XXTEA python

XXTEA python 加密/解密脚本

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normal24里面终于遇到了xxtea了，新加密get，找比较好的解密脚本也花了不少时间。另外还有tea和xtea的，这个写的头疼等摆一会有空补上。

参考http://www.yaowenming.com/A/8Bz8qZ6XJx/

import struct

_DELTA = 0x9E3779B9  #固定常量

def _long2str(v, w):
    n = (len(v) - 1) << 2
    if w:
        m = v[-1]
        if (m < n - 3) or (m > n): return ''
        n = m
    s = struct.pack('<%iL' % len(v), *v)
    return s[0:n] if w else s

def _str2long(s, w):
    n = len(s)
    m = (4 - (n & 3) & 3) + n
    s = s.ljust(m, b"\0")
    v = list(struct.unpack('<%iL' % (m >> 2), s))
    if w: v.append(n)
    return v

def encrypt(str, key): #加密部分
    if str == '': return str
    v = _str2long(str, True)
    k = _str2long(key.ljust(16, b"\0"), False)
    n = len(v) - 1
    z = v[n]
    y = v[0]
    sum = 0
    q = 6 + 52 // (n + 1) #得出轮数
    while q > 0:
        sum = (sum + _DELTA) & 0xffffffff #进行叠加
        e = sum >> 2 & 3 #固定运算
        for p in  range(n):
            y = v[p + 1]
            v[p] = (v[p] + ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z))) & 0xffffffff
            z = v[p]
        y = v[0]
        v[n] = (v[n] + ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[n & 3 ^ e] ^ z))) & 0xffffffff  #xxtea加密的标识性加密步骤
        z = v[n]
        q -= 1
    return _long2str(v, False)

def decrypt(str, key): #解密部分
    if str == '': return str
    v = _str2long(str, False)
    k = _str2long(key.ljust(16, b"\0"), False)
    n = len(v) - 1
    z = v[n]
    y = v[0]
    q = 6 + 52 // (n + 1)
    sum = (q * _DELTA) & 0xffffffff
    while (sum != 0):
        e = sum >> 2 & 3
        for p in range(n, 0, -1):
            z = v[p - 1]
            v[p] = (v[p] - ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z))) & 0xffffffff
            y = v[p]
        z = v[n]
        v[0] = (v[0] - ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[0 & 3 ^ e] ^ z))) & 0xffffffff
        y = v[0]

        sum = (sum - _DELTA) & 0xffffffff

    return _long2str(v, True)

解密使用示例

flag = [188, 165, 206, 64, 244, 178, 178, 231, 169, 18, 157, 18, 174, 16, 200, 91, 61, 215, 6, 29, 220, 112, 248, 220]

x = decrypt(bytes(flag),'flag'.encode()) flag就是四位数的key

print(x)