POJ3233Matrix Power Series

http://poj.org/problem?id=3233

 1 #include 
 2 
 3 using namespace std;
 4 
 5 const int N = 50;
 6 int n, k, mod;
 7 
 8 // 定义结构体
 9 struct mat {
10     int m[N][N];
11     mat () {
12         memset(m, 0, sizeof m);
13         for (int i = 0; i < N; i ++) {
14             m[i][i] = 1;
15         }
16     }
17 };
18 
19 // 矩阵乘法
20 mat multi(mat a, mat b) {
21     mat res;
22     for (int i = 0; i < n; i ++) {
23         for (int j = 0; j < n; j ++ ) {
24             res.m[i][j] = 0;
25             for (int k = 0; k < n; k ++) {
26                 res.m[i][j] = (res.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
27             }
28         }
29     }
30     return res;
31 }
32 
33 // 矩阵相加
34 mat add(mat a, mat b) {
35     mat res;
36     for (int i = 0; i < n; i ++) {
37         for (int j = 0; j < n; j++) {
38             res.m[i][j] = (a.m[i][j] + b.m[i][j]) % mod;
39         }
40     }
41     return res;
42 }
43 
44 // 矩阵快速幂
45 mat fastpow(mat a, int k) {
46     mat res;
47     while (k) {
48         if (k & 1) res = multi(res, a);
49         a = multi(a, a);
50         k >>= 1;
51     }
52     return res;
53 }
54 
55 // 矩阵求和
56 // sum(k) = (E+A^{k/2}) * sum(k/2); // 偶数
57 // sum(k) = (E+A^{k-1}/2) * sum((k - 1)/2) + A^k; // 奇数
58 mat sum(mat a, int k) {
59     mat tmp;
60     if (k == 1) return a;
61     if (k & 1) {
62         mat t = multi(sum(a, (k-1)/2),add(tmp, fastpow(a, (k-1) / 2)));
63         return add(t, fastpow(a, k));
64     } else {
65         return multi(sum(a, k/2),add(tmp, fastpow(a, k / 2)));
66     }
67 }
68 
69 int main() {
70     cin >> n >> k >> mod;
71 
72     // 读矩阵
73     mat a; 
74     for (int i = 0; i < n; i++) {
75         for (int j = 0; j < n; j ++) {
76             cin >> a.m[i][j];
77         }
78     }
79 
80     mat res = sum(a, k);
81 
82     // 输出矩阵
83     for (int i = 0; i < n; i++) {
84         for (int j = 0; j < n; j ++) {
85             cout << res.m[i][j] << " ";
86         }
87         cout << endl;
88     }
89 
90 
91     return 0;
92 }