asm_re
手撕汇编，配合 GPT 分析，加上一点点的猜测。
在汇编推出可以看到加密逻辑：
value = ord(f[1])
value *= 0x50;
value += 0x14;
value ^= 0x4D;
value += 0x1E;
print(value)
已经知道第一个为 f , f 经过加密得到 0x1FD7， l 可以得到 0x21B7。同理，根据数据段可以
还原出flag，代码如下【部分非16进制数手动调整】：
import re
text = """
__const:0000000100003F10 D7 DCB 0xD7
__const:0000000100003F11 1F DCB 0x1F
__const:0000000100003F12 00 DCB 0
__const:0000000100003F13 00 DCB 0
__const:0000000100003F14 B7 DCB 0xB7
__const:0000000100003F15 21 DCB 0x21 ; !
__const:0000000100003F16 00 DCB 0
__const:0000000100003F17 00 DCB 0
__const:0000000100003F18 47 DCB 0x47 ; G
__const:0000000100003F19 1E DCB 0x1E
__const:0000000100003F1A 00 DCB 0
__const:0000000100003F1B 00 DCB 0
__const:0000000100003F1C 27 DCB 0x27 ; '
__const:0000000100003F1D 20 DCB 0x20
__const:0000000100003F1E 00 DCB 0
__const:0000000100003F1F 00 DCB 0
__const:0000000100003F20 E7 DCB 0xE7
__const:0000000100003F21 26 DCB 0x26 ; &
__const:0000000100003F22 00 DCB 0
__const:0000000100003F23 00 DCB 0
__const:0000000100003F24 D7 DCB 0xD7
__const:0000000100003F25 10 DCB 0x10
__const:0000000100003F26 00 DCB 0
__const:0000000100003F27 00 DCB 0
__const:0000000100003F28 27 DCB 0x27 ; '
__const:0000000100003F29 11 DCB 0x11
__const:0000000100003F2A 00 DCB 0
__const:0000000100003F2B 00 DCB 0
__const:0000000100003F2C 07 DCB 0x07
__const:0000000100003F2D 20 DCB 0x20
__const:0000000100003F2E 00 DCB 0
__const:0000000100003F2F 00 DCB 0
__const:0000000100003F30 C7 DCB 0xC7
__const:0000000100003F31 11 DCB 0x11
__const:0000000100003F32 00 DCB 0
__const:0000000100003F33 00 DCB 0
__const:0000000100003F34 47 DCB 0x47 ; G
__const:0000000100003F35 1E DCB 0x1E
__const:0000000100003F36 00 DCB 0
__const:0000000100003F37 00 DCB 0
__const:0000000100003F38 17 DCB 0x17
__const:0000000100003F39 10 DCB 0x10
__const:0000000100003F3A 00 DCB 0
__const:0000000100003F3B 00 DCB 0
__const:0000000100003F3C 17 DCB 0x17
__const:0000000100003F3D 10 DCB 0x10
__const:0000000100003F3E 00 DCB 0
__const:0000000100003F3F 00 DCB 0
__const:0000000100003F40 F7 DCB 0xF7
__const:0000000100003F41 11 DCB 0x11
__const:0000000100003F42 00 DCB 0
__const:0000000100003F43 00 DCB 0
__const:0000000100003F44 07 DCB 0x07
__const:0000000100003F45 20 DCB 0x20
__const:0000000100003F46 00 DCB 0
__const:0000000100003F47 00 DCB 0
__const:0000000100003F48 37 DCB 0x37 ; 7
__const:0000000100003F49 10 DCB 0x10
