2021 ISCC MISC 做题笔记（一）

1.李华的红包

打开题目文件夹，发现了一张图片，双击打开（我设置的是自动使用winhex）未发现与flag有关的信息，果断拖入虚拟机进行binwalk

──(kali㉿kali)-[~]
└─$ binwalk -e /home/kali/桌面/hongbao2.jpg 

在得到的文件夹内获得了一个文档（下位内容）

24,43,13,13,12,21,43

具体操作情况如图:

 

 看到这串字符感觉无从下手，去查资料，关于编码，找到了一个略显靠谱的：敲击码

 

 根据上述解码图进行解码，得出flag：

2.Retrieve_the_passcode

打开题目附件里的文件夹，发现有一个txt文件和一个rar的压缩文件。

打开rar压缩包发现需要密码，打开itxt文件

1:3:1;
1.25:3:1;
1.5:3:1;
1.75:3:1;
2:3:1;
2:2.75:1;
2:2.5:1;
2:2.25:1;
2:2:1;
2:1.75:1;
2:1.5:1;
1:2.25:1;
1.25:2.25:1;
1.5:2.25:1;
1.75:2.25:1;
1:1.5:1;
1.25:1.5:1;
1.5:1.5:1;
1.75:1.5:1;
3:3:1;
3.25:3:1;
3.5:3:1;
3.75:3:1;
4:3:1;
3.25:2.25:1;
3.5:2.25:1;
3.75:2.25:1;
4:2.25:1;
4:2:1;
4:1.75:1;
4:1.5:1;
3:1.5:1;
3.25:1.5:1;
3.5:1.5:1;
3.75:1.5:1;
3:1.75:1;
3:2:1;
3:2.25:1;
3:2.5:1;
3:2.75:1;
5:3:1;
5.25:3:1;
5.5:3:1;
5.75:3:1;
6:3:1;
6:2.25:1;
6:2:1;
6:1.75:1;
6:1.5:1;
5.75:1.5:1;
5.5:1.5:1;
5.25:1.5:1;
5:1.5:1;
5:2.25:1;
5.25:2.25:1;
5.5:2.25:1;
5.75:2.25:1;
5:2.5:1;
5:2.75:1;
7:3:1;
7.25:3:1;
7.5:3:1;
7.75:3:1;
8:3:1;
8:2.75:1;
8:2.5:1;
8:2.25:1;
8:2:1;
8:1.75:1;
8:1.5:1;
9:3:1;
9.25:3:1;
9.5:3:1;
9.75:3:1;
10:3:1;
10:2.75:1;
10:2.5:1;
10:2.25:1;
9.75:2.25:1;
9.5:2.25:1;
9.25:2.25:1;
9:2.25:1;
9:2:1;
9:1.75:1;
9:1.5:1;
9.25:1.5:1;
9.5:1.5:1;
9.75:1.5:1;
10:1.5:1;
11:3:1;
11.25:3:1;
11.5:3:1;
11.75:3:1;
12:3:1;
12:2.75:1;
12:2.5:1;
12:2.25:1;
12:2:1;
12:1.75:1;
12:1.5:1;
11.75:1.5:1;
11.5:1.5:1;
11.25:1.5:1;
11:1.5:1;
11:1.75:1;
11:2:1;
11:2.25:1;
11:2.5:1;
11:2.75:1;
11.25:2.25:1;
11.5:2.25:1;
11.75:2.25:1

通过百度得到一个脚本来生成3D图片：

from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
dot1 = [[1, 3, 1], [1.25, 3, 1], [1.5, 3, 1], [1.75, 3, 1], [2, 3, 1], [2, 2.75, 1], [2, 2.5, 1], [2, 2.25, 1], [2, 2, 1], [2, 1.75, 1], [2, 1.5, 1], [1, 2.25, 1], [1.25, 2.25, 1], [1.5, 2.25, 1], [1.75, 2.25, 1], [1, 1.5, 1], [1.25, 1.5, 1], [1.5, 1.5, 1], [1.75, 1.5, 1], [3, 3, 1], [3.25, 3, 1], [3.5, 3, 1], [3.75, 3, 1], [4, 3, 1], [3.25, 2.25, 1], [3.5, 2.25, 1], [3.75, 2.25, 1], [4, 2.25, 1], [4, 2, 1], [4, 1.75, 1], [4, 1.5, 1], [3, 1.5, 1], [3.25, 1.5, 1], [3.5, 1.5, 1], [3.75, 1.5, 1], [3, 1.75, 1], [3, 2, 1], [3, 2.25, 1], [3, 2.5, 1], [3, 2.75, 1], [5, 3, 1], [5.25, 3, 1], [5.5, 3, 1], [5.75, 3, 1], [6, 3, 1], [6, 2.25, 1], [6, 2, 1], [6, 1.75, 1], [6, 1.5, 1], [5.75, 1.5, 1], [5.5, 1.5, 1], [5.25, 1.5, 1], [5, 1.5, 1], [5, 2.25, 1], [5.25, 2.25, 1], [5.5, 2.25, 1], [5.75, 2.25, 1], [5, 2.5, 1], [5, 2.75, 1], [7, 3, 1], [7.25, 3, 1], [7.5, 3, 1], [7.75, 3, 1], [8, 3, 1], [8, 2.75, 1], [8, 2.5, 1], [8, 2.25, 1], [8, 2, 1], [8, 1.75, 1], [8, 1.5, 1], [9, 3, 1], [9.25, 3, 1], [9.5, 3, 1], [9.75, 3, 1], [10, 3, 1], [10, 2.75, 1], [10, 2.5, 1], [10, 2.25, 1], [9.75, 2.25, 1], [9.5, 2.25, 1], [9.25, 2.25, 1], [9, 2.25, 1], [9, 2, 1], [9, 1.75, 1], [9, 1.5, 1], [9.25, 1.5, 1], [9.5, 1.5, 1], [9.75, 1.5, 1], [10, 1.5, 1], [11, 3, 1], [11.25, 3, 1], [11.5, 3, 1], [11.75, 3, 1], [12, 3, 1], [12, 2.75, 1], [12, 2.5, 1], [12, 2.25, 1], [12, 2, 1], [12, 1.75, 1], [12, 1.5, 1], [11.75, 1.5, 1], [11.5, 1.5, 1], [11.25, 1.5, 1], [11, 1.5, 1], [11, 1.75, 1], [11, 2, 1], [11, 2.25, 1], [11, 2.5, 1], [11, 2.75, 1], [11.25, 2.25, 1], [11.5, 2.25, 1], [11.75, 2.25, 1]]  # 得到五个点
plt.figure()  # 得到画面
ax1 = plt.axes(projection='3d')
ax1.set_xlim(0, 15)  # X轴，横向向右方向
ax1.set_ylim(15, 0)  # Y轴,左向与X,Z轴互为垂直
ax1.set_zlim(0, 15)  # 竖向为Z轴
color1 = ['r', 'g', 'b', 'k', 'm']
marker1 = ['o', 'v', '1', 's', 'H']
i = 0
for x in dot1:
    ax1.scatter(x[0], x[1], x[2], c=color1[1],
                marker=marker1[1], linewidths=4)  # 用散点函数画点
    i += 1
plt.show()

运行脚本后得到：

 

 得到密码口令：365728

 

 在解密后的压缩包内发现pdf，打开后在其中发现两行类似摩斯密码的段落，尝试解码

CONGRATULATIONTHEFLAGICHALLENGEISCCTWOZEROTWOONE

得出flag