题目
#
p = 0xb6ce21ea137206f9ac1a0e9a004457b090a7fd1745026ba230bd6e33d134eebf8498d98e061be16bc5cb3dfa1e24adc9fbf4214af86aaf6100a064dd6bb0d737ff3a38ac274487a1bf2a6a736fd03fa783b84399d65c79d64c19090b9adf629e55b40e735320535e09684d830432c53e740ba62da8498022820b89aec87b8941L
q = 0xf77307cf5c1bc4f5e45b6515d2de47ce64e9bfc1b3362391a8a791063dcd6c7711c5f1397780abcfb8eaa1201ab82f87e5b314c57fefbfb4f7b2b99a42f93918fd12cb039b50e1ba38bf86f8efc40fe60cc2e57eafce3f3597d6ed8b939d988a34dcadac6b394bf447ce5024d2083dc12b7f1ccd73073f0af70943cf2f133defL
e = 0x10001
c = 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
思路
#
答案
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import libnum
p = 0xb6ce21ea137206f9ac1a0e9a004457b090a7fd1745026ba230bd6e33d134eebf8498d98e061be16bc5cb3dfa1e24adc9fbf4214af86aaf6100a064dd6bb0d737ff3a38ac274487a1bf2a6a736fd03fa783b84399d65c79d64c19090b9adf629e55b40e735320535e09684d830432c53e740ba62da8498022820b89aec87b8941
q = 0xf77307cf5c1bc4f5e45b6515d2de47ce64e9bfc1b3362391a8a791063dcd6c7711c5f1397780abcfb8eaa1201ab82f87e5b314c57fefbfb4f7b2b99a42f93918fd12cb039b50e1ba38bf86f8efc40fe60cc2e57eafce3f3597d6ed8b939d988a34dcadac6b394bf447ce5024d2083dc12b7f1ccd73073f0af70943cf2f133def
e = 0x10001
c = 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
n = p*q
phi_n = (p-1)*(q-1)
d = libnum.invmod(e, phi_n)
m = pow(c,d,n)
print(libnum.n2s(m))
# flag{we1c0me_t0_Chainer_s_training_r00m}