西湖论剑——指鹿为马
前几天被hxd拉去看看题,其中MISC的指鹿为马很有意思,我来说说这题我的解法。
首先nc服务器down下了三个信息,首先是python的源代码
1 | import numpy as np |
马的base64
1 | 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 |
附转码后的图片(推荐个转码的网站:https://tool.jisuapi.com/base642pic.html)
鹿的base64
1 | 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 |
附转码后的图片
经过代码审阅,发现题目的要求是要你输入一张图片(这里称为test)的base64编码,通过他的条件就会得到flag。
我们发现在传入马和鹿和test时,load_horse(),load_deer(),load_test(),图片的尾部会被添加一个值,除了马是0,其他两张都是1,这是重点!!
在看输入图片后的判断流程,第一个条件check(),check()的作用是将test和鹿的像素差的绝对值相加,如果大于10w就不通过。第二个条件getNeighbors(),比较test和马和鹿的像素的欧式距离。如果test和马的欧氏距离最近才能通过。getResponse()的作用就是判断图片的尾部是否为0(感觉这代码也是作者从其他地方扒拉过来的,图片对抗)。
所以我们就理清思路了,我们要输入一张test,test的总像素差和鹿不能超过10w,并且和马的欧式距离最近。真是“指鹿为马”啊。
一开始我是想用PS将两个图片叠加再一起的,可是一直过不了,还尝试了几次不同比例叠加,也不行,听说有大佬用PS过了。我还是交给脚本吧···,我的思路是在鹿的基础上将马的像素点一个一个加上去,直到满足两个条件就停止。
1 | from PIL import Image |
然后就得到了一张test的图片(这是我跑完的最后一张),半马半鹿,妙啊。
提交成功得到flag
然后base64转码成txt后发现里面还有图片的base64,再转,才是最后的flag
解题就完成了,如果我讲的有哪里不对,请各位师傅多多包涵,也请评论点出,谢谢!