https://github.com/pandada8/solvingwechall
https://github.com/pandada8/solvingwechall
Last synced: about 2 months ago
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- Host: GitHub
- URL: https://github.com/pandada8/solvingwechall
- Owner: pandada8
- Created: 2015-03-01T11:08:35.000Z (about 10 years ago)
- Default Branch: master
- Last Pushed: 2015-03-01T12:26:17.000Z (about 10 years ago)
- Last Synced: 2025-02-15T18:38:35.036Z (3 months ago)
- Language: Python
- Size: 102 KB
- Stars: 0
- Watchers: 2
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# 个人We Chall 解题记录
## Chall 1
拿筛法筛点素数，然后循环判断就好
## Chall 2
右键查看源代码，最下方就是 Key
## Chall 3
`strings stegano1.bmp` 即为答案
## Chall 4
经典的凯撒加密，直接对每个进行下移位处理就行，找到一个通顺的语句就是结果
## Chall 5
根据提示查看`http://www.wechall.net/robots.txt`,访问其中的 `/challenge/training/www/robots/T0PS3CR3T` 即可
## Chall 6
Python 一行流
```Python
"".join([chr(int(i)) for i in "84, 104, 101, 32, 115, 111, 108, 117, 116, 105, 111, 110, 32, 105, 115, 58, 32, 111, 115, 108, 101, 97, 101, 114, 108, 109, 110, 108, 105".split(', ')])
```
## Chall 7
Python 一行流
```Python
from urllib.parse import unquote
unquote("%59%69%70%70%65%68%21%20%59%6F%75%72%20%55%52%4C%20%69%73%20%63%68%61%6C%6C%65%6E%67%65%2F%74%72%61%69%6E%69%6E%67%2F%65%6E%63%6F%64%69%6E%67%73%2F%75%72%6C%2F%73%61%77%5F%6C%6F%74%69%6F%6E%2E%70%68%70%3F%70%3D%62%67%68%65%65%66%73%6F%6D%61%73%73%26%63%69%64%3D%35%32%23%70%61%73%73%77%6F%72%64%3D%66%69%62%72%65%5F%6F%70%74%69%63%73%20%56%65%72%79%20%77%65%6C%6C%20%64%6F%6E%65%21")
```
访问给出的地址即可
## Chall 8
注意到题目中给出的数据一共441个，是21的平方，脑洞再开大一点，441是7的倍数，而 ASCII 码的最高位留空，刚好为7位，所以写几行 Python 脚本解码如下:
```Python
text = """10101001101000110100111100110100
00011101001100101111100011101000
10000011010011110011010000001101
11010110111000101101001111010001
00000110010111011101100011110111
11100100110010111001000100000110
00011110011110001111010011101001
01011100100000101100111011111110
10111100100100000111000011000011
11001111100111110111110111111100
10110010001000001101001111001101
00000110010111000011110011111100
11110011111010011000011110010111
0100110010111100100101110"""
text = "".join(text.split())
for i in range(int(len(text) / 7)):
print(chr(int(text[i * 7:i*7 + 7], 2)), end="")
print()
```
## Chall 9
使用 requests 库