test_string: # 字符->二进制ascii码， 放到temp_string里
temp_string += bin(ord(i))[2:].zfill(8)
temp_list = list()
if len(test_string) % 3 == 1:
temp_string += "0000"
if len(test_string) % 3 == 2:
temp_string += "00"
while temp_string!='':
temp_list.append(temp_string[0:6])
temp_string=temp_string[6:]
fina_list = list() # 以这些6位的二进制数值查找base表，存到fina_list中
for i in range(0, len(temp_list)):
fina_list.append(dict2[int(temp_list[i], 2)])
fina_string = str()
for i in fina_list: # 转化成字符串fina_string
fina_string += i
if len(test_string) % 3 == 1:
fina_string += "=="
if len(test_string) % 3 == 2:
fina_string += "="
return fina_string
'''print("asfaegqfa123:", base64_encode("asfaegqfa123"), "YXNmYWVncWZhMTIz"==base64_encode("asfaegqfa123"))
print("asfaegqfa12:", base64_encode("asfaegqfa12"), "YXNmYWVncWZhMTI="==base64_encode("asfaegqfa12"))
print("asfaegqfa1231:", base64_encode("asfaegqfa1231"), "YXNmYWVncWZhMTIzMQ=="==base64_encode("asfaegqfa1231"))'''
def base64_decode(string):
#dict2 = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ?!@#$%^&*-+"
dict2="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
list2 = list(dict2)
temp_string=str()
if string[-1] == "=" and string[-2] == "=":
string = string[0 : len(string) - 2]
if string[-1] == "=" and string[-2] != "=":
string = string[0 : len(string) - 1]
for i in string: # 字符转二进制码
temp_string+=str(bin(list2.index(i))[2:].zfill(6))
length=len(temp_string)
if length%8==2:
temp_string=temp_string[0:length-2]
if length%8==4:
temp_string=temp_string[0:length-4]
fina_list=list()
while temp_string!='': #每八位分一组 放入fina_list中
fina_list.append(temp_string[0:8])
temp_string=temp_string[8:]
fina_string=str()
for i in fina_list:
fina_string+=chr(int(i,2))
return fina_string
'''print("asfaegqfa123"==base64_decode("YXNmYWVncWZhMTIz"))
print("asfaegqfa12" == base64_decode("YXNmYWVncWZhMTI="))
print("asfaegqfa1231"==base64_decode(""))'''
`
后来问了问王师傅，基本理解了base的本质其实就是进制转换，每个字符串都对应一串256进制的数字（一个字符都是8bit，用ascii解码方式） 回过头来看题目，这个其实就是base45加密的字符串，思路应该是密文中每个字符对应字典的下标这一串数字是len(字典)进制的，先转成10进制，再转ascii码字符串即可，脚本代码如下：
import libnum
cipher = "E9CV^T+HT5#X36RF4@LAU703+F$E-0N$@68LMXCVDRJJD5@MP#7MUZDTE?WWLG1S#L@+^66H@59KTWYK8TW0RV"
key = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ?!@#$%^&*-+"
sum=0
for i in range(len(cipher)):
sum+=key.index(cipher[i])*pow(len(key),len(cipher)-1-i)
flag=libnum.n2s(sum)
print(flag)
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由这道题引发的思考，可不可以做一个base任意数字的编码，由上述原理写出了一个basexx加解密的脚本，代码如下：
import libnum
def base_xx_decode(cipher):
# key为任意映射表
# key = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ?!@#$%^&*-+"
key = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/" #base64表
sum = 0
for i in range(len(cipher)):
sum += key.index(cipher[i]) * pow(len(key), len(cipher) - 1 - i) #len(key)进制转10进制
flag = libnum.n2s(sum) #ascii值转字符串
return flag
print(base_xx_decode("YWJjMTIz")) # just an example =='abc123'
def base_xx_encode(plaintext):
key = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
beichushu = libnum.s2n(plaintext)
cipher=str()
chushu = len(key) #给chushu yushu shang beichushu赋初值 然后辗转相除法进行10进制转len(key)进制，并直接映射到cipher里
yushu = beichushu % chushu
shang = beichushu // chushu
beichushu = shang
cipher += key[yushu]
while shang != 0:
yushu = beichushu % chushu
shang = beichushu // chushu
beichushu = shang
cipher += key[yushu]
cipher=cipher[::-1]
return cipher
print(base_xx_encode("abc123")) # just an example =='YWJjMTIz'
这个脚本只能加解密无填充规则且ascii编码的那种，比如base64就是无“=”