[GDOUCTF 2023]doublegame wp
一个游戏为贪吃蛇，另一个游戏maze
在string里面能够直接看到迷宫
点击查看代码
000000000000000000000
0 0 0 0 0 0 0
0 0 0 00000 00000 0 0
0 0 0 0
0 000 000 0 000 0 0 0
0 0 0 0 0 0 0 0
0 0 0 00000 000 000 0
0 0 0 0 0 0
0 000 0 0 000 0 0 0 0
0 0 0 0 0 0 0 0 0
0 00000 000 000 0 0 0
0 0 0 0 0
000 0 0 0 000 0 0 0 0
0 0 0 0 0 0 * 0 0 0 0
0 0000000 0 000 00000
@ 0 0 0
0 0 0 0 0 00000000000
0 0 0 0 0
000 0 00000 0 000 000
0 0 0 0 0
000000000000000000000
但是这里面有一个坑
虽然迷宫的地图已经出来了，但是下面还有一个v14[4] = '0';
所以真实的地图应该为
点击查看代码
000000000000000000000
0 0 0 0 0 0 0
0 0 0 00000 00000 0 0
0 0 0 0
0 000 000 0 000 0 0 0
0 0 0 0 0 0 0 0
0 0 0 00000 000 000 0
0 0 0 0 0 0
0 000 0 0 000 0 0 0 0
0 0 0 0 0 0 0 0 0
0 00000 000 000 0 0 0
0 0 0 0 0
000 0 0 0 000 0 0 0 0
0 0 0 0 0 0 * 0 0 0 0
0 0000000 0 000 00000
@ 0 0 0 0
0 0 0 0 0 00000000000
0 0 0 0 0
000 0 00000 0 000 000
0 0 0 0 0
000000000000000000000
但是对于这道题想要获取真是的迷宫还可以用CE调试，将分数修改为大于13371337，就能够直接获取
CE使用学习：https://www.cnblogs.com/LyShark/p/10799926.html
从@走到*救出小猫，然后走到出口
路径脚本
点击查看代码
from collections import deque
# 设置二维四向迷宫, 如果题目是多个小迷宫问题, 拆分多次调用脚本获取路径即可
# 每一排后面要逗号隔开，对齐++++++++++++++++++++++++++
maze = [
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1],
[1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1],
[1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]
path_len = 0x7fffffff # 如果题目未给出终点坐标，则一定会指定路径的长度，在此处修改路径长度，否则请保留path_len的极大值
#已知步长时更改，否则不管+++++++++++++++++++++++++++++++++
# 进行BFS寻找路径
def bfs(start, end, barrier):
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)] # 定义四个方向的移动
for i in range(len(maze)): # 获取起点和终点在列表中的索引
for j in range(len(maze[i])):
if (maze[i][j] == start):
start = (i, j)
if (maze[i][j] == end):
end = (i, j)
# 以下均是bfs算法套路
queue = deque()
queue.append((start, [start])) # (当前位置, 路径)
visited = set()
visited.add(start)
while queue:
position, path = queue.popleft()
if position == end:
return path
elif len(path) == path_len:
return path
for d in directions:
next_position = (position[0] + d[0], position[1] + d[1])
if 0 <= next_position[0] < len(maze) and 0 <= next_position[1] < len(maze[0]) and \
maze[next_position[0]][next_position[1]] != barrier and next_position not in visited:
queue.append((next_position, path + [next_position]))
visited.add(next_position)
return None
# 执行BFS搜索并打印结果
if __name__ == '__main__':
#更改起点和终点的坐标，如果知道path而坐标未知则注释掉+++++++++++++++++++++++++++++
maze[15][0] = 'S' #如果题目给了起点终点的坐标，在这里直接给起点和终点添加特征
maze[7][20] = 'E'#前为横坐标，后为纵坐标
path = bfs('S', 'E', 1) # bfs函数传入参数代表起点、终点、障碍的特征(若题目给出的数据无特征, 手动添加特征即可, 通常障碍是1也有可能是0或其它字符如'#')
print("移动路径坐标：", path)
print("移动路径方位：{", end='')
for i in range(1, len(path)):
x1, y1, x2, y2 = path[i - 1][0], path[i - 1][1], path[i][0], path[i][1]
if (x1 > x2): # 上
print("w", end='')
elif (x1 < x2): # 下
print("s", end='')
elif (y1 > y2): # 左
print("a", end='')
elif (y1 < y2): # 右
print("d", end='')
print('}')
路径为dddssssddwwwwddssddwwwwwwddddssaassddddwwwwddwwwwddd
maze下面有score的检验
同时这个score也是打开门的key
异或
点击查看代码
print(0x1DC4 ^ 13376013)
#13371337
即score为13371337
整个游戏逻辑为：
先玩贪吃蛇玩到13371337进入走迷宫，这个score同时也是打开门的钥匙key，再输入路径救出小猫
flag格式为
HZCTF{md5(path)+score}
NSSCTF{811173b05afff098b4e0757962127eac13371337}
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