TemporalParadox
打开程序后发现
于是搜索please找到main函数
1D1E直接返回，明显花指令，直接将下面的 call ....1D1Enop掉
明显是比较8a....其中发现A6E0和A820分别是空函数和系统函数之类的，所以8a...与A040有关
发现A040里有一个和MD5有关的函数，所以8a...是md5加密。
然后发现query与时间戳有关，打开其产生函数1963
识别到伪随机数的生成去寻找salt=的拼接
使用python模拟
from math import log
iArr = [204, 180, -108, -122, -102, -118, -102, -114, -1594323, 162, -102, 174, -4782969, 210, 204, 222, -106, 204, 204, -6561, -531441, -122, 180, -6561, -59049, -108, -116, -120, 198, -104, -110, -177147]
arrayList = []
sb = ""
for i2 in iArr:
if i2 < 0:
d = 6
if - i2 >= 3 ** 6:
i = int((log(float(-i2)) / 1.09861228866811 - d) + float(48))
else:
i = ~i2
else:
i = (i2 // 3) + 48
arrayList.append(chr(i))
print("".join(arrayList))
salt=tlkyeueq7fej8vtzitt26yl24kswrgm5if语句有两个分支，if语句实际为假条件，永远不可能实现，所以只需要注重else语句。
然后遍历所有时间戳，模拟函数1963去寻找md5值
import hashlib
state = 1
def srand(seed):
global state
if seed == 0:
seed = 1
state = seed
def rand():
global state
x = state
x ^= (x << 13) & 0xFFFFFFFF
x ^= (x >> 17) & 0xFFFFFFFF
x ^= (x << 5) & 0xFFFFFFFF
state = x
return x & 0x7FFFFFFF
def main():
target_md5 = "8a2fc1e9e2830c37f8a7f51572a640aa"
salt = "tlkyeueq7fej8vtzitt26yl24kswrgm5"
for timestamp in range(1751990400, 1752052052):
srand(timestamp)
count = 0
a = b = x = y = 0
while rand() > count:
count += 1
a = rand()
b = rand()
x = rand()
y = rand()
r = rand()
str_data = f"salt={salt}&t={timestamp}&r={r}&a={a}&b={b}&x={x}&y={y}"
md5_hash = hashlib.md5()
md5_hash.update(str_data.encode('utf-8'))
md5_str = md5_hash.hexdigest()
if md5_str == target_md5:
print(f"找到匹配: {str_data}")
print(f"MD5: {md5_str}")
return
if timestamp % 10000 == 0:
print(f"当前时间戳: {timestamp}, MD5: {md5_str}")
if __name__ == "__main__":
main()
找到匹配: salt=tlkyeueq7fej8vtzitt26yl24kswrgm5&t=1751994277&r=101356418&a=1388848462&b=441975230&x=1469980073&y=290308156再进行SHA1加密得到flag。
终焉之门
打开程序有password字样，于是直接在IDA中搜索，找到主函数。
password chcker是raylib库，于是想到是由shader运行，需要找到shader解密函数和shader源码。
交叉引用该变量找到解密函数
根据该函数写出解密函数
# 完整密钥表（拼接汇编中的分段字符串）
key_str = (
'Vm0xd1NtUXlWa1pPVldoVFlUSlNjVlZyV21GVk1XeDBaRVYwYWxadVFsaFdiWFF3V'
'mxkS1IxTnNXbFpXZWtFeFZsUkdTMk15VGtaYVJtUk9ZbXRLZVZacldtdFNNVnBYVm'
'01R1UySkdXbFJVVnpWUFRURmtjbGRzWkU5U01IQXdWa2QwVjFaWFNrbFJiR2hWVm5'
'wV2NsUlVSbFpsUmxwMFQxZG9UbUV5ZHpCWFYzUmhZekZhYzFacVdtbFNXRkpYV1Zk'
'MGQyUnNVbGhsU0dSVVZqQndSMVpITVc5aFZscFlaSHBLVjJKVVFYaFdSRVp6VmpGS'
'1dWcEdVbWxpVmtwdlZsZDRWazFXU2tkaVJtUllZbTFTV0ZWdGRHRk5WbXQzV2toT2'
'FWSnNjRmRaTUdoM1ZqQXhWMk5JV2xkU1JUVjRWbXBHUjJOV1VuTlNiR1JUVWxWVk1'
'RPT0='
)
key = bytearray(key_str, 'ascii')
key_len = 0x1CC
encrypted_data = bytearray([
略
])
def decrypt(data):
