HDU 4920 Matrix multiplication 矩阵相乘。稀疏矩阵

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1775    Accepted Submission(s): 796

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≤10⁹).

 

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

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

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#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

struct Matrix
{
    int mat[801][801];
}hxl,tom,now;
void solve(int n)
{
    int i,j,k;
    for(i=1;i<=n;i++)
    {
        for(k=1;k<=n;k++)
        {
            if(hxl.mat[i][k]==0) continue;
            for(j=1;j<=n;j++)
            {
                now.mat[i][j] = now.mat[i][j] + hxl.mat[i][k]*tom.mat[k][j];
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(j==1)printf("%d",now.mat[i][j]%3);
            else printf(" %d",now.mat[i][j]%3);
        }
        printf("
");
    }
}
int main()
{
    int n;
    int i,j;
    while(scanf("%d",&n)>0)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&hxl.mat[i][j]);
                hxl.mat[i][j]%=3;
                now.mat[i][j] = 0;
            }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&tom.mat[i][j]);
                tom.mat[i][j]%=3;
            }
        solve(n);

    }
    return 0;
}