矩阵乘法 --- hdu 4920 ： Matrix multiplication-优快云博客

本文介绍了一种在大规模数据集上进行矩阵乘法并求模的算法实现，使用了简单的暴力算法来解决矩阵乘法问题，并通过取模操作避免了大整数运算，适用于竞赛编程环境。

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 820 Accepted Submission(s): 328

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

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

Mean:

给你两个矩阵，计算两个矩阵的积。

nalyse:

很多人认为这题用暴力过不了，时间复杂度为O(n^3),800^3=512000000，差不多要接近于10^9了，可能是hdu的评测速度给力吧，再加上这题是单点评测，每个评测点的时限都有2000ms，所以说暴力过了也实属正常。

Time complexity：O(n^3)

Source code：

#include<stdio.h>
int a[800][800],b[800][800],c[800][800],n,i,j,k;
int main(){
    while(scanf("%d",&n)!=EOF){
        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;
        for(i=0;i<n;++i)
            for(j=0;j<i;++j)
                k=b[i][j],b[i][j]=b[j][i],b[j][i]=k;
        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]+=a[i][k]*b[j][k];
                c[i][j]%=3;
            }
        for(i=0;i<n;++i)
            for(j=0;j<n;++j)
                printf(j==n-1?"%d\n":"%d ",c[i][j]);
    }
    return 0;
}