西湖论剑2021WP

西湖论剑2021WP_RE复现

先复现RE的吧，之后可能会有别的

RE

ROP

32位无壳，逻辑很简单。粘上主要实现代码

printf("Input:");
scanf("%40s", (char)flag);
if (strlen(flag) != 40)
    exit(0);
for (i = 0; i < 0x28; i += 8)
{
    for (j = 0; j < 8; ++j)
    {
        v4 = ((v5[j] & flag[i + 3]) << (8 - (3 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 3]) >> ((3 - j) % 8u)) | ((v5[j] & flag[i + 2]) << (8 - (2 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 2]) >> ((2 - j) % 8u)) | ((v5[j] & flag[i + 1]) << (8 - (1 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 1]) >> ((1 - j) % 8u)) | ((v5[j] & (unsigned __int8)flag[i]) << (8 - -j % 8u)) | ((v5[j] & (unsigned int)flag[i]) >> (-j % 8u));
        Buf2[j + i] = byte_405000[(unsigned __int8)(((v5[j] & (unsigned __int8)flag[i + 7]) << (8 - (7 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 7]) >> ((7 - j) % 8u)) | ((v5[j] & flag[i + 6]) << (8 - (6 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 6]) >> ((6 - j) % 8u)) | ((v5[j] & flag[i + 5]) << (8 - (5 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 5]) >> ((5 - j) % 8u)) | ((v5[j] & flag[i + 4]) << (8 - (4 - j) % 8u)) | ((v5[j] & (unsigned int)flag[i + 4]) >> ((4 - j) % 8u)) | v4)];
    }
}
if (memcmp(&unk_405100, Buf2, 0x28u))
{
    puts("Wrong");
    exit(0);
}
puts("Congratulations");
puts("flag is DASCTF{your input}");

但脚本不好写，导致当时没出来，z3不能直接进行位运算，需要加上BitVecs，当时也没怎么看懂，还是脚本写少了，都分析出来了就是出不来。

进行整理v0和buf2

v0 = 
	  ((*(&v2 + j) & *(flag+3 + i)) << (8 - (3 - j) % 8u)) | 
	  ((*(&v2 + j) & *(flag+3 + i)) >> (3 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+2 + i)) << (8 - (2 - j) % 8u)) |
	  ((*(&v2 + j) & *(flag+2 + i)) >> (2 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+1 + i)) << (8 - (1 - j) % 8u)) |
	  ((*(&v2 + j) & *(flag+1 + i)) >> (1 - j) % 8u) |
	  ((*(&v2 + j) & *(flag + i)) << (8 - -j % 8u)) |
	  ((*(&v2 + j) & *(flag + i)) >> -j % 8u);
        
Buf2[j + i] = byte_405000[
	  ((*(&v2 + j) & *(flag+7 + i) << (8 - (7 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+7 + i) >> (7 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+6 + i)) << (8 - (6 - j) % 8u)) |
	  ((*(&v2 + j) & *(flag+6 + i)) >> (6 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+5 + i)) << (8 - (5 - j) % 8u)) |
	  ((*(&v2 + j) & *(flag+5 + i)) >> (5 - j) % 8u) |
	  ((*(&v2 + j) & *(flag+4 + i)) << (8 - (4 - j) % 8u)) |
	  ((*(&v2 + j) & *(flag+4 + i)) >> (4 - j) % 8u) |
	  v0)];  

