HDU 1575 Tr A (matrix fast power), hdu1575

HDU 1575 Tr A (matrix fast power), hdu1575

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

If Problem DescriptionA is A square matrix, Tr A indicates the trace of A (the sum of all items on the main diagonal). Tr (A ^ k) % 9973 is required.

 

The first line of Input data is a T, indicating that T groups of data exist.
The first row of each data group contains n (2 <= n <= 10) and k (2 <= k <10 ^ 9) data. Next there are n rows, each row has n data, and the range of each data is [], which indicates the content of square matrix.

 

Output corresponds to each group of data, and Output Tr (A ^ k) % 9973.

 

Sample Input22 21 00 13 999999991 2 34 5 67 8 9

 

Sample Output22686

 

Authorxhd

 

SourceHDU 2007-1 Programming Contest

 

Recommendlinle | We have carefully selected several similar problems for you: 1757 1588 2256 2604 2254 Matrix Quick power !. We need to create a new matrix to meet the following requirements: the diagonal line is 1, the remaining is 0, and then run the Matrix Quick power.

 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 }