题目链接：

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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤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

题意

输入两个矩阵，然后输出他们的乘积得出的矩阵，并且模3.

题解：

暴力就可以了，刚开始想复杂了，然后写了个类，重载了*号，然后果断超时了，上网查了一下，说是什么在乘的时候有个优化处理，类似这样：

for (int i(0); i

但是我发现不用这个优化也过了，而且就差了几十ms，所以还是个暴力题。。。

for (int i(0); i

代码

#include
    
     #include
     
      using namespace std;int F1[800][800], F2[800][800],F3[800][800];int main() {    int n,t;    while (scanf("%d",&n)!=EOF) {        memset(F3, 0, sizeof F3);        //a.m_n = n;        //b.m_n = n;        for (int i(0); i < n; i++)            for (int j(0); j < n; j++) {                //cin >>t;                scanf("%d", &t);                F1[i][j] = t % 3;            }        for (int i(0); i < n; i++)            for (int j(0); j < n; j++) {                //cin >> t;                scanf("%d", &t);                F2[i][j] = t % 3;            }        for (int i(0); i