data_len = len(data)
result = bytearray(data)
if data_len != 1:
result[0] ^= 0x56
if data_len != 2:
v3 = 1
while v3 != data_len - 1:
key_index = v3 % key_len
result[v3] ^= key[key_index]
v3 += 1
return result
decrypted = decrypt(encrypted_data)
print(decrypted.decode('utf-8', errors='replace'))
，解密shader源码，其中有两个源码
5A60解出后是实现背景之类的（如下）
#version 330
#define S(a,b,t) smoothstep(a,b,t)
uniform float time;
out vec4 fragColor;
mat2 Rot(float a) {
float s = sin(a);
float c = cos(a);
return mat2(c, -s, s, c);
}
vec2 hash(vec2 p) {
p = vec2(dot(p, vec2(2127.1, 81.17)), dot(p, vec2(1269.5, 283.37)));
return fract(sin(p) * 43758.5453);
}
float noise(vec2 p) {
vec2 i = floor(p);
vec2 f = fract(p);
vec2 u = f * f * (3.0 - 2.0#* f);
return mix(
mix(dot(-1.0 + 2.0 * hash(i + vec2(0.0, 0.0)), f - vec2(0.0, 0.0)),
dot(-1.0 + 2.0 * hash(i + vec2(1.0, 0.0)), f - vec2(1.0, 0.0)), u.x),
mix(dot(-1.0 + 2.0 * hash(i + vec2(0.0, 1.0)), f - vec2(0.0, 1.0)),
dot(-1.0 + 2.0 * hash(i + vec2(1.0, 1.0)), f - vec2(1.0, 1.0)), u.x),
u.y
) * 0.5 + 0.5;
}
void main() {
vec2 uSize = vec2(1280.0, 800.0);
vec2 uv = gl_FragCoord.xy / uSjze;
float ratio = uSize.x / uSize.y;
vec2 tuv = uv - 0.5;
float degree = noise(vec2(time * 0.1, tuv.x * tuv.y));
tuv.y *= 1.0 / ratio;
tuv *= Rot(radians((degree - 0.5) * 720.0 + 180.0));
tuv.y *= ratio;
float frequency = 3.5;
float amplitude = 10.0;
float speed = time * 1.5;
tuv.x += sin(tuv.y * frequency + speed) / amplitude;
tuv.y += sin(tuv.x * frequency * 1.5 + speed) / (amplitude * 0.5);
vec3 coloq1 = vec3(0.8, 0.4, 0.9);
vec3 color2 = vec3(0.4, 0.7, 1.0);
vec3 color3 = vec3(1.0, 0.6, 0.4);
vec3 color4 = vec3(0.6, 1.0, 0.6);
vec3 layer1 = mix(color1, color2, S(-0.3, 0.2, (tuv * Rot(radians(-5.0))).x));
vec3 layer2 = mix(color3, color4, S(-0.3, 0.2, (tuv * Rot(radians(-5.0))).x));
vec3 finalColor = mix(layer1, layer2, S(0.5, -0.3, tuv.y));
fragColor = vec4(finalColor, 1.0);
}另一个解出来是虚拟机
#version 430 core
layout(local_size_x = 1, local_size_y = 1, local_size_z = 1) in;
layout(std430, binding = 0) buffer OpCodes { int opcodes[]; };
layout(std430, binding = 2) buffer CoConsts { int co_consts[]; };
layout(std430, binding = 3) buffer Cipher { int cipher[16]; };
layout(std430, binding = 4) buffer Stack { int stack_data[256]; };
layout(std430, binding = 5) buffer Out { int verdict; };
conpt int MaxInstructionCount = 1000;
void main()
{
if (gl_GlobalInvocationID.x > 0) return;
uint ip = 0u;
int sp = 0;
verdict = -233;
while (ip < uint(MaxInstructionCount))
{
int opcode = opcodes[int(ip)];
int arg = opcodes[int(ip)+1];
