UVa 10870 (矩阵快速幂) Recurrences

给出一个d阶线性递推关系，求f(n) mod m的值。

,

求出A^n-dv₀，该向量的最后一个元素就是所求。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 const int maxn = ;

 typedef long long Matrix[maxn][maxn];
 typedef long long Vector[maxn];

 int d, n, m;

 void matrix_mul(Matrix A, Matrix B, Matrix res)
 {
     Matrix C;
     memset(C, , sizeof(C));
     for(int i = ; i < d; i++)
         for(int j = ; j < d; j++)
             for(int k = ; k < d; k++)
                 C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % m;
     memcpy(res, C, sizeof(C));
 }

 void matrix_pow(Matrix A, int n, Matrix res)
 {
     Matrix a, r;
     memcpy(a, A, sizeof(a));
     memset(r, , sizeof(r));
     for(int i = ; i < d; i++) r[i][i] = ;
     while(n)
     {
         if(n&) matrix_mul(r, a, r);
         n >>= ;
         matrix_mul(a, a, a);
     }
     memcpy(res, r, sizeof(r));
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);

     while(cin >> d >> n >> m && d)
     {
         Matrix A;
         memset(A, , sizeof(A));
         Vector a, f;
         for(int i = ; i < d; i++) { cin >> a[i]; a[i] %= m; }
         for(int i = ; i < d; i++) { cin >> f[i]; f[i] %= m; }
         if(n <= d) { cout << f[n-] << "\n"; continue; }
         for(int i = ; i < d-; i++) A[i][i+] = ;
         for(int i = ; i < d; i++) A[d-][i] = a[d-i-];
         matrix_pow(A, n-d, A);
         long long ans = ;
         for(int i = ; i < d; i++) ans = (ans + A[d-][i]*f[i]) % m;
         cout << ans << "\n";
     }

     return ;
 }

代码君