矩阵快速幂 POJ 3070 Fibonacci

题目传送门

 /*
     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
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e3 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = ;
 struct Mat   {
     int m[][];
 };

 Mat multi_mod(Mat a, Mat b)   {
     Mat ret;    memset (ret.m, , sizeof (ret.m));
     for (int i=; i<; ++i) {
         for (int j=; j<; ++j) {
             for (int k=; k<; ++k) {
                 ret.m[i][j] = (ret.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
             }
         }
     }
     return ret;
 }

 int matrix_pow_mod(Mat x, int n)   {
     Mat ret; ret.m[][] = ret.m[][] = ;  ret.m[][] = ret.m[][] = ;
     while (n)   {
         if (n & )  ret = multi_mod (ret, x);
         x = multi_mod (x, x);
         n >>= ;
     }
     return ret.m[][];
 }

 int main(void)  {       //POJ 3070 Fibonacci
     int n;
     while (scanf ("%d", &n) == )   {
         if (n == -)    break;
         Mat x;
         x.m[][] = x.m[][] = x.m[][] = ; x.m[][] = ;
         printf ("%d\n", matrix_pow_mod (x, n));
     }

     return ;
 }