Fast Matrix Calculation HDU - 4965

One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is 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, in which each element is not less than 0 and not greater than 5. With the two 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 but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.

InputThe input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has 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.OutputFor each case, output the sum of all the elements in M’ in a 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

给你一个n*m和一个m*n的矩阵，经过上面的4步之后会得到一个新的矩阵M，求M中所有元素的总和。

n是一个可以到1000的数，但是m巨小，最多到6，矩阵开1000会爆栈，我们可以转化一下：

A*B^(n*n) = A*B*A*B*A*B...*A*B = A*(B*A)^(n*n-1)*B

B*A是一个m*m的矩阵，嘿嘿~~~

//Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define swap(a,b,t) t = a, a = b, b = t
#define CLS(a, v) memset(a, v, sizeof(a))
#define test() cout<<"============"<<endl
#define debug(a)  cout << #a << " = "  << a <<endl
#define dobug(a, b)  cout << #a << " = "  << a << " " << #b << " = " << b << endl
using namespace std;
typedef long long ll;
const int N = ;
const ll MOD=;
const int INF = ( << );
const double PI=atan(1.0)*;
const int maxn = +;
const ll mod = ;
ll n, m, len, ans, sum, v, w, T, num;
int A[maxn][maxn], B[maxn][maxn];
int c1[maxn][maxn], c2[maxn][maxn];
 
struct Matrix {
    long long grid[N][N];
    int row,col;
    Matrix():row(N),col(N) {
        memset(grid, , sizeof grid);
    }
    Matrix(int row, int col):row(row),col(col) {
        memset(grid, , sizeof grid);
    }
 
    //矩阵乘法
    Matrix operator *(const Matrix &b) {
        Matrix res(row, b.col);
        for(int i = ; i<res.row; i++)
            for(int j = ; j<res.col; j++)
                for(int k = ;k<col; k++)
                    res[i][j] = (res[i][j] + grid[i][k] * b.grid[k][j] + MOD) % MOD;
        return res;
    }
 
    //矩阵快速幂
    Matrix operator ^(long long exp) {
        Matrix res(row, col);
        for(int i = ; i < row; i++)
            res[i][i] = ;
        Matrix temp = *this;
        for(; exp > ; exp >>= , temp = temp * temp)
            if(exp & ) res = temp * res;
        return res;
    }
 
    long long* operator[](int index) {
        return grid[index];
    }
 
    void print() {
        for(int i = ; i <row; i++) {
            for(int j = ; j < col-; j++)
                printf("%d ",grid[i][j]);
            printf("%d\n",grid[i][col-]);
        }
    }
};
 
void input(){
    ios_base::sync_with_stdio(false);
    while( cin >> n >> m && (n+m) ) {
        for(int i=; i<n; i++)
            for(int j=; j<m; j++)
                cin >> A[i][j];
        for(int i=; i<m; i++)
            for(int j=; j<n; j++)
                cin >> B[i][j];
        Matrix C(m, m);
        for(int i=; i<m; i++) {
            for(int j=; j<m; j++) {
                C[i][j] = ;
                for(int k=; k<n; k++) {
                    C[i][j] += ( B[i][k]*A[k][j]);
                    C[i][j] %= ;
                }
            }
        }
        C = C^(n*n-);
 
        for(int i=; i<n; i++) {
            for(int j=; j<n; j++) {
                c1[i][j] = ;
                for(int k=; k<m; k++) {
                    c1[i][j] += A[i][k]*C[k][j];
                    c1[i][j] %= ;
                }
            }
        }
        for(int i=; i<n; i++) {
            for(int j=; j<n; j++) {
                c2[i][j] = ;
                for(int k=; k<m; k++) {
                    c2[i][j] += c1[i][k]*B[k][j];
                }
            }
        }
 
        ans = ;
        for(int i=; i<n; i++) {
            for(int j=; j<n; j++) {
                ans += (c2[i][j]%MOD);
            }
        }
        cout << ans << endl;
    }
}
 
int main(){
    input();
    return ;
}