hdu 4920 矩阵相乘

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. InputThe 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,B ij≤109).OutputFor 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

       给你两个矩阵，然后让你计算矩阵的乘积

直接乘就可以了，有0的话直接跳过不要乘，还有不要把取余运算放在三重循环中（会超时）

ac代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;int a[1000][1000],b[1000][1000],c[1000][1000];int main(){int n;while(scanf("%d",&n)!=-1){for(int i=0; i<n; i++)for(int j=0; j<n; j++){scanf("%d",&a[i][j]);a[i][j]=a[i][j]%3;c[i][j]=0;}for(int i=0; i<n; i++)for(int j=0; j<n; j++){scanf("%d",&b[i][j]);b[i][j]=b[i][j]%3;}for(int i=0; i<n; i++)for(int k=0; k<n; k++)if(a[i][k])for(int j=0; j<n; j++)c[i][j]=(c[i][j]+a[i][k]*b[k][j])%3;for(int i=0; i<n; i++){for(int j=0; j<n-1; j++)printf("%d ",c[i][j]);printf("%d\n",c[i][n-1]);}}return 0;}