HDOJ 4920 Matrix multiplication (矩阵相乘)

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3838    Accepted Submission(s): 1576

Problem Description

Given two matrices A and B of size n×n, find the product of them.

bobo hates big integers. So you are only asked to find the result modulo 3.

 

Input

The input consists of several tests. For each tests:

The first line contains n (1≤n≤800). Each of the following n lines contain n integers -- the description of the matrix A. The j-th integer in the i-th line equals A _ij. The next n lines describe the matrix B in similar format (0≤A _ij,B _ij≤10 ⁹).

 

Output

For each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format.

 

Sample Input

   
   
   
   

    
    
    
    1
0
1
2
0 1
2 3
4 5
6 7
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
0 1
2 1
   
   
   
   

 

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 810
#define MOD 10000
#define LL long long
#define INF 0xfffffff
#define fab(a)(a)>0?(a):(-a)
using namespace std; 
int a[MAXN][MAXN];
int b[MAXN][MAXN];
int ans[MAXN][MAXN];
int main()
{
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
			    scanf("%d",&a[i][j]);
			    a[i][j]%=3;
		    }
	    }
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
			    scanf("%d",&b[i][j]);
			    b[i][j]%=3;
		    }
	    }
        memset(ans,0,sizeof(ans));
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                for(k=0;k<n;k++)
                {
                    ans[i][j]=(ans[i][j]+a[i][k]*b[k][j]);//在这里取余的话会TLE
                }
            }
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            printf(j==n-1?"%d\n":"%d ",ans[i][j]%3);
        }
    }
    return 0;
}