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偶是这样做的。。。。In Reply To:就是四川大学上的那个power and power again吧..不过我的方法比较慢... Posted by:achilles at 2005-12-08 01:00:23 llong f(llong a, llong n, llong m)
{
if (m == 0)
{
return 1;
}
if (n == 1)
{
return a % m;
}
int e = euler( m );
int nn = f(a, n-1, e);
if (nn == 0) nn = e;
return mod_pow(a, nn, m);
}
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