z3上，写脚本

from z3 import *
Buf2 = [0x65, 0x55, 0x24, 0x36, 0x9D, 0x71, 0xB8, 0xC8, 0x65, 0xFB, 0x87, 0x7F, 0x9A, 0x9C, 0xB1, 0xDF, 0x65, 0x8F, 0x9D, 0x39, 0x8F, 0x11, 0xF6, 0x8E, 0x65, 0x42, 0xDA, 0xB4, 0x8C, 0x39, 0xFB, 0x99, 0x65, 0x48, 0x6A, 0xCA, 0x63, 0xE7, 0xA4, 0x79]
byte_405000 = [0x65, 0x08, 0xF7, 0x12, 0xBC, 0xC3, 0xCF, 0xB8, 0x83, 0x7B, 0x02, 0xD5, 0x34, 0xBD, 0x9F, 0x33, 0x77, 0x76, 0xD4, 0xD7, 0xEB, 0x90, 0x89, 0x5E, 0x54, 0x01, 0x7D, 0xF4, 0x11, 0xFF, 0x99, 0x49, 0xAD, 0x57, 0x46, 0x67, 0x2A, 0x9D, 0x7F, 0xD2, 0xE1, 0x21, 0x8B, 0x1D, 0x5A, 0x91, 0x38, 0x94, 0xF9, 0x0C, 0x00, 0xCA, 0xE8, 0xCB, 0x5F, 0x19, 0xF6, 0xF0, 0x3C, 0xDE, 0xDA, 0xEA, 0x9C, 0x14, 0x75, 0xA4, 0x0D, 0x25, 0x58, 0xFC, 0x44, 0x86, 0x05, 0x6B, 0x43, 0x9A, 0x6D, 0xD1, 0x63, 0x98, 0x68, 0x2D, 0x52, 0x3D, 0xDD, 0x88, 0xD6, 0xD0, 0xA2, 0xED, 0xA5, 0x3B, 0x45, 0x3E, 0xF2, 0x22, 0x06, 0xF3, 0x1A, 0xA8, 0x09, 0xDC, 0x7C, 0x4B, 0x5C, 0x1E, 0xA1, 0xB0, 0x71, 0x04, 0xE2, 0x9B, 0xB7, 0x10, 0x4E, 0x16, 0x23, 0x82, 0x56, 0xD8, 0x61, 0xB4, 0x24, 0x7E, 0x87, 0xF8, 0x0A, 0x13, 0xE3, 0xE4, 0xE6, 0x1C, 0x35, 0x2C, 0xB1, 0xEC, 0x93, 0x66, 0x03, 0xA9, 0x95, 0xBB, 0xD3, 0x51, 0x39, 0xE7, 0xC9, 0xCE, 0x29, 0x72, 0x47, 0x6C, 0x70, 0x15, 0xDF, 0xD9, 0x17, 0x74, 0x3F, 0x62, 0xCD, 0x41, 0x07, 0x73, 0x53, 0x85, 0x31, 0x8A, 0x30, 0xAA, 0xAC, 0x2E, 0xA3, 0x50, 0x7A, 0xB5, 0x8E, 0x69, 0x1F, 0x6A, 0x97, 0x55, 0x3A, 0xB2, 0x59, 0xAB, 0xE0, 0x28, 0xC0, 0xB3, 0xBE, 0xCC, 0xC6, 0x2B, 0x5B, 0x92, 0xEE, 0x60, 0x20, 0x84, 0x4D, 0x0F, 0x26, 0x4A, 0x48, 0x0B, 0x36, 0x80, 0x5D, 0x6F, 0x4C, 0xB9, 0x81, 0x96, 0x32, 0xFD, 0x40, 0x8D, 0x27, 0xC1, 0x78, 0x4F, 0x79, 0xC8, 0x0E, 0x8C, 0xE5, 0x9E, 0xAE, 0xBF, 0xEF, 0x42, 0xC5, 0xAF, 0xA0, 0xC2, 0xFA, 0xC7, 0xB6, 0xDB, 0x18, 0xC4, 0xA6, 0xFE, 0xE9, 0xF5, 0x6E, 0x64, 0x2F, 0xF1, 0x1B, 0xFB, 0xBA, 0xA7, 0x37, 0x8F]
key = [128,64,32,16,8,4,2,1]

flag = [BitVec(f"flag[{i}]", 8) for i in range(40)]

