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re1

rust，缺少符号表，很难看

使用bindiff恢复符号表，首先需要配置rust环境

Rust语言开发环境搭建详细教程_rust环境搭建-CSDN博客

参考rust恢复符号表 | clev1L’s blog

恢复了一点点吧

根据可疑字符串定位到

关键部分

v29是

v29 = *(_WORD *)v16;

将内存地址 v16 处的前两个字节解释为一个 unsigned short（即 16 位无符号整数），赋值给 v29。

enc = [0x0, 0xA1, 0xFB, 0x53, 0x1C, 0xFA, 0xF0, 0x1B, 0x6, 0x40, 0xD4, 0x8C,
       0x16, 0xF4, 0x90, 0x27, 0x42, 0xB9, 0x8B, 0xF, 0x2, 0xD7, 0x31, 0xB7,
       0x26, 0x12, 0x6, 0x7E, 0xAE, 0xDF, 0xDA, 0x68, 0xAF, 0x35, 0xCC, 0xB7,
       0xB0, 0xD0, 0x9A, 0x59, 0x2B, 0xB]  # length = 42

def rol(val, r_bits, max_bits=8):
    return ((val << r_bits) & (2**max_bits - 1)) | (val >> (max_bits - r_bits))

def decrypt(ciphertext):
    prefix = b"fl"
    result = bytearray(len(ciphertext))
    v29 = int.from_bytes(prefix[:2], "little")
    for i in range(len(ciphertext)):
        v33 = ((v29 >> 2) ^ (v29 >> 3) ^ (v29 >> 1)) & 0xFFFF
        next_v29 = ((v29 >> 1) | (v33 << 15)) & 0xFFFF
        v34 = rol(v29 & 0xFF, 4)
        v35 = ((4 * (v34 & 0x33)) | ((v34 >> 2) & 0x33)) & 0xFF
        xor_val = (i + ((2 * (v35 & 0x55)) | ((v35 >> 1) & 0x55))) & 0xFF
        result[i] = ciphertext[i] ^ xor_val
        v29 = next_v29
    return bytes(result)

ciphertext = bytes(enc)
plaintext = decrypt(ciphertext)
print(plaintext.decode(errors='ignore'))

仔细解释下

next_v29 = ((v29 >> 1) | (v33 << 15)) & 0xFFFF
v34 = rol(v29 & 0xFF, 4)

为什么是v29 & 0XFF,因为rol是byte，而v33的byte是v29

flag{1c98572d-7f7b-4fbf-8750-4a2986c695ce}

re2

# v8: shuffle 掩码（reverse）
shuffle_mask = list(reversed(range(16)))

# v9: 每字节加法掩码
add_mask = bytes([
    0x01, 0x02, 0x03, 0x04,
    0x05, 0x06, 0x07, 0x08,
    0x09, 0x0A, 0x0B, 0x0C,
    0x0D, 0x0E, 0x0F, 0x10
])

# 加密的目标密文
ciphertext = b"cge87k?9<>?@=pss393=>;8@:Cp@DAuH"

def decrypt_block(block: bytes) -> bytes:
    # 先减去加法掩码
    subtracted = bytes((b - m) & 0xFF for b, m in zip(block, add_mask))
    # 再反转（inverse shuffle）
    return bytes(subtracted[i] for i in shuffle_mask)

# 解密两个 block
plaintext = b""
for i in range(0, len(ciphertext), 16):
    block = ciphertext[i:i+16]
    plaintext += decrypt_block(block)

# 输出 flag
print(f"flag{{{plaintext.decode()}}}")

GPT速通

flag{cdb0444318e24beb8f374e9181599072}

re!!!

IOS逆向

看着像密文

rc4 –> base64 –> obfusecateString –> transformString.

