矩阵幂和(SOJ 2919)

SOJ 2919：Matrix Power Series

题意：计算$(\sum_{i=1}^{k}\mathbf{A}^{i})\%m$.

分析：$f_{k+1}=\mathbf{A}+\mathbf{A}^{2}+\cdots+\mathbf{A}^{k+1}=\mathbf{A}+\mathbf{A}(\mathbf{A}+\cdots+\mathbf{A}^{k})=\mathbf{A}+\mathbf{A}\cdot f_{k}$. 据此，我们可得递推式：

$\left(\begin{array}{c}f_{k+1}\\ \mathbf{A}\end{array}\right)=\left(\begin{array}{cc} \mathbf{A} &\mathbf{I}\\ \mathbf{0} &\mathbf{I} \end{array}\right)\left(\begin{array}{c} f_{k}\\ \mathbf{A}\end{array}\right)$

代码：(注意，这里我们计算矩阵的幂跟之前的方法有所不同，从二进制末位开始计算)

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4 int A[31][31];
 5 int m;
 6 struct matrix
 7 {
 8     int P[62][62];
 9 };
10 matrix matMul(matrix P1, matrix P2,int a,int b,int c)
11 {
12     matrix P3;
13     memset(P3.P, 0, sizeof(P3.P));
14     int i, j, k;
15     for (i = 1; i <= a; i++)
16         for (j = 1; j <= c; j++)
17             for (k = 1; k <= b; k++)
18                 P3.P[i][j] = (P3.P[i][j]+P1.P[i][k]*P2.P[k][j])%m;
19     return P3;
20 }
21 matrix matPow(matrix M, int n, int k)
22 {
23     matrix Res;
24     int i;
25     memset(Res.P, 0, sizeof(Res.P));
26     for (i = 1; i <= 2 * n; i++)
27         Res.P[i][i] = 1;
28     while (k)
29     {
30         if (k & 1)
31             Res = matMul(Res,M,2*n,2*n,2*n);
32         k >>= 1;
33         M=matMul(M,M,2*n,2*n,2*n);
34     }
35     return Res;
36 }
37 int main()
38 {
39     int n, k;
40     int i, j;
41     matrix B;
42     while (scanf("%d%d%d", &n, &k, &m) == 3)
43     {
44         memset(B.P, 0, sizeof(B.P));
45         for (i = 1; i <= n; i++)
46             for (j = 1; j <= n; j++)
47             {
48             scanf("%d", &A[i][j]);
49             B.P[i][j] = A[i][j];
50             }
51         for (i = 1; i <= n; i++)
52         {
53             B.P[i][n + i] = 1;
54             B.P[n + i][n + i] = 1;
55         }
56         matrix Fin = matPow(B, n, k-1);
57         matrix Fin1, Fin2;
58         for (i = 1; i <= n; i++)
59             for (j = 1; j <= 2 * n; j++)
60                 Fin1.P[i][j] = Fin.P[i][j];
61         for (i = 1; i <= n; i++)
62             for (j = 1; j <= n;j++)
63             {
64             Fin2.P[i][j] = A[i][j];
65             Fin2.P[n + i][j] = A[i][j];
66             }
67         Fin = matMul(Fin1, Fin2, n, 2 * n, n);
68         for (i = 1; i <= n; i++)
69         {
70             for (j = 1; j < n; j++)
71                 printf("%d ", Fin.P[i][j]);
72             printf("%d\n", Fin.P[i][n]);
73         }
74     }
75     return 0;
76 }

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