Z2kDHCrypto 未解决
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一 血:
晨曦
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解 决:
9
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描 述:
Or, more precisely: $\mathbb{Z}/2^k\mathbb{Z}^*$-Diffie-Hellman
Sick of how slow it is to take the modulus of a large prime number?
Tired of performing exponentiation over some weird group like ed25519?
Just use the integers mod 2^k group!
Efficient, reliable, doesn't require any hard math!
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buukuu2024 8天前
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wctf{P0HL1G_H3LLM4N_$M4LL_pr1M3}
wsmemail 5月前
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思路是什么?求教