uva 10006 快速幂

比较基本的数论题目，不过《挑战程序设计》上说这个题有不用幂运算求解的两种方法，一种复杂度为根号n，一种是O(n)预处理，O(1)判定，我还没有想出来....

快速幂的方法：

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long ll;
const int N = 65001;
bool visit[N];

void sieve( int n )
{
    int r = sqrt( n + 0.5 );
    memset( visit, 0, sizeof(visit) );
    visit[0] = visit[1] = 1;
    for ( int i = 2; i <= r; i++ )
    {
        if ( !visit[i] )
        {
            for ( int j = i * i; j <= n; j += i )
            {
                visit[j] = 1;
            }
        }
    }
}

ll pow_mod( ll a, ll n, ll mod )
{
    ll ans = 1, w = a % mod;
    while ( n )
    {
        if ( n & 1 )
        {
            ans = ans * w % mod;
        }
        w = w * w % mod;
        n >>= 1;
    }
    return ans;
}

int main ()
{
    sieve( N - 1 );
    int n;
    while ( cin >> n, n )
    {
        bool flag = true;
        if ( visit[n] )
        {
            for ( int x = 2; x < n; x++ )
            {
                int tmp = pow_mod( x, n, n );
                if ( tmp != x )
                {
                    flag = false;
                    break;
                }
            }
        }
        else
        {
            flag = false;
        }
        if ( flag )
        {
            cout << "The number " << n << " is a Carmichael number." << endl;
        }
        else
        {
            cout << n << " is normal." << endl;
        }
    }
    return 0;
}