Hdu4920Matrix multiplication(矩阵乘法)

来源：互联网发布：纵横软件好用吗编辑：程序博客网时间：2024/05/16 17:05

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1839 Accepted Submission(s): 818

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

10120 12 34 56 7

Sample Output

00 12 1

题意：两个n*n的矩阵相乘。

直接相乘将第二个矩阵进行倒置。b[i][j]<->b[j][i]。另一种方法是将a的一个数进行判断，如果等于0则不进行运算，否则则进行运算。还有，奇怪的是用G++超时，C++和C都AC。至今不懂为什么会这样。

矩阵相乘优化策略:
矩阵转置：矩阵相乘是矩阵A的一行数据元素与矩阵B的一列的数据元素对应相乘,然后再对其积求和,由于C与C++的二维数组是以行为主序存储的,因此矩阵A的行数据元素是连续存储的,而矩阵B的列数据元素是不连续存储的(N*1的矩阵除外),为了在矩阵相乘时对矩阵B也连续读取数据,根据局部性原理对矩阵B进行转置,矩阵转置是数据元素B[i][j]与B[j][i]进行交换,矩阵转置后,矩阵相乘是矩阵A的一行数据元素与转置后的矩阵B的一行的数据元素对应相乘,然后再对其积求和,这个和就是矩阵C的一个数据元素。

参考资料http://wenku.baidu.com/linkurl=261XeEzHAZkFGPiN63t1nnojoQF50yiuMoviHroGjVXjjRlxFcvWLcws0jgQcmZo4oA9BJcjnPxVreWRu-XXa9zb6r5gUUTxmBXn_qWSsu&qq-pf-to=pcqq.group

#include <stdio.h>#define LL __int64int a[900][900],b[900][900],c[900][900];int main(){    int n,i,j,k;    while (~scanf("%d",&n))    {    int x;        for (i=0;i<n;i++)            for (j=0;j<n;j++)            {                scanf("%d",&x);                a[i][j]=x%3;            }        for (i=0;i<n;i++)            for (j=0;j<n;j++)            {                scanf("%d",&x);                b[j][i]=x%3;            }        for (i=0;i<n;i++)        {        for (j=0;j<n;j++)        {        c[i][j]=0;           for (k=0;k<n;k++)                c[i][j]=c[i][j]+a[i][k]*b[j][k];                printf(j==0?"%d":" %d",c[i][j]%3);            }            printf("\n");        }    }    return 0;}

#include <stdio.h>#define LL __int64int a[810][810],b[810][810],c[810][810],n,i,j,k,x;int main(){while (~scanf("%d",&n)){for (i=0;i<n;i++)for (j=0;j<n;j++){scanf("%d",&x);a[i][j]=x % 3;}for (i=0;i<n;i++)for (j=0;j<n;j++){scanf("%d",&x);b[j][i]=x % 3;}for (i=0;i<n;i++){for (j=0;j<n;j++){c[i][j]=0;for (k=0;k<n;k++)c[i][j]=c[i][j]+a[i][k]*b[j][k];printf(j==0?"%d":" %d",c[i][j]%3);}printf("\n");}}return 0;}

0 0