先看最后一个

根据 n190 的大小分三种情况（单字节、两字节、三字节或四字节）

逆一下:

def decrypt_specialized(hex_str: str) -> str:
    # 1. 清理输入：去掉空白
    hex_str = ''.join(hex_str.split())
    # 2. 验证长度为偶数
    if len(hex_str) % 2 != 0:
        raise ValueError(f"Invalid hex string length {len(hex_str)}, must be even.")
    # 3. 从十六进制解码为字节
    data = bytes.fromhex(hex_str)
    # 4. 按 UTF-8 解码得到中间混淆字符串
    transformed = data.decode('utf-8')

    result_chars = []
    # 5. 逐字符逆向还原
    for i, ch in enumerate(transformed):
        # 偶数位置使用密钥 0xEF，奇数位置使用 0xBE
        key = 0xEF if (i % 2 == 0) else 0xBE
        code = ord(ch)
        # ASCII 范围时先减 1 再异或，否则直接异或
        if code <= 0x7F:
            n190 = code - 1
        else:
            n190 = code
        orig_cp = n190 ^ key
        result_chars.append(chr(orig_cp))

    # 6. 拼接并返回结果
    return ''.join(result_chars)

if __name__ == '__main__':
    # 示例：将下面替换为你的密文 hex 字符串
    cipher_hex = (
        "c2a7c3b9c2acc3a5c2a2c3b6c391c295c2aac38cc28bc38ac2a6c3aec28bc28f"
        "c2a1c3aac287c382c2bfc3b6c282c38ec2b9c3a2c2a13cc28ac3adc2b1c28"
        "0c2b2c384c28dc3bbc283c396c2b03dc28a3bc2b12cc287c3b0c2852bc282"
        "c39ac28432c29320c29d21c29ac392c291c3a1c296c3a06d1866396c25631"
        "0c299c3946931c291c3917a2e470b632a7811730f65c385"
    )
    try:
        plaintext = decrypt_specialized(cipher_hex)
        print(plaintext)
    except Exception as e:
        print("解密失败：", e)

接下来那部分：

这段是核心混淆逻辑：

• 将当前字符值加上迭代器 v6（动态偏移）；
• 再加一个固定偏移 0xDEADBEEF；
• 然后取低字节；
• 如果结果非负，取负值再取反（取补码）形成非线性变换。

if ((v11 & 0xFFFFFF80) != 0)
    v13 = (((v11 & 0x3F) << 8) | ((unsigned int)v11 >> 6)) + 33217;
else
    v13 = v11 + 1;

这是根据字符值决定如何处理的分支：

• 如果是高位字符（非 ASCII），执行位运算变换并加上 33217。
• 否则简单地加一。

import base64

def decrypt_specialized(hex_str: str):
    hex_str = ''.join(hex_str.split())
    if len(hex_str) % 2 != 0:
        raise ValueError(f"Invalid hex string length {len(hex_str)}, must be even.")
    data = bytes.fromhex(hex_str)
    transformed = data.decode('utf-8')
    result_chars = []
    for i, ch in enumerate(transformed):
        key = 0xEF if (i % 2 == 0) else 0xBE
        code = ord(ch)
        if code <= 0x7F:
            n190 = code
        else:
            n190 = code
        orig_cp = n190 ^ key
        result_chars.append(chr(orig_cp))

    # 6. 拼接并返回结果
    return result_chars

if __name__ == '__main__':
    hex_enc = "c2a7c3b9c2acc3a5c2a2c3b6c391c295c2aac38cc28bc38ac2a6c3aec28bc28fc2a1c3aac287c382c2bfc3b6c282c38ec2b9c3a2c2a13cc28ac3adc2b1c280c2b2c384c28dc3bbc283c396c2b03dc28a3bc2b12cc287c3b0c2852bc282c39ac28432c29320c29d21c29ac392c291c3a1c296c3a06d1866396c256310c299c3946931c291c3917a2e470b632a7811730f65c385"
    cipher_hex = (
        "c2a7c3b9c2acc3a5c2a2c3b6c391c295c2aac38cc28bc38ac2a6c3aec28bc28fc2a1c3aac287c382c2bfc3b6c282c38ec2b9c3a2c2a13cc28ac3adc2b1c280c2b2c384c28dc3bbc283c396c2b03dc28a3bc2b12cc287c3b0c2852bc282c39ac28432c29320c29d21c29ac392c291c3a1c296c3a06d1866396c256310c299c3946931c291c3917a2e470b632a7811730f65c385"
    )
    try:
        plaintext = decrypt_specialized(cipher_hex)
        flag_base64 = ''
        for i in range(len(plaintext)):
            flag_base64 += chr((ord(plaintext[i]) - i - 0xDEADBEEF) & 0xFF)
        base64.b64decode(flag_base64).decode()
    except Exception as e:
        print("解密失败：", e)

