Hdu 4920矩阵乘法（内存访问的讲究）

题目链接

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2143 Accepted Submission(s): 967

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

请看完这篇博文，看完就去AC吧。加了输入优化，效果并不明显。

Accepted Code:

 /*************************************************************************

     > File Name: 1010.cpp

     > Author: Stomach_ache

     > Mail: sudaweitong@gmail.com

     > Created Time: 2014年08月05日 星期二 19时22分23秒

     > Propose:

  ************************************************************************/

 #include <cmath>

 #include <string>

 #include <cstdio>

 #include <fstream>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 int n;

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

 int read() {

     int res = ;

     char c = ' ';

     while (c < '' || c > '') c = getchar();

     while (c >= '' && c <= '') res += c - '', c = getchar();

     return res%;

 }

 int main(void) {

       while (~scanf("%d", &n)) {

           for (int i = ; i < n; i++)

               for (int j = ; j < n; j++)

                   a[i][j] = read();

         for (int i = ; i < n; i++)

               for (int j = ; j < n; j++)

                   b[i][j] = read();

         memset(c, , sizeof(c));

         for (int i = ; i < n; i++) {

             for (int k = ; k < n; k++) {

                   for (int j = ; j < n; j++) {

                     c[i][j] += a[i][k] * b[k][j];      //注意这里的循环顺序

                 }

             }

         }

         for (int i = ; i < n; i++)

               for (int j = ; j < n; j++)

                   printf("%d%c", c[i][j]%, j == n- ? '\n' : ' ');

     }

     return ;

 }

巴特西

Hdu 4920矩阵乘法（内存访问的讲究）

Matrix multiplication

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