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摘要： 解码 gcd (x, n) = 1 => gcd(x^c, n) = 1 => gcd(x^c % n, n) = gcd(x^c, n) = 1 => gcd(m, n) = 1 符合欧拉定理的条件， => m ^ φ(n) ≡ 1 (mod n) => if (k·c ≡ 1 (mod φ (n 阅读全文