快速幂取模

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int powermod(int a,int b,int c)  {      int ans=1;      while(b)      {          if(b&1)     //b是奇数，即二进制最后一位为1             ans=(ans*a)%c;          a=(a*a)%c;      //权值不断开方，同时可以模c无影响          b=b>>1;       //不断右移除2      }      return ans;  }

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;}

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