__const:0000000100003F4A 00 DCB 0
__const:0000000100003F4B 00 DCB 0
__const:0000000100003F4C 07 DCB 0x07
__const:0000000100003F4D 11 DCB 0x11
__const:0000000100003F4E 00 DCB 0
__const:0000000100003F4F 00 DCB 0
__const:0000000100003F50 17 DCB 0x17
__const:0000000100003F51 1F DCB 0x1F
__const:0000000100003F52 00 DCB 0
__const:0000000100003F53 00 DCB 0
__const:0000000100003F54 D7 DCB 0xD7
__const:0000000100003F55 10 DCB 0x10
__const:0000000100003F56 00 DCB 0
__const:0000000100003F57 00 DCB 0
__const:0000000100003F58 17 DCB 0x17
__const:0000000100003F59 10 DCB 0x10
__const:0000000100003F5A 00 DCB 0
__const:0000000100003F5B 00 DCB 0
__const:0000000100003F5C 17 DCB 0x17
__const:0000000100003F5D 10 DCB 0x10
__const:0000000100003F5E 00 DCB 0
__const:0000000100003F5F 00 DCB 0
__const:0000000100003F60 67 DCB 0x67 ; g
__const:0000000100003F61 1F DCB 0x1F
__const:0000000100003F62 00 DCB 0
__const:0000000100003F63 00 DCB 0
__const:0000000100003F64 17 DCB 0x17
__const:0000000100003F65 10 DCB 0x10
__const:0000000100003F66 00 DCB 0
__const:0000000100003F67 00 DCB 0
__const:0000000100003F68 C7 DCB 0xC7
__const:0000000100003F69 11 DCB 0x11
__const:0000000100003F6A 00 DCB 0
__const:0000000100003F6B 00 DCB 0
__const:0000000100003F6C C7 DCB 0xC7
__const:0000000100003F6D 11 DCB 0x11
__const:0000000100003F6E 00 DCB 0
__const:0000000100003F6F 00 DCB 0
__const:0000000100003F70 17 DCB 0x17
__const:0000000100003F71 10 DCB 0x10
__const:0000000100003F72 00 DCB 0
__const:0000000100003F73 00 DCB 0
__const:0000000100003F74 D7 DCB 0xD7
__const:0000000100003F75 1F DCB 0x1F
__const:0000000100003F76 00 DCB 0
__const:0000000100003F77 00 DCB 0
__const:0000000100003F78 17 DCB 0x17
__const:0000000100003F79 1F DCB 0x1F
__const:0000000100003F7A 00 DCB 0
__const:0000000100003F7B 00 DCB 0
__const:0000000100003F7C 07 DCB 0x07
__const:0000000100003F7D 11 DCB 0x11
__const:0000000100003F7E 00 DCB 0
__const:0000000100003F7F 00 DCB 0
__const:0000000100003F80 47 DCB 0x47 ; G
__const:0000000100003F81 0F DCB 0x0F
__const:0000000100003F82 00 DCB 0
__const:0000000100003F83 00 DCB 0
__const:0000000100003F84 27 DCB 0x27 ; '
__const:0000000100003F85 11 DCB 0x11
__const:0000000100003F86 00 DCB 0
__const:0000000100003F87 00 DCB 0
__const:0000000100003F88 37 DCB 0x37 ; 7
__const:0000000100003F89 10 DCB 0x10
__const:0000000100003F8A 00 DCB 0
__const:0000000100003F8B 