#!/bin/bash
# Copyright (c) 2026 Alibaba Group and its affiliates

# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at

#     http://www.apache.org/licenses/LICENSE-2.0

# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.


# ============================================================
# Challenge: aliyunctf-2024-Misc-582c5006
# Writeup (from instruction.md) - commented out below.
# ============================================================
#
#
# > 以下为解题 writeup 全文，供参考。
#
# # misc-字
# tl;dr: 康熙部首 + 区位码
#
# exp 如下：
#
# ```python
# hanzi, kangxi = open('kangxi.txt', 'r').read().split('\n')
# enc = open('enc.txt', 'r').read()
#
# # 检查每个字是否有对应的康熙部首，如果有的话：
# #   如果是康熙部首则为 1
# #   如果不是康熙部首则为 0
# binary = []
# for ch in enc:
#     if ch in hanzi:
#         binary.append('0')
#     elif ch in kangxi:
#         binary.append('1')
# binary = ''.join(binary)[:315-7]
#
# # 按照区位码的规则转成 GB2312
# gb = []
# for i in range(0, len(binary), 7):
#     gb.append(f'{int(binary[i:i+7], 2) + 0xA0:02X}')
# gb = ''.join(gb)
#
# # 读取 GB2312
# msg = bytes.fromhex(gb).decode('gb2312')
# print(msg)
# ```
#
# ## Hints
#
# - 依据题目提供的汉字集合建立分类对照表，将密文字符逐一映射为二进制位。
# - 将拼接后的二进制流按固定长度分段，转换为十进制数并叠加预设偏移值。
# - 结合“区位码”的编码结构，将处理后的数值序列对齐为合法的字符编码字节。
# - 匹配正确的字符集对字节流进行解码，提取最终明文。
# - 核心在于逆向推导字符分类规则、位宽划分与编码偏移量之间的映射关系。
#
# ============================================================
# End of Writeup
# ============================================================

# --- Direct flag solution ---
# Decode and write the flag so that running this script solves the challenge.
mkdir -p /logs/agent
echo -n 'YWxpeXVuY3RmezI2MjZfNTU1NH0=' | base64 -d > /logs/agent/flag.txt