矩阵快速幂 POJ 3070 Fibonacci

 1 /*
 2     矩阵快速幂：求第n项的Fibonacci数，转置矩阵都给出，套个模板就可以了。效率很高啊
 3 */
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <cmath>
 8 using namespace std;
 9 
10 const int MAXN = 1e3 + 10;
11 const int INF = 0x3f3f3f3f;
12 const int MOD = 10000;
13 struct Mat   {
14     int m[2][2];
15 };
16 
17 Mat multi_mod(Mat a, Mat b)   {
18     Mat ret;    memset (ret.m, 0, sizeof (ret.m));
19     for (int i=0; i<2; ++i) {
20         for (int j=0; j<2; ++j) {
21             for (int k=0; k<2; ++k) {
22                 ret.m[i][j] = (ret.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
23             }
24         }
25     }
26     return ret;
27 }
28 
29 int matrix_pow_mod(Mat x, int n)   {
30     Mat ret; ret.m[0][0] = ret.m[1][1] = 1;  ret.m[0][1] = ret.m[1][0] = 0;
31     while (n)   {
32         if (n & 1)  ret = multi_mod (ret, x);
33         x = multi_mod (x, x);
34         n >>= 1;
35     }
36     return ret.m[0][1];
37 }
38 
39 int main(void)  {       //POJ 3070 Fibonacci
40     int n;
41     while (scanf ("%d", &n) == 1)   {
42         if (n == -1)    break;
43         Mat x;
44         x.m[0][0] = x.m[0][1] = x.m[1][0] = 1; x.m[1][1] = 0;
45         printf ("%d\n", matrix_pow_mod (x, n));
46     }
47 
48     return 0;
49 }