[afctf][crypto]有趣的题目

本文主要是介绍[afctf][crypto]有趣的题目，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

slove

核心是使用明文字符串自身的元素生成一个“伪随机字符串”与自身异或生成密文。

伪随机字符串的生成规则由i + space < len(plainText) - 1判断

ord(plainText[i + space])，ord(plainText[space])的space的移动由明文对应字符的ascii字符奇偶性决定。

明文长度28字节。

爆破2^28，应至少存在一个bit序列等于当前密文的每字节最后1bit组成的序列。

知道这个bit序列后就知道了space的移动情况，进而可以获得“伪随机字符串”是明文的哪些项的置换。

根据flag的固定格式afctf可以提前获得某些明文部分。经测试至少知道前2个字符即可解密

def encrypt(plainText):space = 10cipherText = ""try:for i in range(len(plainText)):if i + space < len(plainText) - 1:cipherText += chr(ord(plainText[i]) ^ ord(plainText[i + space]))else:cipherText += chr(ord(plainText[i]) ^ ord(plainText[space]))if ord(plainText[i]) % 2 == 0:space += 1else:space -= 1except:return '1'return cipherTextdef brute():for x in range(2**28):print(x)last_bit=bin(x)[2:].zfill(28)assert len(last_bit)==28tostr=''for each in last_bit:tostr+=chr(int(each,2))tmp=encrypt(tostr)tmp2=''for each in tmp:tmp2+=bin(ord(each))[-1]if tmp2=='1010000101101011111111101000':print('found',last_bit)
cipherText = "15120d1a0a0810010a031d3e31000d1d170d173b0d173b0c07060206"
tmp=''
for each in bytes.fromhex(cipherText):tmp+=bin(each)[-1]
print(tmp) #1010000101101011111111101000
brute()# found 1010011010010101111111101001

part2

# 10,9,10,9,10,11,10,9,10,9,10,11,10,11,10,11,10, 9, 8, 7, 6, 5, 4, 3, 4, 3, 4, 5     sapce
# 0 ,1,2 ,3,4 ,5 ,6 ,7, 8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27    index
# 10,10,12,12,14,16,16,16,18,18,20,22,22,24,24,26,26,26,26,26,26,26,26,26,4,3,4,5,  real index
# b'\x15\x12\r\x1a\n\x08\x10\x01\n\x03\x1d>1\x00\r\x1d\x17\r\x17;\r\x17;\x0c\x07\x06\x02\x06'
cipherText = "15120d1a0a0810010a031d3e31000d1d170d173b0d173b0c07060206"
print(bytes.fromhex(cipherText))
cipher=bytes.fromhex(cipherText)space=[10,9,10,9,10,11,10,9,10,9,10,11,10,11,10,11,10, 9, 8, 7, 6, 5, 4, 3, 4, 3, 4, 5 ]
pseudo_array=[10,10,12,12,14,16,16,16,18,18,20,22,22,24,24,26,26,26,26,26,26,26,26,26,4,3,4,5,]
origin=[0 ,1,2 ,3,4 ,5 ,6 ,7, 8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27]
for a,b in zip(pseudo_array,origin):print('%2d'%a,end=',')
print()
for a,b in zip(pseudo_array,origin):print('%2d'%b,end=',')w=['ww']*28
know='afctf'
w[:len(know)]=know
# print(w,len(w))
while 'ww' in w:for p,o in zip(pseudo_array,origin):if w[o] !='ww':w[p]=chr(ord(w[o])^cipher[o])if w[p] !='ww':w[o]=chr(ord(w[p])^cipher[o])print(''.join(w))
# afctf{cryptanalysis_is_hard} 

这篇关于[afctf][crypto]有趣的题目的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---