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:

 1 /*************************************************************************
 2     > File Name: 1010.cpp
 3     > Author: Stomach_ache
 4     > Mail: [email protected]
 5     > Created Time: 2014年08月05日 星期二 19时22分23秒
 6     > Propose:
 7  ************************************************************************/
 8
 9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 int n;
19 int a[802][802], b[802][802], c[802][802];
20
21 int read() {
22     int res = 0;
23     char c = ‘ ‘;
24     while (c < ‘0‘ || c > ‘9‘) c = getchar();
25     while (c >= ‘0‘ && c <= ‘9‘) res += c - ‘0‘, c = getchar();
26     return res%3;
27 }
28
29 int main(void) {
30       while (~scanf("%d", &n)) {
31           for (int i = 0; i < n; i++)
32               for (int j = 0; j < n; j++)
33                   a[i][j] = read();
34         for (int i = 0; i < n; i++)
35               for (int j = 0; j < n; j++)
36                   b[i][j] = read();
37         memset(c, 0, sizeof(c));
38         for (int i = 0; i < n; i++) {
39             for (int k = 0; k < n; k++) {
40                   for (int j = 0; j < n; j++) {
41                     c[i][j] += a[i][k] * b[k][j];      //注意这里的循环顺序
42                 }
43             }
44         }
45         for (int i = 0; i < n; i++)
46               for (int j = 0; j < n; j++)
47                   printf("%d%c", c[i][j]%3, j == n-1 ? ‘\n‘ : ‘ ‘);
48     }
49     return 0;
50 }

Hdu 4920矩阵乘法（内存访问的讲究）