I need some help!

Question

An array of natural numbers is read and it is required to calculate the sum of the elements symmetrically positioned in the array and display them.
for example:
a[0]=7
a[1]=3
a[2]=5
a[3]=4
a[4]=10
a[5]=8
a[6]=2
So 2+7=9
8+3=11
10+5=15
4=4

Answer 1

If you had any question about that feel free and ask.

Answer 2

Here is an example which produces your desired result on a random basis.

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
using namespace std;

int main()
{
srand(static_cast<size_t>(time(0)));

int n = 2 + rand() % (11 - 2); // [2 , 10] length of array
vector<int> arr(n);
bool odd_flag;

for (auto &i : arr) {
i = rand() % 11; // [0, 10] array values range
cout << i << endl;
}

odd_flag = arr.size() % 2 == 0 ? false : true;
cout << endl;

for (size_t i = 0; i < arr.size() / 2 + odd_flag; ++i) {
if (i < arr.size() / 2) {
cout << arr.at(i) << " + " << arr.at(n - (i + 1)) << " = ";
cout << arr.at(i) + arr.at(n - (i + 1)) << endl;
}
else cout << arr.at(i) << " = " << arr.at(i) << endl;
}
}

[https://code.sololearn.com/caNd1bpeQGg0]

Answer 3

In fact, it's a good time for you to expand your knowledge by doing some examples.

size_t is a synonym for unsigned int. I used that because it's necessary to compare the length(size) of a vector type (which is similar to an array but with more features and very easy to use) with an unsigned integral (integer) type.

Instead of the random number generator, you could feed those variable/container elements manually, but as you see it makes your life a lot easier.

Answer 4

I'm all ears to answer all of your questions, my friend. ;D

Answer 5

Thank you Mr. JPM7 to providing us a better solution. @~)~~~~

Answer 6

HINT:- Use loops(yes loop'S') ,preferably, nested loops(loop inside loop)

Try yourself, if you cannot make it, then ask again.

Answer 7

@BabakSheykhan I am a begginer in C++ and since you used other libraries and rand and size_of I'm a bit confused about your example :)) I am looking for something more simple, but thanks for offering help :)

Answer 8

I am blocked at this point, but here is what i wrote Meharban Singh:
for (i=0;i<n/2;i++)
{
for (j=n-1;j>=n/2;j--)
{
s=v[i]+v[j];
}
cout<<s<<endl;
s=0;
}