文章总结： 本文记录使用AI工具（CodeXCLI配合Gpt-5.4模型及JSReverser-MCP）逆向猿人学反混淆练习平台第一题的加密参数m的完整过程。通过抓包定位接口中的加密参数m，利用AI辅助生成Python纯算MD5源码实现参数构造并成功获取数据，整个AI逆向过程约耗时30分钟。作者认为设计好提示词后AI可大幅简化JS逆向工作，但指出该题混淆难度较低，新手亦可手动完成。 综合评分： 58 文章分类： 爬虫,AI安全,逆向分析,WEB安全
AI逆向|逆向反混淆练习平台第一题加密参数并获取数据
原创
悦来客栈的老板 悦来客栈的老板
菜鸟学Python编程
2026年3月26日 21:00 湖北
关注它，不迷路。
- 本文章中所有内容仅供学习交流，不可用于任何商业用途和非法用途，否则后果自负，如有侵权，请联系作者立即删除！
一.题目地址
https://match.yuanrenxue.cn/match/1
二.抓包分析
打开控制台后，抓包分析，看看所要的数据在哪里:
数据在上面这个接口，有个m的加密参数
古法逆向，到此结束。
三.AI工具
我这里使用codeX的cli工具 + Gpt-5.4 xhigh 的AI模型。
使用的MCP则是 JSReverser-MCP
四.提示词
https://t.zsxq.com/5TbAy
如果需要，可以私下和我交流。
五.AI提供的Python纯算源码
import ctypesimport time
import requests
API_URL = "https://match.yuanrenxue.cn/api/question/1"SESSION_ID = "你的ID"USER_AGENT = "yuanrenxue"SEPARATOR = "\u4e28"
# 固定样本来自离线插桩后的 JS 运行结果：# ts = 1710100000000 -> f = 008f6fa3eb5dd36069d29a8ea3d77bb5SAMPLE_TS = "1710100000000"SAMPLE_HASH = "008f6fa3eb5dd36069d29a8ea3d77bb5"
def _to_int32(value: int) -> int: return ctypes.c_int32(value).value
def _urshift(value: int, bits: int) -> int: return (value & 0xFFFFFFFF) >> bits
def _safe_add(a: int, b: int) -> int: low = (a & 0xFFFF) + (b & 0xFFFF) high = (a >> 16) + (b >> 16) + (low >> 16) return _to_int32((high << 16) | (low & 0xFFFF))
def _bit_rol(value: int, bits: int) -> int: return _to_int32((value << bits) | _urshift(value, 32 - bits))
def _md5_cmn(h: int, e: int, d: int, c: int, g: int, f: int) -> int: return _safe_add(_bit_rol(_safe_add(_safe_add(e, h), _safe_add(c, f)), g), d)
def _md5_ff(g: int, f: int, k: int, j: int, e: int, i: int, h: int) -> int: return _md5_cmn((f & k) | ((~f) & j), g, f, e, i, h)
def _md5_gg(g: int, f: int, k: int, j: int, e: int, i: int, h: int) -> int: return _md5_cmn((f & j) | (k & (~j)), g, f, e, i, h)
def _md5_hh(g: int, f: int, k: int, j: int, e: int, i: int, h: int) -> int: return _md5_cmn(f ^ k ^ j, g, f, e, i, h)
def _md5_ii(g: int, f: int, k: int, j: int, e: int, i: int, h: int) -> int: return _md5_cmn(k ^ (f | (~j)), g, f, e, i, h)
def _str2binl(text: str, chrsz: int = 16) -> list[int]: words: list[int] = [] mask = (1 << chrsz) - 1 for bit in range(0, len(text) * chrsz, chrsz): index = bit >> 5 while len(words) <= index: words.append(0) words[index] = _to_int32(words[index] | ((ord(text[bit // chrsz]) & mask) << (bit % 32))) return words
def _binl2hex(words: list[int]) -> str: table = "0123456789abcdef" output: list[str] = [] for i in range(len(words) * 4): value = words[i >> 2] output.append(table[(_urshift(value, (i % 4) * 8 + 4)) & 0xF]) output.append(table[(_urshift(value, (i % 4) * 8)) & 0xF]) return "".join(output)
def _core_md5(words: list[int], bit_length: int) -> list[int]: index = bit_length >> 5 while len(words) <= index: words.append(0) words[index] = _to_int32(words[index] | (128 << (bit_length % 32)))
