暑假第三周

Tree-24 / 2023-07-18 / 原文

[湖湘杯2018]Replace

先脱壳
img
直接定位到加密函数
提取byte_40150,byte_402151和byte_4021A0数据,照着写解密脚本,直接爆破
img

exp

data = [
  0x32, 0x61, 0x34, 0x39, 0x66, 0x36, 0x39, 0x63, 0x33, 0x38, 
  0x33, 0x39, 0x35, 0x63, 0x64, 0x65, 0x39, 0x36, 0x64, 0x36, 
  0x64, 0x65, 0x39, 0x36, 0x64, 0x36, 0x66, 0x34, 0x65, 0x30, 
  0x32, 0x35, 0x34, 0x38, 0x34, 0x39, 0x35, 0x34, 0x64, 0x36, 
  0x31, 0x39, 0x35, 0x34, 0x34, 0x38, 0x64, 0x65, 0x66, 0x36, 
  0x65, 0x32, 0x64, 0x61, 0x64, 0x36, 0x37, 0x37, 0x38, 0x36, 
  0x65, 0x32, 0x31, 0x64, 0x35, 0x61, 0x64, 0x61, 0x65, 0x36, 
  0x00
]
data2 = [0x61, 0x34, 0x39, 0x66, 0x36, 0x39, 0x63, 0x33, 0x38, 
  0x33, 0x39, 0x35, 0x63, 0x64, 0x65, 0x39, 0x36, 0x64, 0x36, 
  0x64, 0x65, 0x39, 0x36, 0x64, 0x36, 0x66, 0x34, 0x65, 0x30, 
  0x32, 0x35, 0x34, 0x38, 0x34, 0x39, 0x35, 0x34, 0x64, 0x36, 
  0x31, 0x39, 0x35, 0x34, 0x34, 0x38, 0x64, 0x65, 0x66, 0x36, 
  0x65, 0x32, 0x64, 0x61, 0x64, 0x36, 0x37, 0x37, 0x38, 0x36, 
  0x65, 0x32, 0x31, 0x64, 0x35, 0x61, 0x64, 0x61, 0x65, 0x36, 
  0x00]
data3 = [
  0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01, 
  0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D, 
  0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 
  0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC, 
  0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7, 
  0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2, 
  0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 
  0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84, 
  0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB, 
  0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB, 
  0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 
  0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5, 
  0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C, 
  0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D, 
  0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 
  0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB, 
  0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3, 
  0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D, 
  0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 
  0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6, 
  0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E, 
  0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9, 
  0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 
  0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF, 
  0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99, 
  0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16
]

flag = ''
for v4 in range(35):
	for j in range(32,127):
		v5 = j
		v6 = (v5 >> 4) % 16
		v7 = (16 *v5 >> 4) % 16
		v8 = data[2 * v4]
		if(v8 < 48 | v8 > 57):
			v9 = v8-87
		else:
			v9 = v8-48
		v10 = data2[2 * v4]
		v11 = 16 * v9
		if(v10 < 48 | v10 > 57):
			v12 = v10 -87
		else:
			v12 = v10 -48
		if(data3[16 * v6 + v7] == ((v11 + v12) ^ 0x19)):
			flag += chr(j)
			break
print(flag)

flag

flag{Th1s_1s_Simple_Rep1ac3_Enc0d3}

findKey

shift+F12定位到关键字符串,交叉引用后发现没办法反汇编
img
继续交叉引用,比如先交叉引用这个loc_401A37
跳转到loc_401948,继续交叉引用这个函数
img
发现这两个jz和jnz跳转是一起的,说明不管值是否满足条件都会跳转
所以这是个绝对跳转,并且我们也发现了这个loc_40191D有问题
IDA帮我们在右上角标红了
img
原因是在绝对跳转后的指令是肯定不会执行的
在他之后有个jmp指令,jmp指令跳转的位置刚好就是loc_40191D这个函数本身
那么他就会一直循环,ida就无法识别,但由于它在绝对跳转后面,
那他就不会影响程序正常执行,那么很明显,这就是花指令
我们直接用keypatch,ctrl+alt+k快捷打开插件
img
img
然后回到最上面第一个标红的地址按p生成函数,花指令就去完了
img
F5反汇编继续跟踪,sub_40401E跟踪一下
img
通过查阅微软官方文档CryptCreateHash第二个参数为ALG_ID值
继续查阅ALG_ID值,0x8003为MD5加密
img
img
继续分析,跟踪一下sub_401005,是个循环异或
img
提取到unk_423030值为
img
那我们只需要求string1,string1可以通过

0kk`d1a`55k222k2a776jbfgd`06cjjb

与SS循环异或后解MD5得到
异或结果是c8837b23ff8aaa8a2dde915473ce0991
解MD5为123321
img

exp

a = "0kk`d1a`55k222k2a776jbfgd`06cjjb"
b = "SS"
s = ''

for i in range(len(a)):
    s += chr(ord(a[i]) ^ ord(b[i % 2]))
print(s)

enc = [0x57, 0x5E, 0x52, 0x54, 0x49, 0x5F, 0x01, 0x6D, 0x69, 0x46, 0x02, 0x6E, 0x5F, 0x02, 0x6C, 0x57, 0x5B, 0x54, 0x4C]
string1 = '123321'
flag = ''
for i in range((len(enc))):
    flag += chr(enc[i] ^ ord(string1[i % len(string1)]))
print(flag)
#c8837b23ff8aaa8a2dde915473ce0991
#flag{n0_Zu0_n0_die}

flag

flag{n0_Zu0_n0_die}