s = Solver()
for i in range(0,0x28,8):
	for j in range(8):
		v0 = ((key[j] & flag[i + 3]) << (8 - (3 - j) % 8)) | ((key[j] & flag[i + 3]) >> ((3 - j) % 8)) | ((key[j] & flag[i + 2]) << (8 - (2 - j) % 8)) | ((key[j] & flag[i + 2]) >> ((2 - j) % 8)) | ((key[j] & flag[i + 1]) << (8 - (1 - j) % 8)) | ((key[j] & flag[i + 1]) >> ((1 - j) % 8)) | ((key[j] & flag[i]) << (8 - -j % 8)) | ((key[j] & flag[i]) >> (-j % 8))
		Buf2_ = ((key[j] & flag[i + 7]) << (8 - (7 - j) % 8)) | ((key[j] & flag[i + 7]) >> ((7 - j) % 8)) | ((key[j] & flag[i + 6]) << (8 - (6 - j) % 8)) | ((key[j] & flag[i + 6]) >> ((6 - j) % 8)) | ((key[j] & flag[i + 5]) << (8 - (5 - j) % 8)) | ((key[j] & flag[i + 5]) >> ((5 - j) % 8)) | ((key[j] & flag[i + 4]) << (8 - (4 - j) % 8)) | ((key[j] & flag[i + 4]) >> ((4 - j) % 8))
		
		s.add(v0 | Buf2_ == byte_405000.index(Buf2[i+j]))	# 算出结果和数组比对
assert s.check() == sat
m = s.model()

flag1 = []
for i in flag:
	flag1.append(chr(m[i].as_long()))
print(''.join(flag1))	

TacticalArmed

程序开始时有反调试，把int 2D patch成int 3之后就可以动调。
抛出异常后会初始化四个常量，调试可以得到4011F0函数处smc执行后的字节码。

smc可以识别，但脱包不熟练

复现不出来脱包过程，之后再试试

只能在静态看看555

查看函数发现TlsCallback_0函数，进入函数，发现startAddress中存在int 2Dh，限制动调，在调试的时候不会去抛出异常，正常执行的时候抛出异常，也就是说需要抛出异常才是正常情况，可能那里藏了什么，跟踪一下

在OD中打开将其设置为int 3h。

继续无脑按F8发现

mov     esp, [ebp+ms_exc.old_esp]
mov     eax, 4
imul    ecx, eax, 0					;imul为有符号数乘法
mov     dword_405000[ecx], 7CE45630h

mov     eax, 4
shl     eax, 0						;shl为逻辑左移一位
mov     dword_405000[eax], 58334908h

mov     eax, 4
shl     eax, 1
mov     dword_405000[eax], 66398867h

mov     eax, 4
imul    ecx, eax, 3
mov     dword_405000[ecx], 0C35195B1h
mov     [ebp+ms_exc.registration.TryLevel], 0FFFFFFFFh

运行到主函数发现每次循环，里面的指令都不一样，这里应该是一个动态解码的过程，分析这个算法应该为tea算法，确定delta为0x81A5692e

反汇编可见 右移5左移4XOR 推测是TEA加密输入值后比较是否一致，写解密脚本：

#include 
#include 

void decrypt(uint32_t* v, uint32_t* k,int z) {

  uint32_t v0 = v[0], v1 = v[1], sum = 0x81A5692e*33*(z+1), i;         
  uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; 

  for (i = 0; i < 33; i++) {  
    v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
    v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
    sum -= 0x81A5692e;
  } 
  v[0] = v0;
  v[1] = v1;
}

int main()
{
	uint32_t key[] = {0x7CE45630, 0x58334908, 0x66398867, 0x0C35195B1};
	int8_t ida_chars[40] ={0xED, 0x1D, 0x2F, 0x42, 0x72, 0xE4, 0x85, 0x14, 0xD5, 0x78, 0x55, 0x03, 0xA2, 0x80, 0x6B, 0xBF, 0x45, 0x72, 0xD7, 0x97, 0xD1, 0x75, 0xAE, 0x2D, 0x63, 0xA9, 0x5F, 0x66, 0x74, 0x6D, 0x2E, 0x29, 0xC1, 0xFC, 0x95, 0x97, 0xE9, 0xC8, 0xB5, 0x0B};

	int i,j;
	for (i = 0,j=0; i < 40,j < 5; i += 8,j++) 
	{
		decrypt((uint32_t*)(ida_chars + i), key , j);
	}

  printf("%s\n",ida_chars);
}

Re4-虚假的粉丝

看程序

FileName[16] = (char)key1 / -24 + 48;
  v15 = (char)(key1 / 100) % 10 + 48;
  v16 = (char)(key1 / 10) % 10 + 48;
  v17 = key1 % 10 + 48;  //字符串转化为数字，预测是编号
  v18 = 'txt.';
  Stream = fopen(FileName, "r");