index = (((bit_length + 64) >> 9) << 4) + 14 while len(words) <= index: words.append(0) words[index] = _to_int32(bit_length)
o, n, m, l = 1732584193, -271733879, -1732584194, 271733878
for g in range(0, len(words), 16): while len(words) < g + 16: words.append(0)
j, h, f, e = o, n, m, l
o = _md5_ff(o, n, m, l, words[g + 0], 7, -680976936) l = _md5_ff(l, o, n, m, words[g + 1], 12, -389564586) m = _md5_ff(m, l, o, n, words[g + 2], 17, 606105819) n = _md5_ff(n, m, l, o, words[g + 3], 22, -1044525330) o = _md5_ff(o, n, m, l, words[g + 4], 7, -176418897) l = _md5_ff(l, o, n, m, words[g + 5], 12, 1200080426) m = _md5_ff(m, l, o, n, words[g + 6], 17, -1473231341) n = _md5_ff(n, m, l, o, words[g + 7], 22, -45705983) o = _md5_ff(o, n, m, l, words[g + 8], 7, 1770035416) l = _md5_ff(l, o, n, m, words[g + 9], 12, -1958414417) m = _md5_ff(m, l, o, n, words[g + 10], 17, -42063) n = _md5_ff(n, m, l, o, words[g + 11], 22, -1990404162) o = _md5_ff(o, n, m, l, words[g + 12], 7, 1804660682) l = _md5_ff(l, o, n, m, words[g + 13], 12, -40341101) m = _md5_ff(m, l, o, n, words[g + 14], 17, -1502002290) n = _md5_ff(n, m, l, o, words[g + 15], 22, 1236535329)
o = _md5_gg(o, n, m, l, words[g + 1], 5, -165796510) l = _md5_gg(l, o, n, m, words[g + 6], 9, -1069501632) m = _md5_gg(m, l, o, n, words[g + 11], 14, 643717713) n = _md5_gg(n, m, l, o, words[g + 0], 20, -373897302) o = _md5_gg(o, n, m, l, words[g + 5], 5, -701558691) l = _md5_gg(l, o, n, m, words[g + 10], 9, 38016083) m = _md5_gg(m, l, o, n, words[g + 15], 14, -660478335) n = _md5_gg(n, m, l, o, words[g + 4], 20, -405537848) o = _md5_gg(o, n, m, l, words[g + 9], 5, 568446438) l = _md5_gg(l, o, n, m, words[g + 14], 9, -1019803690) m = _md5_gg(m, l, o, n, words[g + 3], 14, -187363961) n = _md5_gg(n, m, l, o, words[g + 8], 20, 1163531501) o = _md5_gg(o, n, m, l, words[g + 13], 5, -1444681467) l = _md5_gg(l, o, n, m, words[g + 2], 9, -51403784) m = _md5_gg(m, l, o, n, words[g + 7], 14, 1735328473) n = _md5_gg(n, m, l, o, words[g + 12], 20, -1921207734)
o = _md5_hh(o, n, m, l, words[g + 5], 4, -378558) l = _md5_hh(l, o, n, m, words[g + 8], 11, -2022574463) m = _md5_hh(m, l, o, n, words[g + 11], 16, 1839030562) n = _md5_hh(n, m, l, o, words[g + 14], 23, -35309556) o = _md5_hh(o, n, m, l, words[g + 1], 4, -1530992060) l = _md5_hh(l, o, n, m, words[g + 4], 11, 1272893353) m = _md5_hh(m, l, o, n, words[g + 7], 16, -155497632) n = _md5_hh(n, m, l, o, words[g + 10], 23, -1094730640) o = _md5_hh(o, n, m, l, words[g + 13], 4, 681279174) l = _md5_hh(l, o, n, m, words[g + 0], 11, -358537222) m = _md5_hh(m, l, o, n, words[g + 3], 16, -722881979) n = _md5_hh(n, m, l, o, words[g + 6], 23, 76029189) o = _md5_hh(o, n, m, l, words[g + 9], 4, -640364487) l = _md5_hh(l, o, n, m, words[g + 12], 11, -421815835) m = _md5_hh(m, l, o, n, words[g + 15], 16, 530742520) n = _md5_hh(n, m, l, o, words[g + 2], 23, -995338651)
