HDU 4965 Fast Matrix Calculation 【矩阵】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1017;
#define M  6
struct Matrix
{
    int v[6][6];
};
int  n, m, k;//矩阵大小
int A[MAXN][6],B[6][MAXN],C[MAXN][6],D[MAXN][MAXN];
Matrix T, E;

Matrix mtMul(Matrix A, Matrix B)        // 求矩阵 A * B
{
    int i, j, k;
    Matrix C;
    for(i = 0; i < m; i ++)
        for(j = 0; j < m; j ++)
        {
            C.v[i][j] = 0;
            for(k = 0; k < m; k ++)
                C.v[i][j] = (A.v[i][k] * B.v[k][j] + C.v[i][j]) % M;
        }
    return C;
}

 Matrix mtPow(Matrix origin,int k)  //矩阵快速幂
 {
     int i;
     Matrix res;
     memset(res.v,0,sizeof(res.v));
     for(i=0;i<m;i++)
         res.v[i][i]=1;
     while(k)
     {
         if(k&1)
             res=mtMul(res,origin);
         origin=mtMul(origin,origin);
         k>>=1;
     }
     return res;
 }


void out(Matrix A)
{
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<m;j++)
        cout<<A.v[i][j]<<" ";
        cout<<endl;
    }
    cout<<endl;
}

void init()
{
    memset(T.v,0,sizeof(T.v));
    memset(C,0,sizeof(C));
    memset(D,0,sizeof(D));
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n == 0 && m == 0)
            break;
        init();
        for(int i = 0; i < n; i++)//矩阵A的大小n*m
        {
            for(int j = 0; j < m; j++)
            {
                scanf("%d",&A[i][j]);
            }
        }
        for(int i = 0; i < m; i++)//矩阵B的大小m*n
        {
            for(int j = 0; j < n; j++)
            {
                scanf("%d",&B[i][j]);
            }
        }
        for(int i = 0; i < m; i++)//矩阵T = B*A大小为m*m
        {
            for(int j = 0; j < m; j++)
            {
                for(int k = 0; k < n; k++)
                {
                    T.v[i][j] += B[i][k]*A[k][j];
                    T.v[i][j] %= M;
                }
            }
        }

        //out(T);

        E = mtPow(T,n*n-1);

        for(int i = 0; i < n; i++)//矩阵C = A*E 大小为n*m
        {
            for(int j = 0; j < m; j++)
            {
                for(int k = 0; k < m; k++)
                {
                    C[i][j] += A[i][k]*E.v[k][j];
                    C[i][j] %= M;
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < n; i++)//矩阵D = C*B大小为n*n
        {
            for(int j = 0; j < n; j++)
            {
                for(int k = 0; k < m; k++)
                {
                    D[i][j] += C[i][k]*B[k][j];
                }
                ans += D[i][j]%M;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}