Skip to main content

C++ = Multiplication of 2 matrices of order m*n by taking input by user.

Code:
#include <iostream>
using namespace std;
int main()
{
int rows1,columns1,rows2,columns2;
cout<<"Enter the rows of first column : ";
cin>>rows1;
    cout<<"Enter the columns of first column : ";
    cin>>columns1;
    
    cout<<"Enter the rows of second column : ";
    cin>>rows2;
    cout<<"Enter the columns of second column : ";
    cin>>columns2;
    int matrix1[rows1][columns1],matrix2[rows2][columns2];
    
    if (rows2==columns1)
    {
    cout<<"Enter the elements of first matrix : ";
    for (int i=0;i<rows1;i++)
    {
    for (int j=0;j<columns1;j++)
    {
    cin>>matrix1[i][j];
}
}
    
    cout<<"Enter the elements of second matrix : ";
    for (int i=0;i<rows2;i++)
    {
    for (int j=0;j<columns2;j++)
    {
    cin>>matrix2[i][j];
}
}
int result[rows1][columns2];
for (int i=0;i<rows1;i++)
{
for (int j=0;j<columns2;j++)
{
result[i][j]={0};
for (int k=0;k<rows2;k++)
{
result[i][j]+=matrix1[i][k]*matrix2[k][j];
}
}
}
cout<<"The resultant matrix is "<<endl;
for (int i=0;i<rows1;i++)
{
for (int j=0;j<columns2;j++)
{
cout<<result[i][j]<<"       ";
}
    cout<<endl;
}
}
}
Result:



Comments

Post a Comment

Popular posts from this blog

C++ = Code for checking any digit palindrome number

Code no: 1 = Using only main function to perform task. #include <iostream> using namespace std; int main() { int number,original,digit,reverse=0; cout<<"Enter any number : "; cin>>number; original=number; while (number!=0) { digit=number%10; reverse=(reverse*10)+digit; number=number/10; } if (original==reverse) { cout<<"This "<<original<<" is Palindrome number."; } else { cout<<"The "<<original<<" is not Palindrome number."; } } Result: Code no: 2 = Using two functions to perform same task. #include <iostream> using namespace std; int isprime(int num) { int digit=0,reverse=0; while (num!=0) { digit=num%10; reverse=(reverse*10)+digit;  num/=10; } return reverse; } int main() { int num; cout<<"Enter any digit to find its sum = "; cin>>num;          int original=num;          if (i...
Post a Comment
Read more

C++ = Printing diamond with asterik

 CODE: #include <iostream> using namespace std; int main() {     for (int i=1;i<=5;i++)     {     for (int j=1;j<=5-i;j++)     {     cout<<" "; }     for (int k=1;k<=(2*i)-1;k++)     {     cout<<"*"; }   cout<<endl; } for (int i=4;i>=1;i--)     { for (int j=1;j<=5-i;j++)     {     cout<<" "; }     for (int k=1;k<=(2*i)-1;k++)     {     cout<<"*"; }        cout<<endl; } return 0; } RESULT:
Post a Comment
Read more

C++ = How to sum of digits of any number code.

Code no: 1 = By using array.  #include <iostream> using namespace std; int main()  { int number; cout<<"Enter the n"; cin>>number; int a[number]; int sum=0; cout<<"Enter a digit : "<<endl; for (int i=1;i<=number;i++) { cin>>a[i]; } for (int i=1;i<=number;i++) { sum+=a[i]; } cout<<"The sum of the number : "<<sum; return 0; } Result: Code no: 2 = By using some logic without using array and all the code is in one function. #include <iostream> using namespace std; int main() { int num,origional,sum=0,digit=0; origional=num; cout<<"Enter any no: to finds its sum = "; cin>>num; while(num!=0) { digit=num%10; sum+=digit;  num/=10;  } cout<<"The sum of the digits is "<<sum; return 0; } Result: Code no: 3 = Using same code as of code no: second but using different function to perform same ta...
Post a Comment
Read more