POJ 3233 Matrix Power Series(矩阵快速幂+二分)

Time Limit: 3000MS   Memory Limit: 131072K
Total Submissions: 16986   Accepted: 7233

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3


解题思路：
代码：
 1 # include<iostream>
 2 # include<cstdio>
 3 # include<cstring>
 4 using namespace std;
 5 const int MAX = 32;
 6 
 7 struct Matrix
 8 {
 9     int v[MAX][MAX];
10 };
11 int n, k, M;
12 
13 Matrix mtAdd(Matrix A, Matrix B)        // 求矩阵 A + B
14 {
15     int i, j;
16     Matrix C;
17     for(i = 0; i < n; i ++)
18         for(j = 0; j < n; j ++)
19         {
20             C.v[i][j] = (A.v[i][j] + B.v[i][j]) % M;
21         }
22     return C;
23 }
24 
25 Matrix mtMul(Matrix A, Matrix B)        // 求矩阵 A * B
26 {
27     int i, j, k;
28     Matrix C;
29     for(i = 0; i < n; i ++)
30         for(j = 0; j < n; j ++)
31         {
32             C.v[i][j] = 0;
33             for(k = 0; k < n; k ++)
34             {
35                 C.v[i][j] = (A.v[i][k] * B.v[k][j] + C.v[i][j]) % M;
36             }
37         }
38     return C;
39 }
40 
41 Matrix mtPow(Matrix A, int k)           // 求矩阵 A ^ k
42 {
43     if(k == 0) {
44         memset(A.v, 0, sizeof(A.v));
45         for(int i = 0; i < n; i ++)
46         {
47             A.v[i][i] = 1;
48         }
49         return A;
50     }
51     if(k == 1) return A;
52     Matrix C = mtPow(A, k / 2);
53     if(k % 2 == 0) {
54         return mtMul(C, C);
55     } else {
56         return mtMul(mtMul(C, C), A);
57     }
58 }
59 
60 Matrix mtCal(Matrix A, int k)           // 求S (k) = A + A2 + A3 + … + Ak
61 {
62     if(k == 1) return A;
63     Matrix B = mtPow(A, (k+1) / 2);
64     Matrix C = mtCal(A, k / 2);
65     if(k % 2 == 0) {
66         return mtMul(mtAdd(mtPow(A, 0), B), C);   // 如S(6) = (1 + A^3) * S(3)。
67     } else {
68         return mtAdd(A, mtMul(mtAdd(A, B), C));   // 如S(7) = A + (A + A^4) * S(3)
69     }
70 }
71 
72 int main(void)
73 {
74     int i, j;
75     Matrix A;
76     cin >> n >> k >> M;
77     for(i = 0; i < n; i ++)
78         for(j = 0; j < n; j ++)
79         {
80            scanf("%d",&A.v[i][j]);
81         }
82     A = mtCal(A, k);
83     for(i = 0; i < n; i ++)
84     {
85         for(j = 0; j < n; j ++)
86         {
87             cout << A.v[i][j] << ' ';
88         }
89         cout << endl;
90     }
91     return 0;
92 }

 



posted @ 2015-05-17 00:32  BYYB_0506  阅读(191)  评论(0)  编辑  收藏  举报