快速幂模板

下面是 m^n  % k 的快速幂：
// m^n % k

int quickpow(int m,int n,int k)
{
    int b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
} 

下面是矩阵快速幂：
//HOJ 3493
/*===================================*/
|| 快速幂（quickpow）模板 
|| P 为等比，I 为单位矩阵
|| MAX 要初始化！！！！
||
/*===================================*/
/*****************************************************/

#include <cstdio>
const int MAX = 3;


typedef  struct{
        int  m[MAX][MAX];
}  Matrix;


Matrix P = {5,-7,4,
            1,0,0,
            0,1,0,
           };


Matrix I = {1,0,0,
            0,1,0,
            0,0,1,
           };
           
Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
       int i,j,k;
       Matrix c;
       for (i = 0 ; i < MAX; i++)
           for (j = 0; j < MAX;j++)
             {
                 c.m[i][j] = 0;
                 for (k = 0; k < MAX; k++)
                     c.m[i][j] += (a.m[i][k] * b.m[k][j])%9997;
                 c.m[i][j] %= 9997;
             }
       return c;
}
          
Matrix quickpow(long long n)
{
       Matrix m = P, b = I;
       while (n >= 1)
       {
             if (n & 1)
                b = matrixmul(b,m);
             n = n >> 1;
             m = matrixmul(m,m);
       }
       return b;
}
               /*************************************/


int main()
{
    Matrix re;
    int f[3] = {2,6,19};
    long long n;
    while (scanf("%I64d",&n) && n != 0)
    {
          if (n == 1)
             printf("1\n");
          else if (n <= 4)
                  printf("%d\n",f[n-2]);
               else {
                      re = quickpow(n - 4);
                      printf("%d\n",(((re.m[0][0]*f[2]) 
                             + (re.m[0][1]*f[1]) + (re.m[0][2]*f[0])) %9997 + 9997) % 9997);
                      }
    }
    return 0;
}