memset(Buffer, 0, sizeof(Buffer));
fseek(Stream, Offset, 0);
fread(Buffer, ElementSize, 1u, Stream)
sub_401350("%s\n", Buffer);
  if ( Buffer[0] != 'U' || Buffer[39] != 'S' ) //需要个40位的 首尾是U和S，拿everything搜一波，得到4157
  {
    sub_401350("Sorry! Wrong Key.\n");
    fclose(Stream);
    return 0;
  }

key1为4157

继续分析

fseek(Stream, key2, 0); // key2为offest，偏移量为1118
fread(Buffer, key3, 1u, Stream); // key3为size，大小为40

程序输出对其进行base64解密

UzNDcmU3X0szeSUyMCUzRCUyMEFsNE5fd0FsSzNS -> S3Cre7_K3y%20%3D%20Al4N_wAlK3R得到key:Al4N_wAlK3R

然后在5315中即可得到flag

:ii:,::iiriririi::.,.,,::ii;:::::i::::iiiiiii;irir;;i;::,::i:::::i::,::i:::iir;rii::ir:ii::::ii;i;i;i;i;iiii:
::::@B@,iiririi:@B@B@B@B::i::2@B:::B@q::i:::ii;iririi::B@Bi::B@B:,:@@U:,BBM:iirii,PB@B;::.@B5:i:i:i:iii:i::::
::,@B@BY:iiri;i:B@B@B@B@::.L:@B@.:,@B@.:,7jr,:i;iri;i:L@B@B.,G@@,.B@B@..B@F:irii:;B@B7,:.@B@B,:r:,Lv,:::,ju7,
:,F@@:@B.:iirir::..B@ ..@B@BkS@B.:.B@B.7@@@@@i:i;i;i:.@@,B@u..@BE @@@B:r@B,:ri;iiB@B@B.:@B@BG.iB@B@B@7.B@B@B:
,,@@BUB@B::;iriii:i@Bi.i@@Oi.BB@.,.@B@ @B@B@B@:ii;ii,@B@j@B@..X@B0B1v@@@B2:iirii:,B@2.E@@;B@@i:@@v @@@ @@@O7.
.B@@@B@@@r:iri;ii:7B@i:i@B.,:2@B5iuB@2.B@B80@U:irii:LB@B@B@B@..B@B@..B@@@.:i;i;ii.@B5.@B@@@@@B,B@..B@B.:uB@BU
7@B;...@B@:i:iiiiir@Br:;B@ii::B@B@B@B,,UB@B@B::iii::B@B...v@B7,BB@B,,@B@0::iiiii::B@F:,,..B@G.i@@:.@B@.@B@B@i
:i:::::::i::::::ii:ii:i:i::ii::,;7;,::i:,iL7::::::::;:::::::ri::::::i:::::i::,::::i:::i:i::::::ii:::::::LL:,,
::iii;ii::B@B@B@:iiiiiiiiiiri;ii:i:iiiiiii::::B@B@@@:iiiiiii:iii:iiiiiiiiir@B@B@Mi:iiiiri;iiiiiiiiiiiiii:i:i:

A_True_AW_14ns

挺难认的

gghdl

看函数存在VHPI，还有调用GHDI接口，搜索看不懂，但是动调可以发现就是把输入一一置换然后与常值比较

f = open("./ghdl.txt", "r").readlines() # ghdl为获取的数据
cmp = []
test = list(
    "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!\"#$ %&'()*+,-./:;<=>?@[\\]^_`{|}~0000000000000000000000000000000000000")
dic = {}
for i in range(0, len(f), 2):
    # print(f[i], f[i+1])
    inin = f[i].strip().split('      ')[1]
    cmpcmp = f[i+1].strip().split('      ')[1]
    cmp.append(cmpcmp)
    dic[inin] = test[i // 2]
assert len(cmp) == len(test)
print(dic)
for i in range(len(cmp)):
    val = cmp[i]
    try:
        print(dic[val], end='')
    except:
        print('?', end='')

WP看不懂这道也不怎么会，之后搞吧