o = _md5_ii(o, n, m, l, words[g + 0], 6, -198630844) l = _md5_ii(l, o, n, m, words[g + 7], 10, 11261161415) m = _md5_ii(m, l, o, n, words[g + 14], 15, -1416354905) n = _md5_ii(n, m, l, o, words[g + 5], 21, -57434055) o = _md5_ii(o, n, m, l, words[g + 12], 6, 1700485571) l = _md5_ii(l, o, n, m, words[g + 3], 10, -1894446606) m = _md5_ii(m, l, o, n, words[g + 10], 15, -1051523) n = _md5_ii(n, m, l, o, words[g + 1], 21, -2054922799) o = _md5_ii(o, n, m, l, words[g + 8], 6, 1873313359) l = _md5_ii(l, o, n, m, words[g + 15], 10, -30611744) m = _md5_ii(m, l, o, n, words[g + 6], 15, -1560198380) n = _md5_ii(n, m, l, o, words[g + 13], 21, 1309151649) o = _md5_ii(o, n, m, l, words[g + 4], 6, -145523070) l = _md5_ii(l, o, n, m, words[g + 11], 10, -1120210379) m = _md5_ii(m, l, o, n, words[g + 2], 15, 718787259) n = _md5_ii(n, m, l, o, words[g + 9], 21, -343485551)
o = _safe_add(o, j) n = _safe_add(n, h) m = _safe_add(m, f) l = _safe_add(l, e)
return [o, n, m, l]
def yrx_hex_md5(text: str) -> str: return _binl2hex(_core_md5(_str2binl(text), len(text) * 16))
def build_m() -> str: # 页面源码使用 Date.parse(new Date())，精度只有秒，因此这里取整秒后再补 000。 ts = int(time.time()) * 1000 + 100000000 return f"{yrx_hex_md5(str(ts))}{SEPARATOR}{ts // 1000}"
def self_check() -> None: actual = yrx_hex_md5(SAMPLE_TS) if actual != SAMPLE_HASH: raise RuntimeError(f"local hash mismatch: {actual} != {SAMPLE_HASH}")
def fetch_page(session: requests.Session, page: int) -> list[int]: params = { "page": page, "pageSize": 10, "kw": "", "m": build_m(), } response = session.get(API_URL, params=params, timeout=10) response.raise_for_status() payload = response.json() if "data" not in payload: raise RuntimeError(f"unexpected response: {payload}") return payload["data"]
def main() -> None: self_check()
session = requests.Session() session.headers.update({"User-Agent": USER_AGENT}) session.cookies.set("sessionid", SESSION_ID, domain="match.yuanrenxue.cn")
total = 0 for page in range(1, 6): data = fetch_page(session, page) total += sum(data) print(f"page {page}: {data}")
print(f"total: {total}")
if __name__ == "__main__": main()
结果如下图:
从AI开始分析到结束，大约耗时30分钟左右。
六.后记
不难看出，当你设计好提示词以后，AI会根据你这个提示进行逆向分析。
对于一般的网站逆向，你只需要抓包分析一下加密参数即可，其他的似乎已经不用操心了。
当然，这题目的混淆代码也不难，对于新手朋友还是挺友好的。
还不需要使用AI这么强的工具。
今天的分享就到这里，感谢阅读。
欢迎加入知识星球，学习更多AST和爬虫技巧。
免责声明：
本文所载程序、技术方法仅面向合法合规的安全研究与教学场景，旨在提升网络安全防护能力，具有明确的技术研究属性。
任何单位或个人未经授权，将本文内容用于攻击、破坏等非法用途的，由此引发的全部法律责任、民事赔偿及连带责任，均由行为人独立承担，本站不承担任何连带责任。
本站内容均为技术交流与知识分享目的发布，若存在版权侵权或其他异议，请通过邮件联系处理，具体联系方式可点击页面上方的联系我。
本文转载自：菜鸟学Python编程 悦来客栈的老板 悦来客栈的老板《AI逆向|逆向反混淆练习平台第一题加密参数并获取数据》
版权声明
本站仅做备份收录，仅供研究与教学参考之用。
读者将信息用于其他用途的，全部法律及连带责任由读者自行承担，本站不承担任何责任。
评论