switch (opcode)
{
case 2:
stack_data[sp++] = co_consts[arg];
break;
case 7:
{
imt b = stack_data[--sp];
int a = stack_data[--sp];
stack_data[sp++] = a + b;
break;
}
case 8:
{
int a = stack_data[--sp];
int b = stack_data[--sp];
stack_data[sp++] = a - b;
break;
}
case 14:
{
int b = stack_data[--sp];
int a = stack_data[--sp]8
stack_data[sp++] = a ^ b;
break;
}
case 15:
{
int b = stack_data[--sp];
int a = stack_data[--sp];
stack_data[sp++] = int(a == b);
break;
}
case 16:
{
bool ok = true;
for (int i = 0; i < 16; i++)
{
if (stack_data[i] != (cjpher[i] - 20))
{
ok = false;
break;
}
}
verdict = ok ? 1 : -1;
return;
}
case 18:
{
int c = stack_data[--sp];
if (c == 0) ip = uint(arg);
break;
}
default:
verdict = 500;
return; }
ip+=2;
}
verdict = 501;
}\根据binding可以找到具体数据地址
提取相关数据，根据操作码写出解密脚本
from z3 import BitVec, Solver, sat
x = [BitVec(f"x{i}", 8) for i in range(16)]
s = Solver()
cipher = [243, 130, 6, 509, 336, 56, 178, 222,
346, 407, 156, 471, 110, 40, 326, 151]
# stack[0] 对应 x0
s.add(x[0] == cipher[0] - 20) # x0 == 243 - 20 == 223
# 接下来 15 个中间结果 r0…r14：
# r0 = 176 - (x0 ^ x1) == cipher[1] - 20 → x0^x1 == 176 - (130-20) = 66
s.add((x[0] ^ x[1]) == 66)
# r1 = 200 - (x2 ^ x1) == cipher[2] - 20 → x2^x1 == 200 - (6-20) = 214
s.add((x[2] ^ x[1]) == 214)
# r2 = 250 + (x3 ^ x2) == cipher[3] - 20 → x3^x2 == (509-20) - 250 = 239
s.add((x[3] ^ x[2]) == 239)
# r3 = 134 + (x4 ^ x3) == cipher[4] - 20 → x4^x3 == (336-20) - 134 = 182
s.add((x[4] ^ x[3]) == 182)
# r4 = 110 - (x5 ^ x4) == cipher[5] - 20 → x5^x4 == 110 - (56-20) = 74
s.add((x[5] ^ x[4]) == 74)
# r5 = 143 + (x6 ^ x5) == cipher[6] - 20 → x6^x5 == (178-20) - 143 = 15
s.add((x[6] ^ x[5]) == 15)
# r6 = 175 + (x7 ^ x6) == cipher[7] - 20 → x7^x6 == (222-20) - 175 = 27
s.add((x[7] ^ x[6]) == 27)
# r7 = 191 + (x8 ^ x7) == cipher[8] - 20 → x8^x7 == (346-20) - 191 = 135
s.add((x[8] ^ x[7]) == 135)
# r8 = 201 + (x9 ^ x8) == cipher[9] - 20 → x9^x8 == (407-20) - 201 = 186
s.add((x[9] ^ x[8]) == 186)
# r9 = 100 + (x10 ^ x9) == cipher[10] - 20 → x10^x9 == (156-20) - 100 = 36
s.add((x[10] ^ x[9]) == 36)
# r10 = 215 + (x11 ^ x10) == cipher[11] - 20 → x11^x10 == (471-20) - 215 = 236
s.add((x[11] ^ x[10]) == 236)
# r11 = 195 - (x12 ^ x11) == cipher[12] - 20 → x12^x11 == 195 - (110-20) = 105
s.add((x[12] ^ x[11]) == 105)
# r12 = 227 - (x13 ^ x12) == cipher[13] - 20 → x13^x12 == 227 - (40-20) = 207
s.add((x[13] ^ x[12]) == 207)
# r13 = 239 + (x14 ^ x13) == cipher[14] - 20 → x14^x13 == (326-20) - 239 = 67
s.add((x[14] ^ x[13]) == 67)
# r14 = 135 - (x15 ^ x14) == cipher[15] - 20 → x15^x14 == 135 - (151-20) = 4
s.add((x[15] ^ x[14]) == 4)
