hdu 4965 Fast Matrix Calculation（矩阵乘法）

Problem Description

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.

 

Input

The 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.

 

Output

For 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
  

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

让求a*b的n*n次方，a是[n,k]的矩阵，b是[k,n]的矩阵，n是1000，k只有6，但是a*b的矩阵就是1000*1000的了，再来个n*n次方，肯定超时，所以我们可以把公式转换一下：

(a*b)^(n*n) == a*(b*a)*(b*a)*....*(b*a)*b   == a*(b*a)^(n*n-1)*b

这样就是求一个6*6矩阵的n*n-1次方，这样就不会超时了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;

const int mod = 6;

struct Matrix
{
    long long m[6][6];
    int n;
    Matrix(int x)
    {
        n = x;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                m[i][j] = 0;
    }
    Matrix(int _n,int a[10][10])
    {
        n = _n;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                m[i][j] = a[i][j];
            }
    }
};
Matrix operator *(Matrix a,Matrix b)
{
    int n = a.n;
    Matrix ans = Matrix(n);
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
            {
                ans.m[i][j] += (a.m[i][k]%mod)*(b.m[k][j]%mod)%mod;
                ans.m[i][j] %= mod;
            }
    return ans;
}
Matrix operator ^(Matrix a,int k)
{
    int n = a.n;
    Matrix c(n);
    int i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            c.m[i][j] = (i==j);
    for(;k;k>>=1)
    {
        if(k&1)
            c=c*a;
        a = a*a;
    }
    return c;
}

int a[1000][6],b[6][1100],c[6][6];
int ac[1000][6];
int main(void)
{
    int n,k,i,j,l;
    while(scanf("%d%d",&n,&k)==2)
    {
        if(n == 0 && k == 0)
            break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        memset(ac,0,sizeof(ac));
        for(i=0;i<n;i++)
            for(j=0;j<k;j++)
                scanf("%d",&a[i][j]);
        for(i=0;i<k;i++)
            for(j=0;j<n;j++)
                scanf("%d",&b[i][j]);
        Matrix C(k);
        for(i=0;i<k;i++)
        {
            for(j=0;j<k;j++)
            {
                for(l=0;l<n;l++)
                {
                    C.m[i][j] += b[i][l]*a[l][j];  //b*a
                    C.m[i][j] %= mod;
                }
            }
        }
        C = C^(n*n-1); // b*a的n*n-1次方

        for(i=0;i<n;i++)
        {
            for(j=0;j<k;j++)
            {
                for(l=0;l<k;l++)
                {
                    ac[i][j] += a[i][l]*C.m[l][j]; // a*c
                    ac[i][j] %= mod;
                }

            }
        }
        int sum = 0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                int t = 0;
                for(l=0;l<k;l++)
                {
                    t += ac[i][l]*b[l][j]; //a*c*b
                }
                sum += t%mod;
            }
        }

        printf("%d\n",sum);
    }

    return 0;
}