HOJ1997//Multiplying Matrices

最新推荐文章于 2022-06-25 15:51:37 发布

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最新推荐文章于 2022-06-25 15:51:37 发布

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分类专栏： Data structure and Algorithm 文章标签： matrix each integer input include output

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本文介绍了一个用于计算可相乘矩阵的乘积的算法。通过输入矩阵的维度和元素，程序能够输出乘积矩阵，遵循特定的格式和限制。

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Multiplying Matrices

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	Source : SCU Programming Contest 2005 Final
	Time limit : 1 sec		Memory limit : 32 M

Submitted : 2012, Accepted : 633

Write a program to compute the product of two matrices A and B that are multipliable in the form of AB.

Input

The first line of input is the number of all the test cases that follow. For each test case, there are three positive integer m, n, p (m, n, p <= 100) on the first line, meaning that A's dimension is m rows by n columns while B's is n rows by p columns. Then comes the data of matrices A and B. Matrix A is represented by m lines of integers ranging in [-1000, 1000], with each line containing n integers separated by a space. A line is corresponding to a row of a matrix. The data format for matrix B is much the same except for its potentially different dimensions. Please refer to the section Sample Input.

Output

For each test case, print the product of A multiplied by B in the form described in the section Input, but do not include any dimension description.

Sample Input

Sample Output

7 10
15 22
0

Author: hewei

/*
  题意：两个矩阵的乘法
  思路：算出来打印就行
  注意每行输出的时候最后没有空格
*/
#include <iostream>
#include<stdio.h>
#include<cmath>
#define MAX 1000
int Matrices1[MAX+1][MAX+1],Matrices2[MAX+1][MAX+1],ANS[MAX+1][MAX+1];
using namespace std;
int main()
{
    int N,n,m,p,i,j,k,sum;
    cin >> N;
    while(N--)
    {
        cin >> n >> m >> p;
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                cin >> Matrices1[i][j];
        for(i=0; i<m; i++)
            for(j=0; j<p; j++)
                cin >> Matrices2[i][j];
        for (i=0; i<n; i++)
        {
            for (j=0; j<p; j++)
            {
                sum=0;
                for (k=0; k<m; k++)
                {
                    sum+=Matrices1[i][k]*Matrices2[k][j];
                }
               ANS[i][j]=sum;
            }
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<p-1; j++)
            {
                cout << ANS[i][j] << " ";
            }
            cout << ANS[i][p-1];
            cout << endl;
        }
    }
    return 0;
}