HDU 4965 Fast Matrix Calculation

题目链接：Fast Matrix Calculation

题面：

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1078    Accepted Submission(s): 515

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

 

解题：

    题意：是将n*k和k*n的矩阵的乘积，求n*n次方后求和。

    虽然用了矩阵快速幂，但是依旧会超时，因为n会达到1000计算量仍旧太大，看了一种解法是分解为A*(B*A)^(n*n-1)*B。着实很难想到，但是接触过之后，以后就可以运用了，也算是一种技巧吧。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n,k;
struct matrix
{
	int map[7][7];
};
matrix mult(matrix a,matrix b)
{
	matrix c;
	memset(c.map,0,sizeof(c.map));
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=k;j++)
		{
			for(int m=1;m<=k;m++)
			{
				c.map[i][j]=(c.map[i][j]+a.map[i][m]*b.map[m][j]);
			}
			c.map[i][j]%=6;
		}
	}
	return c;
}
matrix pow_mult(matrix a,int x)
{
	matrix c;
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=k;j++)
		{
			c.map[i][j]=(i==j);
		}
	}
	for(;x;x>>=1)
	{
		if(x&1)c=mult(c,a);
		a=mult(a,a);
	}
	return c;
}
int a[1005][7],b[7][1005],final[7][1005],mid[7][7],last[1005][1005],sum;
int main()
{
	while(scanf("%d%d",&n,&k)) 
	{
	  if(n==0&&k==0)break;	
	  for(int i=1;i<=n;i++)
	  {
  	    for(int j=1;j<=k;j++)
		scanf("%d",&a[i][j]);	
  	  }
  	  for(int i=1;i<=k;i++)
	  {
  	    for(int j=1;j<=n;j++)
		scanf("%d",&b[i][j]);	
  	  }
  	  memset(mid,0,sizeof(mid));
  	  for(int i=1;i<=k;i++)
  	  {
  	  	for(int j=1;j<=k;j++)
  	  	{
	  	   for(int m=1;m<=n;m++)
	       {
 		     mid[i][j]=mid[i][j]+b[i][m]*a[m][j];		
		   }
		   mid[i][j]%=6;  	
  	    }
  	  }
  	  matrix ans;
  	  for(int i=1;i<=k;i++)
  	  {
  	  	for(int j=1;j<=k;j++)
  	  	{
	  	  	ans.map[i][j]=mid[i][j];
  	    }
  	  }
  	  ans=pow_mult(ans,n*n-1);
  	  memset(final,0,sizeof(final));
  	  for(int i=1;i<=k;i++)
	  {
		for(int j=1;j<=n;j++)
		{
			for(int m=1;m<=k;m++)
			{
				final[i][j]=final[i][j]+ans.map[i][m]*b[m][j];
			}
			final[i][j]%=6;
		}
	  }
	  memset(last,0,sizeof(last));
  	  for(int i=1;i<=n;i++)
	  {
		for(int j=1;j<=n;j++)
		{
			for(int m=1;m<=k;m++)
			{
				last[i][j]=(last[i][j]+a[i][m]*final[m][j]);
			}
			last[i][j]%=6;
		}
	  }
	  sum=0;
	  for(int i=1;i<=n;i++)
	  {
  		for(int j=1;j<=n;j++)
  		{
		  	sum+=last[i][j];
	    }
  	  }
  	  cout<<sum<<endl;
	}
	return 0;
}