Matrix multiplication
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3989 Accepted Submission(s): 1624
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≤10 9).
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
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4920
题意:就是两个矩阵相乘。
PS:这题原来最先在微信上看到别人的题解,主要解决超时,看了他的思路来写,还是超时,最后又在网上找了别人的代码看了一下,原来是这题的数据有点多,用cin和cout超时。
AC代码
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define MAXN 805
int a[MAXN][MAXN],b[MAXN][MAXN],c[MAXN][MAXN];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int x;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
//cin>>x;
scanf("%d",&x);
a[i][j]=x%3;
}
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
//cin>>x;
scanf("%d",&x);
b[i][j]=x%3;
}
memset(c,0,sizeof(c));
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
for(int k=0; k<n; k++)
{
c[i][j]+=a[i][k]*b[k][j];
}
}
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(j)
printf(" ");
//cout<<" ";
//cout<<c[i][j]%3;
printf("%d",c[i][j]%3);
}
//cout<<endl;
printf("\n");
}
}
return 0;
}