得到一个T4.pyc
回编译一下
得到下面代码
python
import base64
def encrypt_and_compare(user_input, offset_str, target_base64):
if len(user_input) != 24:
return 'Please enter a string with a length of 24'
encrypted = None
for i, char in enumerate(user_input):
offset = int(offset_str[i])
ascii_val = ord(char)
if i % 2 == 0:
new_ascii = ascii_val + offset
else:
new_ascii = ascii_val - offset
encrypted_char = chr(new_ascii ^ offset)
encrypted.append(encrypted_char)
encrypted_bytes = ''.join(encrypted).encode('utf-8')
encrypted_base64 = base64.b64encode(encrypted_bytes).decode('utf-8')
print('Encrypted result:{}'.format(encrypted_base64))
if encrypted_base64 == target_base64:
return 'Find key'
return None
offset_str = '123456789012345678901234'
target_base64 = 'TWF/c1sse19GMW5gYVRoWWFrZ3lhd0B9'
user_input = input('Please enter a string with a length of 24:')
result = encrypt_and_compare(user_input, offset_str, target_base64)
print(result)
很简单逻辑,你解密出来就是第一段密码(这个我不想再写了,相信各位都会,不会私信我,嘿嘿)
第一段密码拿去解压缩出来
就有这些东西
这是一个处理png文件的模型
我让AI修改一下(我肯定是不会的)
python
import torch
from torch import nn
from torchvision import transforms
from PIL import Image
import os
# 定义网络结构
class Net(nn.Module):
def __init__(self):
super(Net, self).__init__()
self.fc1 = nn.Linear(28 * 28, 128)
self.fc2 = nn.Linear(128, 64)
self.fc3 = nn.Linear(64, 10)
def forward(self, x):
x = x.view(-1, 28 * 28)
x = torch.relu(self.fc1(x))
x = torch.relu(self.fc2(x))
x = self.fc3(x)
return x
# 直接加载整个模型实例
model_instance = torch.load('D:/CTF/problem/iscc/AI/AI-23/confused_digit_recognition_model.pt', map_location='cpu')
# 确保模型处于评估模式
model_instance.eval()
# 定义图像预处理变换
transform = transforms.Compose([
transforms.Grayscale(num_output_channels=1),
transforms.Resize((28, 28)),
transforms.ToTensor(),
transforms.Normalize((0.5,), (0.5,))
])
# 图像文件所在目录
image_folder = "D:\\CTF\\problem\\iscc\\AI\\AI-23\\"
# 遍历编号从1到24的图像文件并进行预测
for i in range(1, 25):
# 构建完整的图像路径
image_path = os.path.join(image_folder, f"{i}.png")
try:
# 加载并预处理图像
image = Image.open(image_path)
image = transform(image)
image = image.unsqueeze(0)
# 模型预测
with torch.no_grad():
outputs = model_instance(image)
_, predicted = torch.max(outputs.data, 1)
# 打印预测结果
print( predicted.item(),end='')
except IOError as e:
print(f"Error opening {image_path}: {e}")
except Exception as e:
print(f"An error occurred during prediction for {image_path}: {e}")
得到第二段密码
这就是对照表的内容啦
然后写个脚本就OK
python
def toint(a):
b=ord(a)-ord('0')
return b
a='384362683985682257091427'
for i in range(24):
enc=toint(a[i])
if(enc==0):
print('@nd',end='')
elif(enc==1):
print('a!',end='')
elif (enc == 2):
print('_',end='')
elif (enc == 3):
print('F',end='')
elif (enc == 4):
print('SSS',end='')
elif (enc == 5):
print('W@',end='')
elif (enc == 6):
print('K',end='')
elif (enc == 7):
print('1',end='')
elif (enc == 8):
print('C',end='')
elif (enc == 9):
print('d',end='')