HDU 1575 Tr A（矩陣快速冪）

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5537    Accepted Submission(s): 4161

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int MAXN=101;
 5 inline void read(int &n){char c='+';bool flag=0;n=0;    
 6 while(c<'0'||c>'9') c=='-'?flag=1,c=getchar():c=getchar();    
 7 while(c>='0'&&c<='9') n=n*10+c-48,c=getchar();flag==1?n=-n:n=n;}
 8 struct matrix
 9 {
10     int m[11][11];matrix(){memset(m,0,sizeof(m));}
11 };
12 matrix ma;
13 int limit;
14 const int mod=9973;
15 matrix mul(matrix a,matrix b)
16 {
17     matrix c;
18     for(int k=0;k<limit;k++)
19         for(int i=0;i<limit;i++)
20             for(int j=0;j<limit;j++)
21                 c.m[i][j]=(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
22     return c;
23 }
24 matrix fast_martix_pow(matrix ma,int p)
25 {
26     matrix bg;
27     for(int i=0;i<limit;i++)
28         for(int j=0;j<limit;j++)
29             bg.m[i][j]=(i==j);
30     while(p)
31     {
32         if(p&1)    bg=mul(bg,ma);
33         ma=mul(ma,ma);
34         p>>=1;
35     }
36     return bg;
37 }
38 int main()
39 {
40     int T;read(T);
41     while(T--)
42     {
43         read(limit);int n;read(n);
44         for(int i=0;i<limit;i++)
45             for(int j=0;j<limit;j++)
46                 read(ma.m[i][j]);
47         matrix ans=fast_martix_pow(ma,n);
48         int out=0;
49         for(int i=0;i<limit;i++)
50             out+=ans.m[i][i]%mod;
51         printf("%d\n",out%mod);
52     }
53     return 0;
54 }