Matrix Different Dimension HCF

 Matrix Different Dimension HCF 

PROBLEM STATEMENT :

The program must accept two integer matrices of size MxN and RxC as the input. The program must print the HCF of the integer values present in the intersecting area of the given two matrices. 

Boundary Condition(s): 

2 <= M, N, R, C <= 50 

1 <= Matrix element value <= 1000 

Input Format:

The first line contains M and N separated by a space. 

The next M lines, each contains N integers separated by a space. 

The (M+2)th line contains R and C separated by a space. 

The next R lines, each contains C integers separated by a space. 

Output Format: 

The lines, each contains the integer values separated by a space. 

Example Input/Output 1: 
Input: () 
2 6 
4 4 3 2 2 4 
8 8 6 3 8 5 
3 3 
4 2 6 
7 2 6 
5 1 8 
Output: 
4 2 3 
1 2 6 

Explanation: 

The integers present in the intersecting area of the first matrix are highlighted below. 

4 4 3 2 2 4 
8 8 6 3 8 5 

The integers present in the intersecting area of the second matrix are highlighted below. 

4 2 6 
7 2 6 
5 1 8 

The HCF of the integer values present in the intersecting area are given below 

4 2 3 
1 2 6 

Example Input/Output 2: 
Input: () 
3 4 
4 8 5 1 
1 8 4 8 
4 6 5 9 
4 5 
1 1 15 5 14 
7 6 16 10 15 
3 9 18 16 1 
7 10 18 12 14 
Output: 
1 1 5 1 
1 2 4 2
1 3 1 1

---

SOLUTION :

C (Programming Language)

      
#include<stdio.h>
#include<stdlib.h>

int hcf(int a,int b)
{
    if(a==0)
        return b;
    return hcf(b%a,a);
}
int main()
{
    int M,N,R,C;
    scanf("%d %d",&M,&N);
    int mat1[M][N];
    for(int i=0;i<M;i++)
    {
        for(int j=0;j<N;j++) scanf("%d ",&mat1[i][j]);
    }
    scanf("%d %d",&R,&C);
    int mat2[R][C];
    for(int i=0;i<R;i++)
    {
        for(int j=0;j<C;j++) scanf("%d ",&mat2[i][j]);
    }
    int minRow = M<R ? M : R;
    int minCol = N<C ? N : C;
    for(int i=0;i<minRow;i++)
    {
        for(int j=0;j<minCol;j++)
        {
            printf("%d ",hcf(mat1[i][j],mat2[i][j]));
        }
        printf("\n");
    }
}
 

C++ (CPP)

       
#include <bits/stdc++.h>
 
using namespace std;

int main(int argc, char** argv)
{
    int r1 , c1 ;
    cin >> r1 >> c1 ;
    int mat1[r1][c1];
    for(int i = 0 ; i<r1 ; i++)
    {
        for(int j = 0 ; j<c1 ; j++)
          cin >> mat1[i][j];
    }
    int r2 , c2 ;
    cin >> r2 >> c2 ;
    int mat2[r2][c2];
    for(int i = 0 ; i<r2 ; i++)
    {
        for(int j = 0 ; j<c2 ; j++) 
          cin >> mat2[i][j];
    }
    for(int i = 0 ; i<min(r1,r2) ; i++)
    {
        for(int j = 0 ; j<min(c1,c2) ; j++)
          cout << __gcd(mat1[i][j] , mat2[i][j]) << " ";
          
        cout << endl;
    }
}
 

JAVA

       
import java.util.*;
public class Hello {
    public static int gcd(int a,int b)
    {
        if (b==0)
        return a;
        else 
        return gcd(b,a%b);
    }

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int r1=sc.nextInt(),c1=sc.nextInt();
		int[][] mat1 =new int[r1][c1];
		for(int i=0;i<r1;i++)
		{
		    for(int j=0;j<c1;j++)
		    {
		        mat1[i][j]=sc.nextInt();
		    }
		}
		int r2=sc.nextInt(),c2=sc.nextInt();
		int[][] mat2=new int [r2][c2];
		for(int i=0;i<r2;i++)
		{
		    for(int j=0;j<c2;j++)
		    {
		        mat2[i][j]=sc.nextInt();
		    }
		}
		for(int i=0;i<Math.min(r1,r2);i++)
		{
		 for(int j=0;j<Math.min(c1,c2);j++)
		 {
		        System.out.print(gcd(mat1[i][j],mat2[i][j])+" ");
		 }
		 System.out.println();
		}
	}
}
 

PYTHON

       
import math
m,n=map(int,input().split())
l1=[list(map(int,input().split())) for _ in range(m)]
r,c=map(int,input().split())
l2=[list(map(int,input().split())) for _ in range(r)]
for i in range(min(m,r)):
    for j in range(min(n,c)):
        print(int(math.gcd(l1[i][j],l2[i][j])),end=' ')
    print()
 

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