[POJ3233] Matrix Power Series（矩阵快速幂）

传送门

 

k <= 109 暴力肯定超时

根据矩阵性质，可以发现

S(4) = A1 + A2 + A2 * (A1 + A2)

S(5) = A1 + A2 + A2 * (A1 + A2) + A5

所以可以递归二分求解，分别判断 Ak，k 为奇数和偶数的情况

 

——代码

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 struct Matrix
 5 {
 6     int a[30][30];
 7     Matrix()
 8     {
 9         memset(a, 0, sizeof(a));
10     }
11 };
12 
13 int n, k, p;
14 Matrix I, P;
15 
16 inline Matrix operator * (const Matrix x, const Matrix y)
17 {
18     Matrix ans;
19     int i, j, k;
20     for(i = 0; i < n; i++)
21         for(j = 0; j < n; j++)
22             for(k = 0; k < n; k++)
23                 ans.a[i][j] = (ans.a[i][j] + x.a[i][k] * y.a[k][j]) % p;
24     return ans;
25 }
26 
27 inline Matrix operator + (const Matrix x, const Matrix y)
28 {
29     Matrix ans;
30     int i, j;
31     for(i = 0; i < n; i++)
32         for(j = 0; j < n; j++)
33             ans.a[i][j] = (ans.a[i][j] + x.a[i][j] + y.a[i][j]) % p;
34     return ans;
35 }
36 
37 inline Matrix operator ^ (Matrix x, int y)
38 {
39     Matrix ans = I;
40     while(y)
41     {
42         if(y & 1) ans = ans * x;
43         x = x * x;
44         y >>= 1;
45     }
46     return ans;
47 }
48 
49 inline Matrix work(Matrix x, int y)
50 {
51     if(!y) return P;
52     Matrix ans = work(x, y >> 1);
53     ans = ans + (x ^ (y >> 1)) * ans;
54     if(y & 1) ans = ans + (x ^ y);
55     return ans; 
56 }
57 
58 int main()
59 {
60     int i, j;
61     Matrix ans;
62     for(i = 0; i < 30; i++) I.a[i][i] = 1;
63     while(~scanf("%d %d %d", &n, &k, &p))
64     {
65         for(i = 0; i < n; i++)
66             for(j = 0; j < n; j++)
67                 scanf("%d", &ans.a[i][j]);
68         ans = work(ans, k);
69         for(i = 0; i < n; i++)
70         {
71                 for(j = 0; j < n; j++) printf("%d ", ans.a[i][j]);
72                 puts("");
73            }
74     }
75     return 0;
76 }
View Code

 

posted @ 2017-05-12 14:32  zht467  阅读(197)  评论(0)  编辑  收藏  举报