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2018安恒杯9月misc1 crc32碰撞

ctf
字数统计: 248阅读时长: 1 min
2018/09/23 Share

压缩包打开看crc32

前面三个文本合起来就是压缩包密码

通过crc去还原文本内容

每一个都有很多种可能

通过python整合一下

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list1=[
"05J728",
"0EvF7h",
"2ysXnu",
"3y2iul",
"R9DrOf",
"WQkoQX",
"avuKGt",
"dO875V",
"dSwk4B",
"forum_",
"go3DvF",
"ldpDP2",
"r6wKtc",
"s66zoz",
"yQGfVS"
]
list2=[
"2VSYDo",
"5OTgnD",
"7sQy7Y",
"91ctf_",
"AVfsVk",
"N5K_u8",
"OYyCje",
"PgLPQi",
"aYUJmn",
"c425Xo",
"cePT4s",
"d1zWsP",
"pt05kx",
"rTzw3q"
]
list3=[
"1Atmmb",
"6XsSGI",
"EXFyUM",
"KWYIiC",
"Qm0jH5",
"Spkdxd",
"TilZRO",
"Uub7HB",
"ZfsjnA",
"cN3O_z",
"com_66",
"dW4quQ",
"kXjpRF",
"lAmNxm",
"n0EmLx",
"r24Q5p",
"rcV0Yl",
"tf_ciJ"
]
for i in list1:
	for j in list2:
		for n in list3:
			print(i+j+n)

然后去爆破:

压缩包打开:
convert.txt里面一串二进制:
用jpocketknife转换成ascii:

是个二维码,扫一下就是flag

原文作者：醉梦半醒

原文链接：https://www.iloveflag.com/2018/09/23/%E5%AE%89%E6%81%92%E6%9D%AF9%E6%9C%88misc1%20crc32%E7%A2%B0%E6%92%9E/

发表日期：September 23rd 2018, 4:01:41 am

更新日期：August 24th 2020, 3:44:34 pm

版权声明：本文采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可

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