2014 Multi-University Training Contest 5 HDOJ 4920 Matrix multiplication

题目：

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≤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

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解题思路：

矩阵乘法优化，这题数据量800,用一般的矩阵乘法会超时，我们可以对齐进行优化。C=A*B，在读入的时候将B逆置，相乘的时候，c[i][j] = A的第i行×B的第j行。之前我是直接用经典的矩阵乘法做的，TLE，用斯特拉森矩阵算法也是TLE，哭瞎，这种思路还是听了tsm学弟说的，orz学弟。ps：这种做法要用C编译器提交。

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代码：

#include 

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