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低指数加密广播攻击 e选取的很小 对同一个明文 M 进行了多次加密 且知道 Ni 和 Ci 用中国剩余定理 结合 ni ci 算出 m^e 然后用小指数 e 开方即可 from Crypto.Util.number import * import primefac import gmpy2 e = 阅读全文
posted @ 2023-09-17 13:19
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An interesting problem. First we are come up with this .txt: 72143238992041641000000.000000, 77135357178006504000000000000000.000000, 1125868345616435 阅读全文
posted @ 2023-09-15 16:48
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It is a problem to learn how to solve AES's problem. In AES, we use key and vector iv and the mode(CBC)to encrypt the flag. We should first search for 阅读全文
posted @ 2023-09-15 16:15
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The source code: import sympy from gmpy2 import gcd, invert from random import randint from Crypto.Util.number import getPrime, isPrime, getRandomNBit 阅读全文
posted @ 2023-09-14 21:26
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We have n,e,dp,c. Assume that it is quite difficult to factorize n, but it can be easily divided in fact. What can we do with dp leak? dp ≡ d mod (p-1 阅读全文
posted @ 2023-09-14 14:37
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This is the problem that requires some math skills. We have p,q,dp,dq,c. Now let's start to solve it. dp ≡ d mod(p-1) dq ≡ d mod(q-1) m ≡ c^d mod n m 阅读全文
posted @ 2023-09-14 13:37
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The solution is based on the fact that we have got the source code.However, how could I get the source code? So, this one is to broaden my web knowled 阅读全文
posted @ 2023-09-14 09:00
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Assume that we have 2 groups of data, we find that n is the same and m also is. What can we do to work out m without factorize n?(actually we cannot f 阅读全文
posted @ 2023-09-13 23:01
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To review some types of problems solved today. 1. Just using brute force to factorize the n, as long as we find the p and q, we can easily calculate t 阅读全文
posted @ 2023-09-12 22:19
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当e很小 n很大的时候 直接枚举开方即可 from Crypto.Util.number import * import gmpy2 from gmpy2 import * import primefac def modinv(a,n): return primefac.modinv(a,n) % 阅读全文
posted @ 2023-09-12 08:11
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