Even and Odd vector position

Question

So I am trying to change my int vec1 that is populated with all zeroes. What I am doing for every even number I would make it to a 10 and for every odd would be an 11. But I am unsure how to do that and unsure how to print it out. What I thought was to use the % when finding out indices like vec1[0], vec1[2], and so forth are even and would turn them to 10. basically, a hint on how to look into the indices themselves.

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

int main()
{
    vector<int> v;

    int vec1(10);
    for (int i = 0; i <= 10; ++i)
    {
        vec2.push_back(i);
        cout << vec1(i) <<" ";//keep getting error for this but unsure how to print it out
    }


    return 0;
}

Answer 1

You can achieve this like this:

First you have to loop over the v , to check if we are on even index or odd index, we simply take mod with 2 of i . If it is zero then it pushes 10, if not then 11. And in the second loop it simply print the array.

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

int main()
{
    int size = 10;
    vector<int> v(size, 0);

    for (int i = 0; i < size; ++i)
       if( i % 2 == 0)
            v[i] = 10;
       else
            v[i] = 11;
 
    for( auto i: v){
         cout << i << " " ;
     }

    return 0;
}

You can get size via cin >> size .