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License: Other
Write-ups for hackergame 2019
License: Other
也许可以通过 sagemath 的 nth_root() 实现?
的确,ffmpeg可以提取好几张关键帧,但是最后一张八成是提取不出来的
怎么办呢?
推测最后一张在视频结尾附近,于是整点Vegas之类的视频编辑软件,然后看结尾时段的缩略图就行了
因为这种软件的缩略图选取算法非常精彩,所以高概率会直接把白底黑字那一帧挑出来显示
至少咱是这么找到的
等一下,主要贡献者之一也是救世主?????
ffmpeg -i output.mp4 -vf "select='if(gt(scene,0.01),st(1,t),lte(t-ld(1),1))',setpts=N/FRAME_RATE/TB" out2.mp4
(我不会ffmpeg,这个是百度的如何去除重复帧的。。
v = -1900000000000000000
不等于64位溢出-(2**64+1) =-18446744073709551617
用Chrome打开cat.png
生成均值图像用的是np.ndarray.astype(np.uint8)
,做的是向下取整。而题解中却用round
四舍五入来解释
题目中拟合的形式是y=k0*x0+k1*x1+...+k49999*x49999
,不存在偏置项b
,那么用下面不带偏置项的回归器应该更科学一点
reg = linear_model.Lasso(alpha=1, positive=True, fit_intercept=False)
但是,实际效果却变差了很多,通不过官方给的测试。运行一个更弱的条件,发现有四张正确图片的系数是0
[x for x in choices if x not in np.argwhere(reg.coef_>0).reshape(-1)] # [4303, 24496, 36462, 39326]
这真是一个玄学题目😂
把这行
user = authenticate(request, username=username, password=password)
改成
user = User.objects.get(username='admin')
即可以任意用户名密码登录成admin
解这道题大概用了一秒吧(逃
深色主题的windows10,打开直接就看到flag了23333
虽然的确简单,但是wp里一句话略过还是有点点草率
这里提供一个闭着眼睛两分钟敲出来的解题脚本的核心部分
char temp_password[9];
char Upassword[] = {68, 0, 2, 65, 67, 71, 16, 99, 0};
temp_password[8] = '\x00';
int i;
for (i = 7; i >= 0; --i) {
for(int ii = 48;ii <= 123;ii++) {
temp_password[i] = ii;
temp_password[i] =
((temp_password[i] | temp_password[i + 1]) & ~(temp_password[i] & temp_password[i + 1]) | i) &
~((temp_password[i] | temp_password[i + 1]) & ~(temp_password[i] & temp_password[i + 1]) & i);
if (temp_password[i] == Upassword[i]) {
temp_password[i] = ii;
printf("%c", ii);
break;
}
}
}
那么它最主要地 有什么不好呢?
输出的结果是反着的
win10 1903,当时想直接改时间的时候发现只能设置到2049年😂
三教奇妙夜用了ffmpeg做的,十二个小时的视屏,解帧到11小时30分左右突然开始变慢,一帧一帧解,不是关键帧,就停了
用Duplicate Cleaner找出了所有不同的文件,最后只剩一百多张了,flag找的也差不多了,
后来把最后三十分钟的视频剪出来又继续解帧,
(有几百万张图片吧,解了两个多小时,心疼我的硬盘.....)
嗯,全程大概三个小时吧
愚蠢解法
之前没有看到body上的style,就直接把图片添加负片效果了😂
img {
filter: invert(1)
}
然后白猫就变成一副很诡异的样子😂
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