Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3647    Accepted Submission(s): 1522


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 ​​

 


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直接矩阵乘 O(n^3) 800^3=5120,0000 TLE

所以利用mod3的性质 稍加优化


#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)|
#define MAXN (800+10)
#define F (3)
#define pb push_back
#define mp make_pair 
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,a[MAXN][MAXN],b[MAXN][MAXN],c[MAXN][MAXN];
int main()
{
//  freopen("j.in","r",stdin);
  
  while(cin>>n) {
    For(i,n) For(j,n) scanf("%d",&a[i][j]),a[i][j]%=3;
    For(i,n) For(j,n) scanf("%d",&b[i][j]),b[i][j]%=3,c[i][j]=0;
    
    
    
    For(i,n) For(k,n)
    if (a[i][k]) For(j,n) c[i][j]=(c[i][j]+a[i][k]*b[k][j])%3;
    
    For(i,n) {
      For(j,n-1) printf("%d ",c[i][j]);
      printf("%d\n",c[i][n]);
    }
    
  }
  
  return 0;
}