/* The following C / C++ program require user to enter the elements 
of arrays having equal number of array elements. Equal number of array 
elements are required in this program. This program can be extended to 
any of the data type & to the 'n' number of arrays simultaneously*/
Code:
#include<iostream>
#include<conio.h>
using namespace std;
void main ()
{
int long a[100],b[100],c[100];
int d=0,n;
cout<<"Number of Elements in Arrays: ";
cin>>n;
cout<<"Enter For Array 1"<<endl;
while (d<n)
{
cout<<"Enter the element number "<<d+1<<" : ";
cin>>a[d];
d=d+1;
}
d=0;
cout<<"Enter For Array 2"<<endl;
while (d<n)
{
cout<<"Enter the element number "<<d+1<<" : ";
cin>>b[d];
d=d+1;
}
d=0;
while (d<n)
{
c[d]=a[d]*b[d];
d=d+1;
}
d=0;
cout<<"\nResult of element wise product of array elements:\n";
while (d<n)
{
cout<<c[d]<<"\t";
d=d+1;
}
getche();
}
Output
  
Code:
#include<iostream>
#include<conio.h>
using namespace std;
void main ()
{
int long a[100],b[100],c[100];
int d=0,n;
cout<<"Number of Elements in Arrays: ";
cin>>n;
cout<<"Enter For Array 1"<<endl;
while (d<n)
{
cout<<"Enter the element number "<<d+1<<" : ";
cin>>a[d];
d=d+1;
}
d=0;
cout<<"Enter For Array 2"<<endl;
while (d<n)
{
cout<<"Enter the element number "<<d+1<<" : ";
cin>>b[d];
d=d+1;
}
d=0;
while (d<n)
{
c[d]=a[d]*b[d];
d=d+1;
}
d=0;
cout<<"\nResult of element wise product of array elements:\n";
while (d<n)
{
cout<<c[d]<<"\t";
d=d+1;
}
getche();
}
Output
![]()  | 
| Output of C / C++ Program to calculate element wise product of 2 arrays : element wise multiplication | 



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