CTF-JarvisOJ-Crypto-[61dctf]bbencode

flag = open("flag", "r").read().strip()
assert len(flag) == 32
def str2num(s):
    return int(s.encode('hex'), 16)
def bbencode(n):
    a = 0
    for i in bin(n)[2:]:
        a = a << 1
        if (int(i)):
            a = a ^ n
        if a >> 256:
            a = a ^ 0x10000000000000000000000000000000000000000000000000000000000000223L
    return a
print bbencode(str2num(flag))
result:61406787709715709430385495960238216763226399960658358000016620560764164045692

由于异或本身是可逆运算，且分析算法可知每位异或的值还是一样的，所以直接用bbencode函数再异或一次，就可以逆回原码

python代码如下

def bbencode(n):
    a = 0
    for i in bin(n)[2:]:
        a = a << 1
        if (int(i)):
            a = a ^ n
        if a >> 256:
            a = a ^ 0x10000000000000000000000000000000000000000000000000000000000000223L
    return a
result = 61406787709715709430385495960238216763226399960658358000016620560764164045692
for i in range(10000):
    result = bbencode(result)
    if("666c6167" == str(hex(result))[2:10]):
        print i
        print hex(result)[2:-1].decode('hex')