lisanduishu

  设p是素数，a是p的本原根，即a1,a2,a3,…,ap-1在mod p下产生1到p-1的所有值，所以对任何b属于{1,…,p-1},有唯一的i属于{1,…,p-1}使得ai mod p 等于p。称i为模p下以a为底b的离散对数。(P is a prime number based, a is the primitive root p, that is, a1, a2, a3, ..., ap-1 in the mod p under 1 to p-1 for all values, so for any b belongs to (1, ..., p-1), there are only i belong to (1, ..., p-1) makes ai mod p equivalent to p. Said i was under the modulus p to a for b at the end of the discrete logarithm.)