00 DCB 0
__const:0000000100003F8C 47 DCB 0x47 ; G
__const:0000000100003F8D 1E DCB 0x1E
__const:0000000100003F8E 00 DCB 0
__const:0000000100003F8F 00 DCB 0
__const:0000000100003F90 37 DCB 0x37 ; 7
__const:0000000100003F91 10 DCB 0x10
__const:0000000100003F92 00 DCB 0
__const:0000000100003F93 00 DCB 0
__const:0000000100003F94 D7 DCB 0xD7
__const:0000000100003F95 1F DCB 0x1F
__const:0000000100003F96 00 DCB 0
__const:0000000100003F97 00 DCB 0
__const:0000000100003F98 07 DCB 0x07
__const:0000000100003F99 11 DCB 0x11
__const:0000000100003F9A 00 DCB 0
__const:0000000100003F9B 00 DCB 0
__const:0000000100003F9C D7 DCB 0xD7
__const:0000000100003F9D 1F DCB 0x1F
__const:0000000100003F9E 00 DCB 0
__const:0000000100003F9F 00 DCB 0
__const:0000000100003FA0 07 DCB 0x07
__const:0000000100003FA1 11 DCB 0x11
__const:0000000100003FA2 00 DCB 0
__const:0000000100003FA3 00 DCB 0
__const:0000000100003FA4 87 DCB 0x87
__const:0000000100003FA5 27 DCB 0x27 ; '
__const:0000000100003FA6 00 DCB 0
__const:0000000100003FA7 00 DCB 0
"""
pattern = re.compile(r'DCB\s+(0x[\da-fA-F]{2})')
matches = pattern.findall(text)
result = []
# 打印所有匹配到的十六进制数字
for match in matches:
result.append(match)
for i in range (0, len(result),2):
temp = result[i+1]+result[i][2:]
value = int(temp, 16)
value -= 0x1E
value ^= 0x4D
value -= 0x14;
print(chr(value // 0x50),end="")
结果：flag{67e9a228e45b622c2992fb5174a4f5f5}
androidso_re
动态调试一下拿到DES的Key和IV【下面是十进制，化为 16 进制后解密即可】
Key: 4138556457616571
IV: 576633444c757073
CBC模式解密即可JqslHrdvtgJrRs2QAp+FEVdwRPNLswrnykD/sZMivmjGRKUMVIC/rw==
gdb_debug
逻辑并不难，加密的过程：设置随机数种子 -> 异或随机数 -> 置换[随机数生成的] -> 异或随机数 -> 得到v31
最后一遍异或的在内存中可以得到：
s2 = "congratulationstoyoucongratulationstoy"
print(len(s2))
v32 = [191, 215, 46, 218, 238, 168, 26, 16, 131, 115, 172,241, 6, 190, 173, 136, 4, 215, 18, 254, 181, 226, 97,183, 61, 7, 74, 232, 150, 162, 157, 77, 188, 129, 140,233, 136, 120]
print(len(v32))
v31 = []
for i in range(0, len(s2)):
v31.append(v32[i]^ord(s2[i]))
print(v31)
接着前面还有三个加密过程 设置随机数种子 -> 异或随机数 -> 置换[随机数生成的] -> 异或随机数，本来我想随机数种子范围有限，想要暴破，不过后来发现，不需要暴破直接按照原来的逻辑就可以，代码如下【注意，由于 linux 和 windows 的随机数生成不同，所以代码要在 linux 上运行】：
#include <iostream>
#include <string>
int main(int argc, const char** argv, const char** envp) {
// 加密过程 异或随机数 -> 置换[随机数生成的] -> 异或随机数 -> 得到v31
unsigned char v18; // [rsp+55h] [rbp-DBh]
char v19; // [rsp+56h] [rbp-DAh]
