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 ;
 }