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[原创]看雪.深信服 2021 KCTF 春季赛 第二题 南冥神功 WP
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发表于: 2021-5-11 15:45
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此题应该属于迷宫类的题目。要求是从开始处走遍预定的路径，且走过的路径点不能重复走，相邻两次的位置行和列的变化均不能超过1格。每个输入字符可以走迷宫两步。思路是通过找到迷宫路径，然后反推输入，最后得到flag。
迷宫是9*10的：
0x53是开始处，0的表示可以走的路径点。
不难看出，有效路径为：
反算就简单了：
0x53, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01,
0x01, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00,
0x00, 0x00, 0x01, 0x00, 0x01, 0x01, 0x01, 0x01, 0x01, 0x00,
0x00, 0x01, 0x01, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00,
0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01,
0x01, 0x01, 0x00, 0x01, 0x01, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x00, 0x01, 0x01, 0x01, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x01, 0x01, 0x00, 0x00, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01, 0x00, 0x00
0x53, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01,
0x01, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00,
0x00, 0x00, 0x01, 0x00, 0x01, 0x01, 0x01, 0x01, 0x01, 0x00,
0x00, 0x01, 0x01, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00,
0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01,
0x01, 0x01, 0x00, 0x01, 0x01, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x00, 0x01, 0x01, 0x01, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x01, 0x01, 0x00, 0x00, 0x01, 0x00, 0x01, 0x00, 0x01,
0x00, 0x00, 0x00, 0x01, 0x00, 0x00, 0x01, 0x01, 0x00, 0x00
table = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
row = 0
col = 0
path = [(0,1),(1,2),(2,1),(2,0),(3,0),(4,0),(4,1),(5,2),(6,1),(6,0),(7,0),(8,0),(8,1),(8,2),(7,3),(7,4),(8,4),(8,5),(7,6),(6,6),(5,6),(4,6),(3,6),(3,5),(4,4),(4,3),(3,3),(2,3),(1,3),(0,3),(0,4),(1,5),(1,6),(0,6),(0,7),(1,8),(1,9),(2,9),(3,9),(3,8),(4,7),(5,8),(6,8),(7,8),(8,8),(8,9),]
def get_choice(r,c):
global row,col
if r-row == 0:
if c-col == -1:
row,col = r,c
return 4
elif c-col == 1:
row,col = r,c
return 1
elif r-row == -1:
if c == col and row&1 == 0:
row,col = r,c
return 5
elif c == col and row&1 == 1:
row,col = r,c
return 0
elif c != col and row&1 == 1:
row,col = r,c
return 5
elif c != col and row&1 == 0:
row,col = r,c
return 0
elif r-row == 1:
if c == col and row&1 == 1:
row,col = r,c
return 2
elif c == col and row&1 == 0:
row,col = r,c
return 3
elif c != col and row&1 == 0:
row,col = r,c
return 2
elif c != col and row&1 == 1:
row,col = r,c
return 3
else:
print 'err'
def run(idx,n1,n2):
for i in range(36):
if (idx+i/6)%6 == n2 and 5-((idx+i)%6) == n1:
return table[i]
def main():
l = []
for r,c in path:
l.append(get_choice(r,c))
flag = ''
for i in range(len(l)/2):
flag += run(i,l[2*i],l[2*i+1])
print flag
if __name__ == '__main__':
main()
table = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
row = 0
col = 0
path = [(0,1),(1,2),(2,1),(2,0),(3,0),(4,0),(4,1),(5,2),(6,1),(6,0),(7,0),(8,0),(8,1),(8,2),(7,3),(7,4),(8,4),(8,5),(7,6),(6,6),(5,6),(4,6),(3,6),(3,5),(4,4),(4,3),(3,3),(2,3),(1,3),(0,3),(0,4),(1,5),(1,6),(0,6),(0,7),(1,8),(1,9),(2,9),(3,9),(3,8),(4,7),(5,8),(6,8),(7,8),(8,8),(8,9),]
def get_choice(r,c):
global row,col
if r-row == 0:
if c-col == -1:
row,col = r,c
return 4
elif c-col == 1:
row,col = r,c
return 1
elif r-row == -1:
if c == col and row&1 == 0:
row,col = r,c
return 5
elif c == col and row&1 == 1:
row,col = r,c
return 0
elif c != col and row&1 == 1:
row,col = r,c
return 5
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