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HDU 4920 （Matrix multiplication）

編輯：關於C++

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≤10⁹).
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

題意:兩個n*n矩陣相乘，輸出相乘後的矩陣。思路：普通方法一般來說是超時的，我們可以先將一個矩陣轉置。然後再相乘。注意：結果要%3。所以一開始直接先%3.後來再%3.

#include 
#include 
#include 
#include
#include 
#include 
#include 
#include
#include 
#include 
#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