uva 10006 Carmichael

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：

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cmath>
 4 using namespace std;
 5
 6 typedef long long ll;
 7 const int N = 65001;
 8 bool visit[N];
 9
10 void sieve( int n )
11 {
12     int r = sqrt( n + 0.5 );
13     memset( visit, 0, sizeof(visit) );
14     visit[0] = visit[1] = 1;
15     for ( int i = 2; i <= r; i++ )
16     {
17         if ( !visit[i] )
18         {
19             for ( int j = i * i; j <= n; j += i )
20             {
21                 visit[j] = 1;
22             }
23         }
24     }
25 }
26
27 ll pow_mod( ll a, ll n, ll mod )
28 {
29     ll ans = 1, w = a % mod;
30     while ( n )
31     {
32         if ( n & 1 )
33         {
34             ans = ans * w % mod;
35         }
36         w = w * w % mod;
37         n >>= 1;
38     }
39     return ans;
40 }
41
42 int main ()
43 {
44     sieve( N - 1 );
45     int n;
46     while ( cin >> n, n )
47     {
48         bool flag = true;
49         if ( visit[n] )
50         {
51             for ( int x = 2; x < n; x++ )
52             {
53                 int tmp = pow_mod( x, n, n );
54                 if ( tmp != x )
55                 {
56                     flag = false;
57                     break;
58                 }
59             }
60         }
61         else
62         {
63             flag = false;
64         }
65         if ( flag )
66         {
67             cout << "The number " << n << " is a Carmichael number." << endl;
68         }
69         else
70         {
71             cout << n << " is normal." << endl;
72         }
73     }
74     return 0;
75 }