char ptr[38] = { 0 };
int v31[] = { 220, 184, 64, 189, 156, 201, 110, 101, 239, 18, 216, 152, 105, 208, 222, 252, 107, 174, 125, 139, 214, 141, 15, 208, 79, 102, 62, 157, 250, 195, 233, 36, 211, 239, 255, 157, 231, 1 };
int v28[38] = { 0 };
int len = 38;
int v3 = time(0LL);
srand(v3 & 0xF0000000);
// 存储随机数，之后使用
int arr1[38] = { 0 };
for (int i = 0; i < 38; i++) {
arr1[i] = rand();
}
// ptr 生成
for (int j = 0LL; ; ++j) {
if (j >= 38)
break;
*((char*)ptr + j) = j;
}
// 随机置换ptr数组的元素
for (int k = len - 1; k > 0; --k) {
v18 = rand() % (k + 1);
v19 = *((char*)ptr + k);
*((char*)ptr + k) = *((char*)ptr + v18);
*((char*)ptr + v18) = v19;
}
// 异或
for (int i = 0; i < 38; i++) {
v31[i] ^= rand();
}
// 置换恢复
for (int m = 0; m < 38; m++) {
v28[ptr[m]] = v31[m];
}
// 异或
for (int i = 0; i < 38; i++) {
v28[i] ^= arr1[i];
}
for (int i = 0; i < 38; i++) {
printf("%c", v28[i]);
}
printf("\n");
}
结果：flag{78bace5989660ee38f1fd980a4b4fbcd}
whereThel1b
所实话，依旧是不理解。半蒙半猜吧。
先看 python 文件：
import whereThel1b
flag = input("where is my flag:")
flag = flag.encode()
encry = [108, 117, 72, 80, 64, 49, 99, 19, 69, 115, 94, 93, 94, 115, 71, 95, 84, 89, 56, 101, 70, 2, 84, 75, 127, 68, 103, 85, 105, 113, 80, 103, 95, 67, 81, 7, 113, 70, 47, 73, 92, 124, 93, 120, 104, 108, 106, 17, 80, 102, 101, 75, 93, 68, 121, 26]
whereThel1b.whereistheflag(flag) # 加密逻辑
ret = whereThel1b.trytry(flag) # 加密逻辑
if ret == encry:
print("rrrrrrrrrrrright")
else:
print("wwwwwwwwwwwwwwwrong")
可以看到调用了 whereThel1b 中的 whereistheflag 函数和 trytry 函数，接着 IDA 分析 .os 文件，硬读 Cpython。
到 CreateStringTabAndInitStrings 看看初始化以及一些关键函数的调用：
可以看到有 random ，randint 和 seed
还可以看到 base64 的加密函数：
在_pyx_pymod_exec_whereThel1b，可以看到属性的设置。
base64，randint 在 whereistheflag1 中：
【_pyx_pw_11whereThel1b_5whereistheflag1】
seed 的调用在 trytry 中。
整理下逻辑应该是先对加密后的字符串进行b64encode操作，然后通过seed设置随机数种子后调用random函数获取随机数字符串对明文进行加密然后。
import random
import base64
encry = [108, 117, 72, 80, 64, 49, 99, 19, 69, 115, 94, 93, 94, 115, 71, 95, 84, 89, 56, 101, 70, 2, 84, 75, 127, 68, 103, 85, 105, 113, 80, 103, 95, 67, 81, 7, 113, 70, 47, 73, 92, 124, 93, 120, 104, 108, 106, 17, 80, 102, 101, 75, 93, 68, 121, 26]
random.seed(0)
for i in range(len(encry)):
encry[i] = encry[i] ^ random.randint(0, len(encry))
flag = base64.b64decode(bytes(encry))
print(flag)
rust_baby
这一题挺有意思的，没有想到还有 z3 的神奇解法。
加密逻辑：
其中加密函数：
import base64
