HDU5950（矩阵快速幂）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5950

题意：f(n) = f(n-1) + 2*f(n-2) + n^4，f(1) = a , f(2) = b，求f(n)

思路：对矩阵快速幂的了解仅仅停留在fib上，重现赛自己随便乱推还一直算错，快两个小时才a还wa了好几次....

主要就是构造矩阵：(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1

|1   2   1   4   6   4   1|     |   f(n+1)   |           |    f(n+2)    |

|1   0   0   0   0   0   0|     |     f(n)     |           |    f(n+1)    |

|0   0   1   4   6   4   1|     | (n+1)^4  |           |  (n+2)^4   |

|0   0   0   1   3   3   1|  * | (n+1)^3  |     =    |  (n+2)^3   |

|0   0   0   0   1   2   1|     | (n+1)^2  |           |  (n+2)^2  |

|0   0   0   0   0   1   1|     |    n+1     |           |     n+2      |

|0   0   0   0   0   0   1|     |      1       |           |       1        |

 #include<cstdio>
 using namespace std;
 typedef long long ll;
 const ll mod = ;
 ll n,a,b;
 struct Matrix
 {
     ll m[][];
     void init1()
     {
         m[][] = b,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = a,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
     }
     void init2()
     {
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
         m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ,m[][] = ;
     }
     Matrix operator * (Matrix t)
     {
         Matrix res;
         for (int i = ; i < ; i++)
         {
             for (int j = ; j < ; j++)
             {
                 res.m[i][j] = ;
                 for (int k = ;k < ; k++)
                     res.m[i][j] = (res.m[i][j] + (m[i][k] % mod) * (t.m[k][j] % mod) % mod) % mod;
             }
         }
         return res;
     }
     Matrix operator ^ (int k)
     {
         Matrix res,s;
         res.init2();
         s.init2();
         while(k)
         {
             if(k & )
                 res = res * s;
             k >>= ;
             s = s * s;
         }
         return res;
     }
 };
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%lld %lld %lld",&n,&a,&b);
         if(n == )
         {
             printf("%lld\n",a % mod);
             continue;
         }
         if(n == )
         {
             printf("%lld\n",b % mod);
             continue;
         }
         Matrix ans,t;
         ans.init1();
         t.init2();
         ans = (t^(n-)) * ans;
         printf("%lld\n",ans.m[][]);
     }
     return ;
 }