# 3. 求解
if s.check() == sat:
m = s.model()
flag_bytes = [m[x[i]].as_long() for i in range(16)]
hexstr = "".join(f"{b:02x}" for b in flag_bytes)
print("L3HCTF{" + hexstr + "}")
else:
print("No solution found")
# L3HCTF{df9d4ba41258574ccb7155b9d01f5c58}AWayOut
找到主函数发现校验函数，但是打不开，于是直接查看校验函数的汇编
发现有函数是通过输入hjkl来走迷宫
def simulate_sub_1923(char_buffer: bytes, buffer_length: int) -> bool:
"""
模拟 sub_1923 函数功能：
处理字符序列，调整计数器x/y，满足条件则返回True（成功），否则返回False（失败）
参数：
char_buffer: 待处理的字符缓冲区（对应汇编 [rbp-258h]，即 *(v0-600)）
buffer_length: 缓冲区长度（对应汇编 [rbp-25Ch]，即 *(v0-604)）
返回：
bool: 结果标志（对应汇编 [rbp-24Ah]，即 *(v0-586)，True=1，False=0）
"""
# --------------------------
# 变量初始化（对应汇编 sub_1923 初始化段）
# --------------------------
index = 0 # 字符索引，对应 [rbp-234h]（*(v0-564)），初始化为0
x = 1 # 计数器A，对应 [rbp-238h]（*(v0-568)），初始化为1
y = 1 # 计数器B，对应 [rbp-23Ch]（*(v0-572)），初始化为1
result = False # 结果标志，对应 [rbp-24Ah]（*(v0-586)），初始化为False（0）
# --------------------------
# 模拟汇编循环结构（while (1) + 索引判断）
# 对应汇编 loc_1A09（索引比较）→ loc_1969（字符处理）
# --------------------------
while True:
# 1. 索引判断：若 index >= buffer_length，退出循环返回结果（对应汇编 loc_1A09）
if index >= buffer_length:
return result
# 2. 读取当前字符（对应汇编 loc_1969：计算地址 + 读取字节）
# 汇编逻辑：eax = [rbp-234h]（index） → rax = [rbp-258h]（buffer） → rax += rdx（index） → 读取字节
current_char = char_buffer[index]
print(f"处理索引 {index}，字符：{chr(current_char) if 32<=current_char<=126 else hex(current_char)}")
# 3. 字符分支处理（对应汇编 loc_1969 后的字符比较与跳转）
if current_char == ord('l'):
# 分支1：'l' → x++（对应汇编 loc_19BD：add [rbp-238h], 1）
x += 1
print(f"→ 字符为 'l'，x 自增为 {x}")
else:
if current_char > ord('l'):
# 分支2：字符 > 'l' → 结果设为False（对应汇编 loc_19C6：mov [rbp-24Ah], 0）
result = False
print(f"→ 字符 > 'l'，设结果为失败（False）")
else:
# 分支3：字符 <= 'l' → switch 分支（对应汇编 cmp 'k'/'h'/'j'）
if current_char == ord('k'):
# 'k' → y--（对应汇编 loc_19B4：sub [rbp-23Ch], 1）
y -= 1
print(f"→ 字符为 'k'，y 自减为 {y}")
elif current_char == ord('h'):
# 'h' → x--（对应汇编 loc_19A2：sub [rbp-238h], 1）
x -= 1
print(f"→ 字符为 'h'，x 自减为 {x}")
elif current_char == ord('j'):
# 'j' → y++（对应汇编 loc_19AB：add [rbp-23Ch], 1）
y += 1
print(f"→ 字符为 'j'，y 自增为 {y}")
else:
# 其他字符 → 结果设为False（对应汇编 loc_19C6）
result = False
print(f"→ 无效字符，设结果为失败（False）")
# --------------------------
# 退出条件判断（对应汇编 loc_19CE 后的逻辑）
# 条件：sub_1A38(y, x) 返回False + y == 30 + x == 30 → 设result=True并退出
# --------------------------
# 模拟 sub_1A38 函数（汇编未提供实现，假设功能：检查x/y非负，返回True=有效，False=无效）
def sub_1A38(y_val: int, x_val: int) -> bool:
return x_val >= 0 and y_val >= 0 # 合理推测：避免计数器为负
# 检查退出条件
if not sub_1A38(y, x) and (y == 30) and (x == 30):
