直接爆破

# from flag import flag
# assert flag.startswith("flag{")
# assert flag.endswith("}")
# assert len(flag) == 27
def lfsr(R, mask):
    output = (R << 1) & 0xffffff
    i = R & mask & 0xffffff
    lastbit = 0
    while i != 0:
        lastbit ^= (i & 1)
        i = i >> 1
    output ^= lastbit
    return (output, lastbit)
# R = int(flag[5:-1], 2)
mask = 0x100002
def encrypt(R):
    ret = ''
    for i in range(12):
        tmp = 0
        for j in range(8):
            (R, out) = lfsr(R, mask)
            tmp = (tmp << 1) ^ out
        ret += chr(tmp)
    return ret
def brute_force():
    with open('key', 'rb') as r:
        cipher = r.read()
    for i in range(2**20, 2**21):
        print i
        if encrypt(i) == cipher:
            raw_input()

flag{110111100101001101001}

StreamGame4

1.本题给了一个加密脚本和一个加密过后的key。

2.脚本将flag做为二进制读入，做为密钥。然后循环1024*1024次，每次循环8次nlfsr函数，将结果变换写入key中。所以可以进行爆破，每次检测key的一位，从第一位开始，如果正确在检测下一位。脚本如下：

def nlfsr(R,mask):
    output = (R << 1) & 0xffffff
    i=(R&mask)&0xffffff
    lastbit=0
    changesign=True
    while i!=0:
        if changesign:
            lastbit &= (i & 1)
            changesign=False
        else:
            lastbit^=(i&1)
        i=i>>1
    output^=lastbit
    return (output,lastbit)
f=open("key","rb").read()
mask=0b110110011011001101110
flag = False
for x in range(0,pow(2,27-6)-1):
    R=x
    for i in range(1024*1024):  #此处可以把1024*1024改成一个较小的数
        tmp=0
        for j in range(8):
            (R,out)=nlfsr(R,mask)
            tmp=(tmp << 1)^out
        if f[i]!=tmp:
            break
        elif i==1024*1024-1:   #此处也要同时更改 
            flag=True
            break
    if x%100000==0:
        print('x =',x)
    if flag==True:
        flag=x
        break
print("flag{"+bin(x)[2:]+"}"