Example:4
Write a C++ program to take input for “n” element using an array further display all the elements. Also check and print the lowest array element. Allocate the memory dynamically.
#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
int n,i,m;
char ch;
do
{
cout<<"Enter total elements ";
cin>>n;
//dynamic memory allocation
int *p=new int[n];
//input
for(i=0;i<n;i++)
{
cout<<"Enter "<<i<<" element ";
cin>>p[i];
}
//display
m=p[0];
//or
//m=*p;
for(i=0;i<n;i++)
{
cout<<p[i]<<endl;
if(p[i]<m)
m=p[i];
}
//release memory
delete []p;
cout<<"Min element = "<<m<<endl;
cout<<"Lke to cont .... (y/n) ";
cin>>ch;
}while(ch=='y' || ch=='Y');
return(0);
}
Output:
Enter total elements 5
Enter 0 element 24
Enter 1 element 6
Enter 2 element 32
Enter 3 element 5
Enter 4 element 95
24
6
32
5
95
Min element = 5
Lke to cont …. (y/n) y
Enter total elements 2
Enter 0 element 54
Enter 1 element 25
54
25
Min element = 25
Lke to cont …. (y/n) n
Example:5
Write a C++ program to take input for “n” element using an array further display all the elements. Also check and print the following:
1. lowest array element.
2. Largest array element.
3. Difference between largest and lowest array element.
Allocate the memory dynamically.
Sol:
#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
int n,i,ma,mi,d;
char ch;
do
{
cout<<"Enter total elements ";
cin>>n;
//dynamic memory allocation
int *p=new int[n];
//input
for(i=0;i<n;i++)
{
cout<<"Enter "<<i<<" element ";
cin>>p[i];
}
//display
ma=p[0];
mi=p[0];
//or
//ma=*p;
//mi=*p;
for(i=0;i<n;i++)
{
cout<<p[i]<<endl;
//max no
if(p[i]>ma)
ma=p[i];
//min no
if(p[i]<mi)
mi=p[i];
}
//release memory
delete []p;
cout<<"Max element = "<<ma<<endl;
cout<<"Min element = "<<mi<<endl;
d=ma-mi;
cout<<"Diff = "<<d<<endl;
cout<<"Lke to cont .... (y/n) ";
cin>>ch;
}while(ch=='y' || ch=='Y');
return(0);
}
Output:
Enter total elements 5
Enter 0 element 24
Enter 1 element 63
Enter 2 element 2
Enter 3 element 85
Enter 4 element 6
24
63
2
85
6
Max element = 85
Min element = 2
Diff = 83
Lke to cont …. (y/n) y
Enter total elements 3
Enter 0 element 25
Enter 1 element 6
Enter 2 element 45
25
6
45
Max element = 45
Min element = 6
Diff = 39
Lke to cont …. (y/n) n