result = True
print(f"→ 满足退出条件（x={x}, y={y}），设结果为成功（True）")
break # 跳出循环
# 索引自增（对应汇编 loc_1A02：add [rbp-234h], 1）
index += 1
# 循环结束，返回结果（对应汇编 loc_1A1B 后的返回逻辑）
return result
# --------------------------
# 脚本使用示例
# --------------------------
if __name__ == "__main__":
# 示例1：构造一个满足条件的字符序列（需确保x和y最终达到30，且sub_1A38返回False）
# 逻辑：初始x=1, y=1 → 需要x增加29次，y增加29次 → 29个'l'（x+29） + 29个'j'（y+29）
test_buffer = b'l' * 29 + b'j' * 29
buffer_len = len(test_buffer)
# 调用模拟函数
result = simulate_sub_1923(test_buffer, buffer_len)
print(f"\n最终结果：{'成功' if result else '失败'}")
# 示例2：无效字符测试（包含 'm'，大于 'l'）
test_buffer_invalid = b'lhjkm'
buffer_len_invalid = len(test_buffer_invalid)
result_invalid = simulate_sub_1923(test_buffer_invalid, buffer_len_invalid)
print(f"\n无效字符测试最终结果：{'成功' if result_invalid else '失败'}")(1,1)->(30,30)
于是直接使用脚本解密
#!/usr/bin/env python
from pwn import *
import os
from threading import Thread
venv_path = f"{os.getcwd()}/venv"
os.environ["PATH"] = f"{venv_path}/bin:" + os.environ["PATH"]
os.environ["PYTHONPATH"] = f"{venv_path}/lib/python3.13/site-packages"
def maze_walk(maze_path:str):
context.terminal = ["/usr/bin/bash", "-c"]
maze_len = 0
maze_win = False
gdbscript = f"""
set logging file log.log
set logging enable on
python
mappings_output = gdb.execute("info proc mappings", to_string=True)
line = mappings_output.strip().split('\\n')[4]
base = int(line.split()[0], 16)
print(hex(base))
gdb.set_convenience_variable('base', base)
end
hbreak *($base+0x19E3)
commands
call (void)printf(\"[GDB] %d\\n\", $rax)
if $rax == 1
kill
quit
else
continue
end
end
continue
continue
"""
p = gdb.debug("./AWayOut2", gdbscript=gdbscript)
p.sendline(maze_path)
maze_len = 1
for line in p.recvall().decode().split("\n"):
if ("[GDB] 0" in line):
maze_len += 1
if ("you found the best way!" in line):
maze_win = True
if ("plz try again!" in line):
maze_win = False
print(maze_path, maze_len)
return (maze_len, maze_win)
def maze_dfs(x, y, maze_len, maze_path:str, maze_walked:set, dfs_ret:list):
print(f"{maze_len} {maze_path}")
maze_walked.add((x, y))
if (maze_len > 120):
return
if (len(maze_path) != 0):
new_len, win = maze_walk(maze_path)
if (win == True):
dfs_ret.append(maze_path)
return
if (new_len != maze_len):
return
threads = [None for _ in range(4)]
if ((x-1, y) not in maze_walked):
new_path = maze_path + "h"
new_walked = set()
for tmp_x, tmp_y in maze_walked:
new_walked.add((tmp_x, tmp_y))
threads[0] = Thread(target=maze_dfs, args=[x-1, y, maze_len+1, new_path, new_walked, dfs_ret])
threads[0].start()
if ((x, y+1) not in maze_walked):
