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[802][802], b[802][802], c[802][802];

int read() {
    int res = 0;
    char c = ' ';
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') res += c - '0', c = getchar();
    return res%3;
}

int main(void) {
      while (~scanf("%d", &n)) {
          for (int i = 0; i < n; i++) 
              for (int j = 0; j < n; j++)
                  a[i][j] = read();
        for (int i = 0; i < n; i++)
              for (int j = 0; j < n; j++)
                  b[i][j] = read();
        memset(c, 0, sizeof(c));
        for (int i = 0; i < n; i++) {
            for (int k = 0; k < n; k++) {
                  for (int j = 0; j < n; j++) {
                    c[i][j] += a[i][k] * b[k][j];      //注意这里的循环顺序
                }
            }
        }
        for (int i = 0; i < n; i++) 
              for (int j = 0; j < n; j++)
                  printf("%d%c", c[i][j]%3, j == n-1 ? '
' : ' ');
    }
    return 0;
}