Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3376 Accepted Submission(s): 1414

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

Sample Output

0
0 1
2 1

Author Xiaoxu Guo (ftiasch)
Source 2014 Multi-University Training Contest 5

题意:两个n*n矩阵相乘，输出相乘后的矩阵。思路：普通方法一般来说是超时的，我们可以先将一个矩阵转置。然后再相乘。注意：结果要%3。所以一开始直接先%3.后来再%3.

#include 
  
   
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         #include 
        
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             #include 
            
              using namespace std; #define inf 0x6f6f6f6f #define mod 10 int a[805][805]; int b[805][805]; int c[805][805]; int main() { int n,i,j,k; while(cin>>n) { for(i=1; i<=n; i++) for(j=1; j<=n; j++) { scanf("%d",&a[i][j]); a[i][j]%=3; } for(i=1; i<=n; i++) for(j=1; j<=n; j++) { scanf("%d",&b[j][i]); b[j][i]%=3; } for(i=1; i<=n; i++) for(j=1; j<=n; j++) { c[i][j]=0; for(k=1; k<=n; k++) { c[i][j]+=a[i][k]*b[j][k]; } if(j

HDU 4920 （Matrix multiplication）

Matrix multiplication