adbe2979358798308c911dd4647a2f6a13c042cc324597e1edbc089a9e9015fba9

还有个RC4，key找了一圈发现几个init

跟进下

处理下大小端序的问题

Haruhikage

import base64

def decrypt_specialized(hex_str: str):
    hex_str = ''.join(hex_str.split())
    if len(hex_str) % 2 != 0:
        raise ValueError(f"Invalid hex string length {len(hex_str)}, must be even.")
    data = bytes.fromhex(hex_str)
    transformed = data.decode('utf-8')
    result_chars = []
    for i, ch in enumerate(transformed):
        key = 0xEF if (i % 2 == 0) else 0xBE
        code = ord(ch)
        if code <= 0x7F:
            n190 = code
        else:
            n190 = code
        orig_cp = n190 ^ key
        result_chars.append(chr(orig_cp))

    # 6. 拼接并返回结果
    return result_chars

def rc4_decrypt(data: bytes, key: str) -> bytes:
    # KSA (Key Scheduling Algorithm)
    S = list(range(256))
    j = 0
    key_bytes = key.encode('utf-8')
    for i in range(256):
        j = (j + S[i] + key_bytes[i % len(key_bytes)]) % 256
        S[i], S[j] = S[j], S[i]

    # PRGA (Pseudo-Random Generation Algorithm)
    i = j = 0
    out = bytearray()
    for byte in data:
        i = (i + 1) % 256
        j = (j + S[i]) % 256
        S[i], S[j] = S[j], S[i]
        K = S[(S[i] + S[j]) % 256]
        out.append(byte ^ K)
    return bytes(out)

if __name__ == '__main__':
    hex_enc = "c2a7c3b9c2acc3a5c2a2c3b6c391c295c2aac38cc28bc38ac2a6c3aec28bc28fc2a1c3aac287c382c2bfc3b6c282c38ec2b9c3a2c2a13cc28ac3adc2b1c280c2b2c384c28dc3bbc283c396c2b03dc28a3bc2b12cc287c3b0c2852bc282c39ac28432c29320c29d21c29ac392c291c3a1c296c3a06d1866396c256310c299c3946931c291c3917a2e470b632a7811730f65c385"
    cipher_hex = (
        "c2a7c3b9c2acc3a5c2a2c3b6c391c295c2aac38cc28bc38ac2a6c3aec28bc28fc2a1c3aac287c382c2bfc3b6c282c38ec2b9c3a2c2a13cc28ac3adc2b1c280c2b2c384c28dc3bbc283c396c2b03dc28a3bc2b12cc287c3b0c2852bc282c39ac28432c29320c29d21c29ac392c291c3a1c296c3a06d1866396c256310c299c3946931c291c3917a2e470b632a7811730f65c385"
    )
    try:
        plaintext = decrypt_specialized(cipher_hex)
        flag_base64 = ''
        for i in range(len(plaintext)):
            flag_base64 += chr((ord(plaintext[i]) - i - 0xDEADBEEF) & 0xFF)
        rc4_enc = base64.b64decode(flag_base64)
        rc4_enc_bytes = bytes.fromhex(rc4_enc.decode())
        key = 'Haruhikage'
        decrypted = rc4_decrypt(rc4_enc_bytes, key)
        print(decrypted)
    except Exception as e:
        print("解密失败：", e)

flag{N4nd3_H4ruhik4g3_Y4tt4n0?!!}