第八题 WTF

给出的字符忘了截图了，是一长串base64，解码得到

通过观察发现是二维码的数字串，一共65536个字符，正好256*256.
用processing工具转换成二维码，扫码得到flag。

15 证明自己吧

定位到关键代码，a1是程序要求输入的flag，也就是v5每个字节-5 等于 a1与上0x20的值。注意IDA里的大端和小端的问题就好了

v5的内容为

image.png

buffer = [0x68,0x57,0x19,0x48,0x50,0x6e,0x58,0x78,0x54,0x6a,0x19,0x58,0x5e,0x6]
res = []
ls = ''
for i in buffer:
    te = (i-5)^0x20
    res.append(te)
    ls += chr(te)
print(res)
print(ls)

百度看这是道原题，大佬们用动态调试直接就出来了

py逆向

给了个pyc文件
直接百度pyc 反编译在线反编译一下得到源码
写个脚本就跑出来了，注意加减和位于的优先级问题就行

def encrypt(key, seed, string):
    rst = []
    for v in string:
        rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
        seed = (seed + 1) % len(key)
    
    return rst

if __name__ == '__main__':
    print 'Welcome to python crackme'
    flag = ''
    KEY1 = 'Maybe you are good at decryptint Byte Code, have a try!'
    KEY2 = [111,52,24,28,120,50,37,62,67,52,48,6,1,122,3,22,72,1,1,14,46,27,22]
    en_out = encrypt(KEY1, 5, KEY2)
    if KEY2 == en_out:
        print 'You Win,FLAG IS flag{%s}' % flag
    else:
        print 'Try Again !'
------------------
seed = 5
key1 = "Maybe you are good at decryptint Byte Code, have a try!"
key2 = [111,52,24,28,120,50,37,62,67,52,48,6,1,122,3,22,72,1,1,14,46,27,22]
res = ''
n = len(key1)
l = len(key2)
print(n)
res = []
for i in range(l):
    tem = ((key2[i] ^ ord(key1[seed]))-seed)  % 255

    res.append(tem)
    seed = (seed + 1) % n

print(res)
ls = ""
for i in res :
    ls += chr(i)
print(ls)

REIF

源程序找不到了，先放脚本了

v3 = [3765,3301,7803,10323,5712,3845,5218,1968,4237,1696,704,7713,4746,6517,2339,4192,354,6454,6137,5015,6396,942,2825,8129,448,1760,9863,778,6617,432]

v33 = [297,169,431,229,423,257,465,288,437,86,364,123,390,485,232,352,244,469,54,359,16,110,355,244,256,94,187,19,242,432]

v63 = [34,29,76,98,43,52,97,24,40,23,4,69,44,52,43,80,1,63,79,48,55,8,26,95,4,34,82,11,51,81]

res = []
print(len(v3),len(v33),len(v63))
for i in range(len(v3)):

    tem = (v3[i] - v33[i]) / v63[i]
    res.append(tem)

ls = ''
for i in res:
    ls += chr(int(i))
print(res)
print(ls)