HDU-4920 Matrix multiplication

矩阵相乘，采用一行的去访问，比采用一列访问时间更短，根据数组是一行去储存的。神奇小代码。

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1476    Accepted Submission(s): 650

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

1

2

0 1

2 3

4 5

6 7

Sample Output

0 1

2 1

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[][],b[][],c[][];
int main()
{
    int k,i,n,j,x;
    while(~scanf("%d",&n))
    {
        memset(c,,sizeof(c));
        for(i=; i<=n; i++)
            for(j=; j<=n; j++)
            {
                scanf("%d",&x);
                a[i][j]=x%;
            }
        for(i=; i<=n; i++)
            for(j=; j<=n; j++)
            {
                scanf("%d",&x);
                b[i][j]=x%;
            }
        for(k=; k<=n; k++)
        {
            for(j=;j<=n; j++)
            {
                for(i=;i<=n; i++)
                {
                    c[k][i]+=a[k][j]*b[j][i];
                }
            }
        }
        for(i=;i<=n; i++)
        {
            for(j=;j<n; j++)
                printf("%d ",c[i][j]%);
            printf("%d\n",c[i][n]%);
        }
    }
    return ;
}