2014 Multi-University Training Contest 5 HDOJ 4920 Matrix multiplication

題目：

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 384    Accepted Submission(s): 138

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

1 0 1 2 0 1 2 3 4 5 6 7

 

Sample Output

0 0 1 2 1

傳送門：點擊打開鏈接

解題思路：

矩陣乘法優化，這題數據量800,用一般的矩陣乘法會超時，我們可以對齊進行優化。C=A*B，在讀入的時候將B逆置，相乘的時候，c[i][j] = A的第i行×B的第j行。之前我是直接用經典的矩陣乘法做的，TLE，用斯特拉森矩陣算法也是TLE，哭瞎，這種思路還是聽了tsm學弟說的，orz學弟。ps：這種做法要用C編譯器提交。

相關論文：點擊打開鏈接

代碼：

#include <stdio.h>

int a[805][805], b[805][805], c[805][805];

int main()
{
    int n, i, j, k, x;
    while(~scanf("%d", &n))
    {
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                scanf("%d", &x);
               a[i][j] = x % 3;
            }

        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
               scanf("%d", &x);
                b[j][i] = x % 3;
            }

        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                c[i][j] = 0;
                for(k=0; k<n; k++)
                    c[i][j] = c[i][j] + a[i][k]*b[j][k];
                printf( j==0 ? "%d" : " %d",c[i][j]%3);
            }
            printf("\n");
        }
    }
    return 0;
}