poj3233 Matrix Power Series

思路：

递推 + 快速幂优化。

实现：

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 
 5 typedef vector<vector<int> > matrix;
 6 
 7 matrix multiply(matrix & a, matrix & b, int MOD)
 8 {
 9     matrix c(a.size(), vector<int>(b[0].size()));
10     for (int i = 0; i < a.size(); i++)
11     {
12         for (int k = 0; k < a[0].size(); k++)
13         {
14             for (int j = 0; j < b[0].size(); j++)
15             {
16                 c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
17             }
18         }
19     }
20     return c;
21 }
22 
23 matrix pow(matrix & a, int n, int m)
24 {
25     matrix res(a.size(), vector<int>(a[0].size()));
26     for (int i = 0; i < a.size(); i++)
27     {
28         res[i][i] = 1;
29     }
30     while (n > 0)
31     {
32         if (n & 1)
33             res = multiply(res, a, m);
34         a = multiply(a, a, m);
35         n >>= 1;
36     }
37     return res;
38 }
39 
40 int main()
41 {
42     int n, k, m;
43     cin >> n >> k >> m;
44     matrix a(n * 2, vector<int>(n, 0)), b(n * 2, vector<int>(n * 2, 0));
45     for (int i = 0; i < n; i++)
46     {
47         for (int j = 0; j < n; j++)
48         {
49             cin >> b[i][j];
50         }
51     }
52     for (int i = 0; i < n; i++)
53     {
54         a[i][i] = b[n + i][i] = b[n + i][n + i] = 1;
55     }
56     matrix b_k = pow(b, k + 1, m);
57     matrix ans = multiply(b_k, a, m);
58     for (int i = 0; i < n; i++)
59     {
60         for (int j = 0; j < n; j++)
61         {
62             int tmp = ans[n + i][j];
63             if (i == j) tmp = (tmp - 1 + m) % m;
64             cout << tmp << " ";
65         }
66         cout << endl;
67     }
68     return 0;
69 }