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不一定就是lg(b)了。举个例子(虽然是特殊情况)当gcd(a,m)=1时,a^b mod m = a^(b%(phi(m))) mod m。这样只需要lg(m)的复杂度In Reply To:Re:可以的,复杂度和输入的运符个数有关,和怎么组合无关 Posted by:xwbsw at 2007-10-07 21:01:16 > 求 a^b 本身复杂度就是lg(b)的,前面又要乘大数运算的系数,怎么能说完全无关? Followed by:
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