Answer:
#include <iostream>
using namespace std;
int * reverse(int a[],int n)//function to reverse the array.
{
int i;
for(i=0;i<n/2;i++)
{
int temp=a[i];
a[i]=a[n-i-1];
a[n-i-1]=temp;
}
return a;//return pointer to the array.
}
int main() {
int array[50],* arr,N;//declaring three variables.
cin>>N;//taking input of size..
if(N>50||N<0)//if size greater than 50 or less than 0 then terminating the program..
return 0;
for(int i=0;i<N;i++)
{
cin>>array[i];//prompting array elements..
}
arr=reverse(array,N);//function call.
for(int i=0;i<N;i++)
cout<<arr[i]<<endl;//printing reversed array..
cout<<endl;
return 0;
}
Output:-
5
4 5 6 7 8
8
7
6
5
4
Explanation:
I have created a function reverse which reverses the array and returns pointer to an array.I have also considered edge cases where the function terminates if the value of the N(size) is greater than 50 or less than 0.
Answer:
21
Explanation:
The values of c that make it into the loop are 1, 4, 7.
The values that are added to sum are 3 higher, i.e., 4,7 and 10.
The sum of those is 21.
p.s. why did you not run the program yourself?
(My opinion) Naruto! (then Boruto is on it's way to becoming good)
but other people might say my hero academia or any of the dragon ball saga
