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 + … + A^k.

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 }