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:

 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 }