http://acm.hdu.edu.cn/showproblem.php?pid=1575
题意：A为一方阵，求(A^k)73得到的矩阵的主对角线的和。
思路：矩阵快速幂。模板题。
1/* ***********************************************
2Author :111qqz
3Created Time :2016年02月21日 星期日 10时28分33秒
4File Name :code/hdu/1575.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=12;
34const int MOD = 9973;
35struct Mat
36{
37 int mat[N][N];
38 void clear()
39 {
40 ms(mat,0);
41 }
42}A;
43int n,k;
44
45Mat operator * (Mat a,Mat b)
46{
47 Mat c;
48 c.clear();
49 for ( int i = 0 ; i < n ; i++)
50 for ( int j = 0 ; j < n ; j++)
51 for (int k = 0 ; k < n ; k++)
52 c.mat[i][j] =(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%MOD;
53
54 return c;
55
56}
57Mat operator ^ (Mat a,int b)
58{
59 Mat c;
60 for ( int i = 0 ; i < n ; i++)
61 for ( int j = 0 ; j < n ; j++ )
62 c.mat[i][j]=(i==j);
63 while (b)
64 {
65 if (b&1) c = c * a;
66 b = b>>1;
67 a = a * a;
68 }
69 return c;
70}
71int main()
72{
73 #ifndef ONLINE_JUDGE
74 freopen("code/in.txt","r",stdin);
75 #endif
76
77 int T;
78 scanf("%d",&T);
79 while (T--)
80 {
81 scanf("%d %d",&n,&k);
82 A.clear();
83 for ( int i = 0 ; i < n ; i++)
84 for ( int j = 0 ; j < n; j ++)
85 scanf("%d",&A.mat[i][j]);
86
87 Mat res;
88 res.clear();
89 res = A^k;
90
91 int ans = 0 ;
92 for ( int i = 0 ; i < n ;i++) ans = (ans +res.mat[i][i])%MOD;
93
94 printf("%d\n",ans);
95
96 }
97
98
99 #ifndef ONLINE_JUDGE
100 fclose(stdin);
101 #endif
102 return 0;
103}