CTFshow萌新赛-抱我

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CTFshow萌新赛-抱我

condor2048

题目附件 flag.py

cstring = 'abcdefghijklmnopqrstuvwxyz{}_0123456789'
key = 'flag{********}'
length = 300

def encode():
    res = ''
    for i in range(1, length):
        c = random.randint(0, 36)
        res += cstring[c]
        for n in range(10):
            c = random.randint(0, len(key) - 1)
            res += key[c]
    return res

分析加密代码可知：
输出串长(299) * (10 + 1)
res[i]为混淆码,(i % 11==0)
res[i]为flag内字符,(i % 11 != 0)

思路：
先根据下标筛掉混淆码
res内剩余的字符串可认为是完全无序的flag，所以直接上集合统计字符集
运行程序获得结果：gla3f{6}d

脑补模式，启动！
flag{36d}

代码如下

src = 'qdfl33(省略......)g{6lgf3f'
flag_random = ''
for i in range(len(src)):
    if i % 11 != 0:
        flag_random += src[i]

st = set()

for i in range(len(flag_random)):
    st.add(flag_random[i])

for i in st:
    print(i, end='')