hdu 4920 Matrix multiplication （矩阵乘法）

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

      矩阵乘法就三个for循环而已，没什么说的。不过此题好坑，要小心，如何坑的，请看代码。

   

#include<cstdio>
#include<cstring>
using namespace std;
int n,a[804][804],b[804][804],c[804][804];
int main()
{
    int i,j,k;
    while (~scanf("%d",&n))
    {
        for (i=1;i<=n;i++)
        for (j=1;j<=n;j++) c[i][j]=0;
        for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
        {
            scanf("%d",&a[i][j]);
            a[i][j]%=3;//每输入都要%3，让数据变小，下面才可不%3。
        }
        for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
        {
            scanf("%d",&b[i][j]);
            b[i][j]%=3;
        }
        for (k=1;k<=n;k++) //这个循环不能换在j循环下面，不然会超时。什么原因暂时不清楚。
        for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
            c[i][j]+=a[i][k]*b[k][j]; //这里不能%3，不然会超时，因为多算了一步。
        for (i=1;i<=n;i++)
        {
            for (j=1;j<n;j++)
                printf("%d ",c[i][j]%3);
            printf("%d
",c[i][j]%3);
        }
    }
    return 0;
}