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。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 const int maxn = 20;
 7
 8 typedef long long Matrix[maxn][maxn];
 9 typedef long long Vector[maxn];
10
11 int d, n, m;
12
13 void matrix_mul(Matrix A, Matrix B, Matrix res)
14 {
15     Matrix C;
16     memset(C, 0, sizeof(C));
17     for(int i = 0; i < d; i++)
18         for(int j = 0; j < d; j++)
19             for(int k = 0; k < d; k++)
20                 C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % m;
21     memcpy(res, C, sizeof(C));
22 }
23
24 void matrix_pow(Matrix A, int n, Matrix res)
25 {
26     Matrix a, r;
27     memcpy(a, A, sizeof(a));
28     memset(r, 0, sizeof(r));
29     for(int i = 0; i < d; i++) r[i][i] = 1;
30     while(n)
31     {
32         if(n&1) matrix_mul(r, a, r);
33         n >>= 1;
34         matrix_mul(a, a, a);
35     }
36     memcpy(res, r, sizeof(r));
37 }
38
39 int main()
40 {
41     //freopen("in.txt", "r", stdin);
42
43     while(cin >> d >> n >> m && d)
44     {
45         Matrix A;
46         memset(A, 0, sizeof(A));
47         Vector a, f;
48         for(int i = 0; i < d; i++) { cin >> a[i]; a[i] %= m; }
49         for(int i = 0; i < d; i++) { cin >> f[i]; f[i] %= m; }
50         if(n <= d) { cout << f[n-1] << "\n"; continue; }
51         for(int i = 0; i < d-1; i++) A[i][i+1] = 1;
52         for(int i = 0; i < d; i++) A[d-1][i] = a[d-i-1];
53         matrix_pow(A, n-d, A);
54         long long ans = 0;
55         for(int i = 0; i < d; i++) ans = (ans + A[d-1][i]*f[i]) % m;
56         cout << ans << "\n";
57     }
58
59     return 0;
60 }

代码君