import claripy #一个类似z3的求解库
enc = "igdydo19TVE13ogW1AT5DgjPzHwPDQle1X7kS8TzHK8S5KCu9mnJ0uCnAQ4aV3CSYUl6QycpibWSLmqm2y/GqW6PNJBZ/C2RZuu+DfQFCxvLGHT5goG8BNl1ji2XB3x9GMg9T8Clatc="
enc = list(base64.b64decode(enc))
# 最后面的一次异或
# 通过动态调试找到 xor_box，就是原来的v96
xor_box = [220, 95, 32, 34, 194, 121, 25, 86, 53, 218, 139, 71, 211, 25, 252, 85, 20, 205, 210, 123, 88, 89, 9, 66, 222, 44, 180, 72, 217, 242, 27, 169, 64, 225, 166, 251, 255, 56, 193, 213, 226, 232, 119, 120, 111, 34, 4, 230, 22, 62, 12,
53, 82, 92, 253, 193, 229, 89, 28, 208, 174, 90, 178, 221, 25, 248, 66, 230, 44, 137, 89, 229, 17, 156, 200, 123, 129, 112, 127, 111, 188, 111, 2, 143, 247, 244, 200, 112, 174, 2, 248, 91, 226, 114, 8, 9, 111, 191, 75, 57, 181, 208, 30, 163]
for i in range(len(enc)):
enc[i] ^= xor_box[i]
# 在内存中直接得到
key1 = [0x0E71675B493928150, 0x37C65D4C7BA24118, 0x2F6E0584C3B26920,
0x0C74625ECF3321978, 0x9FFE15F4FB724940, 0x27C69D1453F2E1B0]
# 动态调试得到【好臭的数字】
key2 = [1, 1, 4, 5, 1, 4, 1, 9, 1, 9, 8, 1, 0]
# 模拟程序执行逻辑，用claripy求解
# 程序执行逻辑：
""""
v8 = *((_BYTE *)&v13 + 2 * v7 + 1);
v9 = (*((_BYTE *)&v13 + 2 * v7) ^ *((_BYTE *)&retaddr + 2 * v7) | v8 ^ *((_BYTE *)&retaddr + 2 * v7 + 1)) & 1;
*((_BYTE *)input + 2 * v7) = v7 + *((_BYTE *)input + (*((_BYTE *)&v13 + 2 * v7) & 7)) - v9 + 4;
*((_BYTE *)input + 2 * v7 + 1) = v7 + *((_BYTE *)input + (v8 & 7)) - v9 + 4;
"""
# res 是加密后的数据
# a2 , a3 是 key 不变【注意类型】
def rev(res, a2, a3):
# 初始化一个长度为8的符号数组，每个元素为8位的符号变量，用于代表未知输入a1的每一位
# 简单点，就是参数
a1_arr = [claripy.BVS(f"inp_{i}", 8) for i in range(8)]
# 将输入a2/a3转换为十六进制字符串，反转字节序后，再转换为整数列表表示
# 原因：输入的是内存中的值，原始程序中是_int64，需要反转字节序【即内存序转_int64】
a2_arr = [int(i) for i in list(bytes.fromhex(hex(a2)[2:].zfill(16)))[::-1]]
a3_arr = [int(i) for i in list(bytes.fromhex(hex(a3)[2:].zfill(16)))[::-1]]
# 初始化返回值数组
ret_arr = []
# 遍历4个块，每个块处理a2和a3的两个字节
for i in range(4):
# 计算v9，表示两个对应字节异或结果的最低位
v9 = ((a2_arr[i*2] ^ a3_arr[i*2]) |
(a2_arr[i*2+1] ^ a3_arr[i*2+1])) & 1
# 根据a2的字节值索引a1数组，并应用变换计算ret_arr的元素
ret_arr.append(i-v9+a1_arr[a2_arr[i*2] & 7])
ret_arr.append(i-v9+a1_arr[a2_arr[i*2+1] & 7])
# 对ret_arr中的每个元素执行逆向异或操作0x33，模拟原始编码的逆过程
for i in range(len(ret_arr)):
ret_arr[i] ^= 0x33
# 创建一个求解器实例，用于解决约束满足问题
s = claripy.Solver()
# 添加约束条件，确保计算后的ret_arr与给定的res数组中的元素一一对应
for i in range(len(res)):
s.add(ret_arr[i] == res[i])
# 执行求解，寻找满足条件的a1_arr的具体值，这里只求解一组解
result = s.batch_eval(a1_arr, 1)[0]
