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reverse_re3-入土为安的第十天

news 2025/8/4 0:34:06

一个迷宫题

shirt+f12发现flag字样

点进去发现迷宫

675 应该是3 * 15*15

即有三个迷宫

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import numpy as np# 你的数据（长度应为 225 的倍数）
dword_202020 = [1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,3,1,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,1,1,1,1,1,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,0,0,0,0,0,0,0,0,4,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,3,1,1,1,1,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1,1,1,1,1,0,0,1,1,0,1,1,0,0,0,0,0,0,0,1,0,0,1,1,0,1,1,0,0,0,0,0,0,0,1,0,0,1,1,0,1,1,0,0,0,0,0,1,1,1,1,0,1,1,0,1,1,0,0,0,0,0,1,0,0,1,0,1,1,0,1,1,0,0,0,0,0,1,0,0,0,0,1,1,0,1,1,1,1,1,1,0,1,0,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0,0,4,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,1,1,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0]# 矩阵的大小
n = 15
matrix_size = n * n# 检查数组长度
length = len(dword_202020)# 如果数据长度不足，填充到最近的矩阵大小的整数倍
if length < matrix_size:print(f"数据长度不足 ({length}), 将数据填充至 {matrix_size}")dword_202020 = dword_202020 + [0] * (matrix_size - length)  # 用 0 填充到最近的矩阵大小
elif length % matrix_size != 0:# 如果数据长度超过要求，截取到最近的矩阵大小的整数倍new_length = (length // matrix_size) * matrix_sizeprint(f"数据长度超过 ({length}), 截取至 {new_length}")dword_202020 = dword_202020[:new_length]# 转换为 numpy 数组
array = np.array(dword_202020)# 计算每个矩阵的起始索引
matrices = [array[i:i + matrix_size].reshape(n, n) for i in range(0, len(array), matrix_size)]# 打印矩阵
for idx, mat in enumerate(matrices):print(f"Matrix {idx + 1}:\n{mat}\n")

输出：

Matrix 1:
[[1 1 1 1 1 0 0 0 0 0 0 0 0 0 0]
[1 1 1 1 1 0 3 1 1 0 0 0 0 0 0]
[1 1 1 1 1 0 0 0 1 0 0 0 0 0 0]
[1 1 1 1 1 0 0 0 1 0 0 0 0 0 0]
[1 1 1 1 1 0 0 0 1 1 1 1 1 0 0]
[1 1 1 1 1 0 0 0 0 0 0 0 1 0 0]
[1 1 1 1 1 0 0 0 0 0 0 0 1 0 0]
[1 1 1 1 1 0 0 0 0 0 0 0 1 1 0]
[1 1 1 1 1 0 0 0 0 0 0 0 0 1 0]
[1 1 1 1 1 0 0 0 0 0 0 0 0 4 0]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]]

Matrix 2:
[[1 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 1 0 3 1 1 1 1 1 0 0 0 0 0 0]
[1 1 0 1 1 0 0 0 1 0 0 0 0 0 0]
[1 1 0 0 0 0 0 0 1 0 0 0 0 0 0]
[1 1 0 1 1 0 0 0 1 1 1 1 1 0 0]
[1 1 0 1 1 0 0 0 0 0 0 0 1 0 0]
[1 1 0 1 1 0 0 0 0 0 0 0 1 0 0]
[1 1 0 1 1 0 0 0 0 0 1 1 1 1 0]
[1 1 0 1 1 0 0 0 0 0 1 0 0 1 0]
[1 1 0 1 1 0 0 0 0 0 1 0 0 0 0]
[1 1 0 1 1 1 1 1 1 0 1 0 1 1 0]
[1 1 0 1 1 1 1 1 1 1 1 1 1 1 0]
[1 1 0 0 0 0 0 0 0 0 0 0 0 4 0]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]]

Matrix 3:
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 3 1 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 1 0 1 1 1 0 0 0 0 0 0 0]
[0 0 0 1 1 1 0 1 0 0 0 0 0 0 0]
[0 0 0 0 1 0 0 1 0 0 0 0 0 0 0]
[0 1 1 0 1 0 0 1 0 0 0 0 0 0 0]
[0 0 1 1 1 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 1 1 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 1 1 1 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 4 0]]
3是起点，4是终点

上下左右 wsad

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32小位