Hdu4965--fast matrix calculation (Fast Matrix power optimization)

Problem Description
One day, Alice and Bob felt bored again, Bob knows Alice was a girl who loves math and are just learning something about mat Rix, so he decided-make a crazy problem for her.

Bob has a six-faced dice which have numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he'll choose a number N (4 <= n <=), and for N times, he keeps throwing the He dice for K times (2 < ; =k <= 6) and writes down their number on the top face to make an n*k matrix A, in which each element was not less than 0 And not greater than 5. Then he does similar thing again with a bit difference:he keeps throwing his dice for N times and each time repeat it for K Times to write down a k*n matrix B, with which each element was not less than 0 and not greater than 5. With the both Matrix A and B formed, Alice's task is to perform the following 4-step calculation.

Step 1:calculate a new n*n matrix C = a*b.
Step 2:calculate M = c^ (n*n).
Step 3:for each element x in M, calculate x% 6. All the remainders form a new matrix M '.
Step 4:calculate The sum of all the elements in M '.

Bob just made this problem for kidding and he sees Alice taking it serious, so he also wonders what's the answer is. And then Bob turn-to-because he is not good at math.

Input
The input contains several test cases. Each test case is starts with the integer n and K, indicating the numbers n and K described above. Then N lines follow, and each of the line have K integers between 0 and 5, representing Matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing Matrix B.

The end of input is indicated by N = K = 0.

Output
For each case, output the sum of the "elements in M '" line.

Sample Input
4 2
5 5
4 4
5 4
0 0
4 2 5 5
1 3 1 5
6 3
1 2 3
0 3 0
2 3 4
4 3 2
2 5 5
0 5 0
3 4 5 1 1 0
5 3 2 3 3 2
3 1 5 4 5 2
0 0

Sample Output
14
56

At least two points were noted when looking at the topic:
One is that the scale of the matrix has 1000*1000, too large, and another number m too small, suspicious
Second, the steps are written too detailed, it is not possible to follow the direct walk can be
However, I still cannot control myself to make the hair to explode the stack ....

The topic is the request (A*B) ^ (n*n), is actually a*b*a*b*a*b*a*b*a*b ..., the middle of the b*a proposed to become a A * (b*a) ^ (n*n-1) *b, and B*a is the matrix of 6*6, operation time dropped down a lot

#include <iostream> #include <cstring> #include <string> #include <vector> #include <queue > #include <cstdio> #include <set> #include <math.h> #include <algorithm> #include <queue&
Gt
#include <iomanip> #define INF 0x3f3f3f3f #define MAXN 10000005 #define MOD 10000007 using namespace std;
const int N = 10;
Long Long m,n;
struct Matrix {int mat[n][n];};
    Matrix Mul (Matrix A,matrix b) {matrix res;
            for (int i=0, i<m; ++i) for (int j=0; j<m; ++j) {res.mat[i][j]=0;
                for (int k=0; k<m; ++k) {res.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
            res.mat[i][j]%=6;
}} return res;
    } Matrix Pow_matrix (Matrix A,long long k) {matrix res;
    memset (res.mat,0,sizeof (Res.mat));
    for (int i=0; i<m; ++i) res.mat[i][i]=1;
        while (k!=0) {if (k%2) Res=mul (res,a);
   A=mul (A,a);     k>>=1;

} return res;
} int a[1005][1005],b[1005][1005],c1[1005][1005],c2[1005][1005];
    int main () {Matrix tmp;
        while (~SCANF ("%i64d%i64d", &n,&m)) {if (n==0&&m==0) break;
        memset (tmp.mat,0,sizeof (Tmp.mat));
        for (int. i=0; i<n; ++i) for (int j=0; j<m; ++j) scanf ("%d", &a[i][j]);
        for (int. i=0; i<m; ++i) for (int j=0; j<n; ++j) scanf ("%d", &b[i][j]);
                for (int i=0, i<m; ++i) for (int j=0; j<m; ++j) {tmp.mat[i][j]=0;
                    for (int k=0; k<n; ++k) {tmp.mat[i][j]+=b[i][k]*a[k][j];
                tmp.mat[i][j]%=6;
        }} Matrix P=pow_matrix (tmp,n*n-1);
                for (int i=0, i<n; ++i) for (int j=0; j<m; ++j) {c1[i][j]=0; for (int k=0; k<m;++K) {c1[i][j]+=a[i][k]*p.mat[k][j];
                c1[i][j]%=6; }} for (int i=0, i<n; ++i) for (int j=0; j<n; ++j) {c2[i
                ][j]=0;
                    for (int k=0; k<m; ++k) {c2[i][j]+=c1[i][k]*b[k][j];
                c2[i][j]%=6;
        }} long long ans=0;
        for (int i=0, i<n; ++i) for (int j=0; j<n; ++j) ans+=c2[i][j];
    printf ("%i64d\n", ans);
} return 0;
 }