首页 > 代码库 > Hdu 4920矩阵乘法（内存访问的讲究）

Hdu 4920矩阵乘法（内存访问的讲究）

2024-07-16 11:19:44   220人阅读

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 }

 


联系
我们