由于寒假比较闲，所以找点比赛打。由于需要上交 wp，所以是英文的。 # Crypto ## RSA is easy #1 Since $$N$$ is known, we can comp... 2020-04-07T09:46:09+08:00 Typecho /feed/atom/archives/278/ /archives/278/#comment-118 2020-04-07T09:46:09+08:00 2020-04-07T09:46:09+08:00 h1st https://h1st.github.io/

sto driver

]]> /archives/278/#comment-112 2020-03-10T12:29:23+08:00 2020-03-10T12:29:23+08:00 wxh010910

orz mcfx god

]]> /archives/278/#comment-111 2020-03-04T00:45:10+08:00 2020-03-04T00:45:10+08:00 Trung

I didn't know A* algo, check it rn! Thank you so much!

]]> /archives/278/#comment-109 2020-03-01T04:24:02+08:00 2020-03-01T04:24:02+08:00 admin http://www.sroof.pw

2. Since we can find these matched results, such bfs(or maybe A*) algorithm will lead to a low score situation, that means y is very large, and that's what we need for the answer.

]]> /archives/278/#comment-108 2020-03-01T04:22:14+08:00 2020-03-01T04:22:14+08:00 admin http://www.sroof.pw

In u(x,y), x is current length of input diff, y is current length of matched result, and u(x,y) gives a resonable score of them - to make the result match as much as possible.

]]> /archives/278/#comment-107 2020-02-17T16:12:58+08:00 2020-02-17T16:12:58+08:00 Trung

Hi there, I have read baby_bear writeup, but I can't understand somethings on your exploit code:
1.You use u(x,y) to calculate for the order of the heap, what is the base of that ?
2.How could you use bitxor to find next bytes? (algorithms or somethings else?)
Thanks for your help!
Sorry for those previous noised cmt.

]]> /archives/278/#comment-106 2020-02-17T15:52:29+08:00 2020-02-17T15:52:29+08:00 Trung

Hi there, I have read baby_bear writeup, but I can't understand somethings on your exploit code:
1. def u(x,y):
return x-y*2.5
You use u(x,y) to calculate for the order of the heap, what is the base of that ?
2. def bitxor(s,x):
xt=x>>3
return s[:xt]+bytes([s[xt]^(1 ]]> /archives/278/#comment-105 2020-02-17T15:37:20+08:00 2020-02-17T15:37:20+08:00 Trung

Hi there, I have read baby_bear writeup, but I can't understand somethings on your exploit code:
1. def u(x,y):
return x-y*2.5
You use u(x,y) to calculate for the order of the heap, what is the base of that ?
2. def bitxor(s,x):
xt=x>>3
return s[:xt]+bytes([s[xt]^(1 ]]>