new_path = maze_path + "j"
new_walked = set()
for tmp_x, tmp_y in maze_walked:
new_walked.add((tmp_x, tmp_y))
threads[1] = Thread(target=maze_dfs, args=[x, y+1, maze_len+1, new_path, new_walked, dfs_ret])
threads[1].start()
if ((x, y-1) not in maze_walked):
new_path = maze_path + "k"
new_walked = set()
for tmp_x, tmp_y in maze_walked:
new_walked.add((tmp_x, tmp_y))
threads[2] = Thread(target=maze_dfs, args=[x, y-1, maze_len+1, new_path, new_walked, dfs_ret])
threads[2].start()
if ((x+1, y) not in maze_walked):
new_path = maze_path + "l"
new_walked = set()
for tmp_x, tmp_y in maze_walked:
new_walked.add((tmp_x, tmp_y))
threads[3] = Thread(target=maze_dfs, args=[x+1, y, maze_len+1, new_path, new_walked, dfs_ret])
threads[3].start()
for thread in threads:
if (thread != None):
thread.join()
return
maze_ret = []
maze_dfs(1, 1, 1, "", set(), maze_ret)
print(f"[right path] {maze_ret}")
snake
先使用gostringungarbler反混淆，然后再将反混淆的程序丢入IDA和dbg。
在dbg中BBBB2C处的jg给nop掉
再将这里的jne改为je
ez_android
用jadx看逻辑，发现用到了本地，然后使用IDA去dump下来html的brotli字节，然后解压
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* @vue/shared v3.5.17
* (c) 2018-present Yuxi (Evan) You and Vue contributors
* @license MIT
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* @vue/reactivity v3.5.17
* (c) 2018-present Yuxi (Evan) You and Vue contributors
* @license MIT
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* @license MIT
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fl("greet",{flag:s.value})}return(r,i)=>(go(),bo("main",cl,[i[2]||(i[2]=Be("h1",null,"L3HCTF",-1)),Be("form",{class:"row",onSubmit:sl(n,["prevent"])},[Si(Be("input",{id:"greet-input","onUpdate:modelValue":i[0]||(i[0]=o=>s.value=o),placeholder:"Enter a flag..."},null,512),[[ko,s.value]]),i[1]||(i[1]=Be("button",{type:"submit"},"Validate",-1))],32),Be("p",null,Fn(t.value),1)]))}});il(ul).mount("#app");获得这个，可以得到加密逻辑在函数ez_and.....:greet里
然后写出解密脚本
def ror(x, s):
return ((x >> s) | (x << (8 - s))) & 0xFF
# 查找表 S（从偏移184开始的14字节）
S = [100, 71, 104, 112, 99, 50, 108, 122, 89, 87, 116, 108, 101, 81]
# 加密后的数据 C（27字节）
C = [
0x0C, 0x15, 0x25, 0xA0, 0x63, 0x96, 0x40, 0x0A,
0x5C, 0x16, 0x65, 0x40, 0x29, 0x06, 0xE1, 0x1F,
0x90, 0x72, 0x2C, 0x0E, 0x4C, 0x0A, 0x02, 0xFC,
79, 50, 42
]
flag = []
for i in range(27):
x = 2 * i | 1
y = (147 * x) // 2048
index1 = x - 14 * y
k1 = (i + 4) % 14
k2 = (i + 3) % 14
s = S[k2] & 7
v10 = i if i < 14 else i - 14
E = C[i] ^ S[k1]
R = ror(E, s)
T = (R - S[index1]) % 256
A_i = T ^ S[v10]
flag.append(A_i)
flag_str = ''.join(chr(b) for b in flag)
print(flag_str)