/archives/278/ 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... /archives/278/#comment-118 2020-04-07T09:46:09+08:00 sto driver /archives/278/#comment-112 2020-03-10T12:29:23+08:00 orz mcfx god /archives/278/#comment-111 2020-03-04T00:45:10+08:00 I didn't know A* algo, check it rn! Thank you so much! /archives/278/#comment-109 2020-03-01T04:24:02+08:00 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 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 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 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 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