print("".join(chr(c) for c in result), end="")
# 循环调用rev函数，依次解密8个块
for i in range(8):
rev(enc[i*8:(i+1)*8], key1[i%6], key2[i])
# flag{6e2480b3-4f02-4cf1-9bc0-123b75f9a922}
GoReverse
ciscn24-GoReverse 速览_哔哩哔哩_bilibili
加密套娃。分析后的加密流程如下：
通过远程连接得到结果，这里复现，就本地创建 flag 文件，假设加密的结果为：VPAFNU3PTHPAUQTCYUBTVJY6TGBWY3NGGKZ6OFKZ74JJBPF7LSQ3ZYO4IZLOXPVEIE7KZS46VJKNDJGRBJPUTVKTNFLZRQBLLZUD7VI=
base32 解密，AES 的 CBC 模式解密：
得到：69124cadc128ffe15752488b318e78c68f359d69fb2b8e4bc35ae98ee05493ca31ceb32174201bb091c9406e56430d53c046d1d128a0de5862ee5e6d04040404
下面的 sm4 好麻烦：
sm4并不是对上一步的xxtea密文加密
而是先随机生成16字节(叫做randbytes1)
然后把randbytes1最后1字节加1得到randbytes2，再加1得到randbytes3
sm4加密的密文即为randbytes1+ randbytes2+ randbytes3
得到的密文与上一步的xxtea密文异或
实际上就是SM4 CTR模式。
(CBC 得到的结果的前面就是前 16 位就是 randbytes1）【调试一下】
模拟加密流程，可以得到：66e63d01bbe3a9fdeaf58ff981f941d2a18ca83e76ba547317c13bd0552cd43717415b81d2aebc08f20f6380cb19f21526bde0afa274aa6e0cf8ffdc016763f311bae241451ce9f6ea515263f5415b29
将sm4密文(去掉padding部分)与上面enc除前16字节的部分异或即可得到xxtea密文
import struct
# SM4密文的16进制字符串转换为字节列表
enc = list(bytes.fromhex(
"69124cadc128ffe15752488b318e78c68f359d69fb2b8e4bc35ae98ee05493ca31ceb32174201bb091c9406e56430d53c046d1d128a0de5862ee5e6d04040404"))
# 分割密文，前16字节（SM4块大小）作为SM4解密的明文部分
sm4_plain = enc[:len(enc)//4]
# 打印SM4明文部分的16进制表示
print(bytes(sm4_plain).hex())
# SM4解密后的明文用于异或的密文，从给定的16进制字符串转换
sm4_plain_enc = list(bytes.fromhex("66e63d01bbe3a9fdeaf58ff981f941d2a18ca83e76ba547317c13bd0552cd43717415b81d2aebc08f20f6380cb19f215"))
# 对剩余的密文部分与SM4解密明文异或操作
inp_enc = enc[len(enc)//4:]
for i in range(len(inp_enc)):
inp_enc[i] ^= sm4_plain_enc[i]
# 将异或后的结果转换为字节串
inp_enc = bytes(inp_enc)
# 按每4字节分组，转换为小端整数并打印
for i in range(0, len(inp_enc), 4):
print(hex(struct.unpack("<I", inp_enc[i:i+4])[0]), end=",")
# 0x68a0d3e9,0xb627c840,0x7766af29,0x18d2ad61,0xbce30834,0x8c6889,0xf57d3732,0x9f6a71b2,0xcf62c39b,0xd5579840,0xb2429b2b,0xad81ba1a
xxtea 解密 【有被魔改】：
#include <stdio.h>
#include <stdint.h>
#define DELTA 0x7FAB4CAD
#define MX (((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y)) + ((key[(p & 3) ^ e] ^ z)))
void btea(uint32_t *v, int n, uint32_t const key[4])
{
uint32_t y, z, sum;
unsigned p, rounds, e;
//加密
if (n > 1)
{
rounds = 6 + 52 / n;
sum = 0;
z = v[n - 1];
do
{
sum += DELTA;
e = (sum >> 2) & 3;
for (p = 0; p < n - 1; p++)
{
y = v[p + 1];
z = v[p] += MX;
}
y = v[0];
z = v[n - 1] += MX;
