hdu 4920 Matrix multiplication（矩阵乘法）2014多培训学校5现场

2017-11-22 989

版权

本文内容由阿里云实名注册用户自发贡献，版权归原作者所有，阿里云开发者社区不拥有其著作权，亦不承担相应法律责任。具体规则请查看《阿里云开发者社区用户服务协议》和《阿里云开发者社区知识产权保护指引》。如果您发现本社区中有涉嫌抄袭的内容，填写侵权投诉表单进行举报，一经查实，本社区将立刻删除涉嫌侵权内容。

简介：

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

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≤10 ⁹).

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
 

题意：给出两个n*n的矩阵，求这两个矩阵的乘积。结果对3取余。

分析：拿到题先用了经典的矩阵相乘的方法，提交以后果断超时了。后来在网上搜了一下矩阵相乘优化，找到了一个优化方法，仅仅可惜如今我还没有理解是怎么优化的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 805;
int a[N][N], b[N][N], ans[N][N];
void  Multi(int n)
{
    int  i, j, k, L, *p2;
    int  tmp[N], con;
    for(i = 0; i < n; ++i)
    {
        memset(tmp, 0, sizeof(tmp));
        for(k = 0, L = (n & ~15); k < L; ++k)
        {
            con = a[i][k];
            for(j = 0, p2 = b[k]; j < n; ++j, ++p2)
                tmp[j] += con * (*p2);
            if((k & 15) == 15)
            {
                for(j = 0; j < n; ++j) tmp[j] %= 3;
            }
        }

        for( ; k < n; ++k)
        {
            con = a[i][k];
            for(j = 0, p2 = b[k]; j < n; ++j, ++p2)
                tmp[j] += con * (*p2);
        }
        for(j = 0; j < n; ++j)
            ans[i][j] = tmp[j] % 3;
    }
}
int main()
{
    int n, i, j, k;
    while(~scanf("%d",&n))
    {
        for(i = 0; i < n; i++)
            for(j = 0; j < n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j] %= 3;
            }
        for(i = 0; i < n; i++)
            for(j = 0; j < n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j] %= 3;
            }
        Multi(n);
        for(i = 0; i < n; i++)
        {
            for(j = 0; j < n-1; j++)
                printf("%d ", ans[i][j]);
            printf("%d\n", ans[i][n-1]);
        }
    }
    return 0;
}

http://blog.csdn.net/gogdizzy/article/details/9003369这里面解说了矩阵相乘的优化方法。

以下这样的方法也能够过：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 805;
int a[N][N], b[N][N], ans[N][N];
int main()
{
    int n, i, j, k;
    while(~scanf("%d",&n))
    {
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j] %= 3;
            }
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j] %= 3;
            }
        memset(ans, 0, sizeof(ans));
        for(k = 1; k <= n; k++) //经典算法中这层循环在最内层，放最内层会超时。可是放在最外层或者中间都不会超时。不知道为什么
            for(i = 1; i <= n; i++)
                for(j = 1; j <= n; j++)
                {
                    ans[i][j] += a[i][k] * b[k][j];
                    //ans[i][j] %= 3;   //假设在这里对3取余，就超时了
                }
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j < n; j++)
                printf("%d ", ans[i][j] % 3);
            printf("%d\n", ans[i][n] % 3);
        }
    }
    return 0;
}

本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4756900.html，如需转载请自行联系原作者