HDU 4920 （Matrix multiplication）

最新推荐文章于 2022-02-25 19:38:17 发布

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最新推荐文章于 2022-02-25 19:38:17 发布

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分类专栏：矩阵文章标签： ACM算法 c math math.h printf

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本文介绍了一个矩阵乘法问题，要求计算两个n*n矩阵的乘积并取模3。通过先转置一个矩阵来优化计算效率，最终实现了快速矩阵乘法。

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3376 Accepted Submission(s): 1414

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

Sample Output

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

题意:两个n*n矩阵相乘，输出相乘后的矩阵。

思路：普通方法一般来说是超时的，我们可以先将一个矩阵转置。然后再相乘。

注意：结果要%3。所以一开始直接先%3.后来再%3.

#include <stdio.h>
#include <cmath>
#include <vector>
#include <map>
#include <time.h>
#include <cstring>
#include <set>
#include<iostream>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define inf 0x6f6f6f6f
#define mod 10
int a[805][805];
int b[805][805];
int c[805][805];
int main()
{
    int n,i,j,k;
    while(cin>>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[j][i]);
                b[j][i]%=3;
            }
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                c[i][j]=0;
                for(k=1; k<=n; k++)
                {
                    c[i][j]+=a[i][k]*b[j][k];
                }
                if(j<n)  //一个一个输出，减少时间。
                    printf("%d ",c[i][j]%3);
                else
                    printf("%d\n",c[i][j]%3);
            }
    }
    return 0;
}