} while (--rounds);
}
//解密
else if (n < -1)
{
n = -n;
rounds = 6 + 52 / n;
sum = rounds * DELTA;
y = v[0];
do
{
e = (sum >> 2) & 3;
for (p = n - 1; p > 0; p--)
{
z = v[p - 1];
y = v[p] -= MX;
}
z = v[n - 1];
y = v[0] -= MX;
sum -= DELTA;
} while (--rounds);
}
printf("sum==0x%x\n", sum);
}
//打印数据 hex_or_chr: 1-hex 0-chr
void dump_data(uint32_t *v, int n, bool hex_or_chr)
{
if (hex_or_chr)
{
for (int i = 0; i < n; i++)
{
printf("0x%x,", v[i]);
}
}
else
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < sizeof(uint32_t) / sizeof(uint8_t); j++)
{
printf("0x%x,", (v[i] >> (j * 8)) & 0xFF);
}
}
}
printf("\n");
return;
}
int main()
{
// v为要加解密的数据
uint32_t v[] = {0x68a0d3e9, 0xb627c840, 0x7766af29, 0x18d2ad61, 0x5c673dae, 0x1b929a85, 0x5d5bb8f6, 0x5e795a38, 0x67de5300, 0x3414e420, 0x54c5910b};
// k为加解密密钥，4个32位无符号整数，密钥长度为128位
uint32_t k[4] = {0x385e7342, 0x345a772a, 0x6f38756c, 0x6b402652};
int n = sizeof(v) / sizeof(uint32_t);
/*
printf("加密前明文数据：");
dump_data(v, n, 1);
btea(v, n, k);
*/
printf("加密后密文数据：");
dump_data(v, n, 1);
btea(v, -n, k);
printf("解密后明文数据：");
dump_data(v, n, 1);
printf("解密后明文字符：");
dump_data(v, n, 0);
return 0;
}
// 0x22,0x5b,0x23,0x2d,0x37,0x40,0x2e,0x5f,0xc,0x77,0x53,0x45,0x57,0x49,0x34,0x71,0x5b,0x56,0x1e,0x7d,0x77,0x59,0x7a,0x5c,0x6c,0x63,0x55,0x45,0x73,0x35,0x9,0xe,0x27,0x53,0x0,0x5e,0x59,0x1e,0x0,0x72,0x4,0xc,0x0,0x0
异或：
enc = [0x22, 0x5b, 0x23, 0x2d, 0x37, 0x40, 0x2e, 0x5f, 0xc, 0x77, 0x53, 0x45, 0x57, 0x49, 0x34, 0x71, 0x5b, 0x56, 0x1e, 0x7d, 0x77,
0x59, 0x7a, 0x5c, 0x6c, 0x63, 0x55, 0x45, 0x73, 0x35, 0x9, 0xe, 0x27, 0x53, 0x0, 0x5e, 0x59, 0x1e, 0x0, 0x72, 0x4, 0xc, 0x0, 0x0]
box = [0x44, 0x37, 0x42, 0x4A, 0x4C, 0x73, 0x4F, 0x6B, 0x39, 0x40,
0x66, 0x26, 0x31, 0x64, 0x57, 0x49, 0x6E, 0x35, 0x33, 0x49,
0x44, 0x6C, 0x4A, 0x71, 0x55, 0x53, 0x36, 0x24, 0x5E, 0x57,
0x68, 0x6B, 0x41, 0x6B, 0x32, 0x6B, 0x6B, 0x2A, 0x32, 0x47,
0x61, 0x71, 0x6D, 0x4C, 0x77, 0x69, 0x4C, 0x58, 0x5E, 0x62,
0x47, 0x47, 0x45, 0x24, 0x26, 0x64, 0x6D, 0x71, 0x52, 0x5E,
0x67, 0x35, 0x62, 0x4C, 0x33, 0x6C, 0x43, 0x41, 0x35, 0x5E,
0x48, 0x47, 0x4B, 0x24, 0x39, 0x71, 0x6F, 0x35, 0x54, 0x40,
0x42, 0x77, 0x6F, 0x6D, 0x39, 0x76, 0x45, 0x58, 0x79, 0x61,
0x30, 0x48, 0x41, 0x56, 0x33, 0x4C, 0x72, 0x57, 0x57]
for i in range(len(enc)):
enc[i] ^= box[i]
print(bytes(enc))
# flag{3a4575cf-c85c-4350-90ca-baef8252425e}