UVA Recurrences 矩阵相乘+快速幂

题目大意：

　　f(n) = a₁ f(n - 1) + a₂ f(n - 2) + a₃ f(n - 3) + ... + a_d f(n - d)，已给递推公式，求f(n)的大小。

解题思路：

　　n很大，所以我们就要构造矩阵，运用矩阵快速幂来求解。//题目描述上口口声声说int范围内，但是大家一定不要天真！！！！！！

 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <cstring>
 using namespace std;

 #define LL long long
 const int maxn = ;
 LL d, m;
 struct mat
 {
     LL p[maxn][maxn];
 };

 mat mul (mat a, mat b);
 mat pow (LL n, mat a, mat b);

 int main ()
 {
     LL n;
     mat a, b;

     while (scanf ("%lld %lld %lld", &d, &n, &m), n+m+d)
     {
         memset (a.p, , sizeof(a.p));
         memset (b.p, , sizeof(b.p));

         for (int i=; i<d; i++)//构造矩阵
         {
             scanf ("%lld", &a.p[i][]);
             a.p[i][] %= m;
             a.p[i][i+] = ;
         }
         for (int i=; i<d; i++)//这个矩阵要反过来输入！！！！！！
             {
                 scanf ("%lld", &b.p[][d-i-]);
                 b.p[][i] %= m;
             }

         if (d < n)
         {
             b = pow (n-d, a, b);
             printf ("%lld\n", b.p[][]);
         }
         else
             printf ("%lld\n", b.p[][d-n]);
     }
     return ;
 }

 mat mul (mat a, mat b)
 {
     mat c;
     memset (c.p, , sizeof(c.p));
     for (int i=; i<d; i++)
         for (int j=; j<d; j++)
         {
             for (int k=; k<d; k++)
                 c.p[i][j] = (c.p[i][j] + a.p[i][k] * b.p[k][j]) % m;
         }
         return c;
 }
 mat pow (LL n, mat a, mat b)
 {
     while (n)
     {
         if (n % )
             b = mul (b, a);
         a = mul (a, a);
         n /= ;
     }
     return b;
 }