XCTF crypto x_xor_md5
一天一道CTF题目，能多不能少
拿到文件，用winhex打开，发现是一串数字，根据题目的意思猜测是异或，发现后面的数字有一串是一样的！！
把这一串16进制的数字copy下来，与每一行进行异或得到：
num = ['68', '4d', '4d', '4d', '0c', '00', '47', '4f', '4f', '44', '00', '4a', '4f', '42', '0c', '2a',
'54', '48', '45', '00', '46', '4c', '41', '47', '00', '49', '53', '00', '4e', '4f', '54', '00',
'72', '63', '74', '66', '5b', '77', '45', '11', '4c', '7f', '44', '10', '4e', '13', '7f', '3c',
'55', '54', '7f', '57', '48', '14', '54', '7f', '49', '15', '7f', '0a', '4b', '45', '59', '20',
'5d', '2a', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '72', '6f', '69', '73']
再转化为ASCII码，发现思路没错，出现rctf字样：
接着，发现异或完的数字有许多是0x00隔开的，猜测是不是空格，空格是0x20，如果0x00是空格的话，那么就是0x00^0x20，
整体数据都异或0x20得到：
num2 = ['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c', '0a',
'74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f', '74', '20',
'52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e', '33', '5f', '36',
'75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a', '6b', '65', '79', '2a',
'7d', '0a', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '52', '4f', '49', '53']
转化成字符串形式得到：
看这情况八九不离十是正确的思路，，，，，
突然发现有些地方字符不对劲，，
*key后面是个空格，还有ut前面，猜测key被两个*包括，即*key*，所以就是0x00^0x2a,所以不对劲的地方1c^2a得到36，所以正确的列表如下:
num2 = ['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c', '0a',
'74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f', '74', '20',
'52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e', '33', '5f', '36',
'75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a', '6b', '65', '79', '2a',
'7d', '0a', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '52', '4f', '49', '53']
转化成字符串形式得到：
嗯，差不多就是这样了，，，
提交，不对，，，，，，，，，，，，，
回头看题目提示，说有MD5值，我们异或的就是MD5值吧，32位，刚刚好
编写脚本：
m = ['01','78','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53']
zz = []
for i in range(0,len(m)):
x = int(m[i],16)
zz.append(hex(x^32)[2:])
print(zz)
得到：['21', '58', '2c', '6c', '30', 'be', '12', '17', '32', '2c', 'db', '9a', 'eb', ' af', '4a', '73']
经过解码得到that
把*key*替换成that
也就是提交RCTF{We1l_d0n3_6ut_wh4t_i5_that}，Right！！！！！
最后把做题的整个脚本贴上，有兴趣就看一看，可能有些冗余！！
整体脚本：
n = ['69','35','41','01','1C','9E','75','78','5D','48','FB','F0','84','CD','66','79',
'55','30','49','4C','56','D2','73','70','12','45','A8','BA','85','C0','3E','53',
'73','1B','78','2A','4B','E9','77','26','5E','73','BF','AA','85','9C','15','6F',
'54','2C','73','1B','58','8A','66','48','5B','19','84','B0','80','CA','33','73',
'5C','52','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53',
'01','78','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53',
'01','78','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53',
'01','78','0C','4C','10','9E','32','37','12','0C','89','D5','A2','FC']
m = ['01','78','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53']
j = 0
s = ""
num = []
for i in range(0,len(n)):
x = int(n[i],16)
if j >= 16:
j = 0
y = int(m[j],16)
j += 1
num.append(hex(x^y)[2:])
s += chr(x^y)
print(num)
print(s)
num = ['68', '4d', '4d', '4d', '0c', '00', '47', '4f', '4f', '44', '00', '4a', '4f', '42', '0c', '2a',
'54', '48', '45', '00', '46', '4c', '41', '47', '00', '49', '53', '00', '4e', '4f', '54', '00',
'72', '63', '74', '66', '5b', '77', '45', '11', '4c', '7f', '44', '10', '4e', '13', '7f', '3c',
'55', '54', '7f', '57', '48', '14', '54', '7f', '49', '15', '7f', '0a', '4b', '45', '59', '20',
'5d', '2a', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '72', '6f', '69', '73']
ss = ""
num2 = []
for i in range(0,len(num)):
x = int(num[i],16)
num2.append(hex(x^32)[2:])
ss += chr(x^32)
print(num2)
print(ss)
#查看字符串，发现有个不对劲的地方，猜测key被两个*包括，即*key*，所以就是00^2a,所以不对劲的地方1c^2a得到36，所以正确的列表如下
num2 = ['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c', '0a',
'74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f', '74', '20',
'52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e', '33', '5f', '36',
'75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a', '6b', '65', '79', '2a',
'7d', '0a', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '52', '4f', '49', '53']
xx = ""
for i in range(0,len(num2)):
xx += chr(int(num2[i],16))
print(xx)
zz = []
for i in range(0,len(m)):
x = int(m[i],16)
zz.append(hex(x^32)[2:])
print(zz)