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Hdu 4920矩阵乘法(内存访问的讲究)

时间:2014-08-08 15:33:26      阅读:271      评论:0      收藏:0      [点我收藏+]

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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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤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: sudaweitong@gmail.com
 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),布布扣,bubuko.com

Hdu 4920矩阵乘法(内存访问的讲究)

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原文地址:http://www.cnblogs.com/Stomach-ache/p/3899133.html

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