hdu 2035 快速幂

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int pow_mod( int a, int n, int m )
 5 {
 6     int ans = 1;
 7     a = a % m;
 8     while ( n )
 9     {
10         if ( n & 1 )
11         {
12             ans = ans * a % m;
13         }
14         a = a * a % m;
15         n = n >> 1;
16     }
17     return ans;
18 }
19 
20 int main ()
21 {
22     int a, b;
23     while ( cin >> a >> b, a + b )
24     {
25         cout << pow_mod( a, b, 1000 ) << endl;
26     }
27     return 0;
28 }