Pikachu vs Team Meowstic and Helping Hand

Pikachu loves battling with other Pokemon. This time he has a team of 

$N$ Meowstic to fight, $i^{th}$ of which has strength $a_i$. He wants to fight with all of them $K$ times. Team Meowstic came to know about this and now they have devised a strategy to battle against the mighty Pikachu.

All the $N$ Meowstic stand in a straight line numbered from $1$ to $N$. Before every round of battle, they simultaneously use a move, called Helping Hand. It changes the attacking power of Team Meowstic as follows:

• The attacking power of first Meowstic remains $a_1$.
• The attacking power of the remaining Meowstic changes as $a_i=a_i|a_{i-1}$ where $2\le i\le N$ and $A|B$ represents the bitwise OR of $A$ and $B$ 

For example, if the current attacking powers are $[2,1,4,6,3]$, after using the Helping Hand, the powers change to $[2,1|2,4|1,6|4,3|6]$, or $[2,3,5,6,7]$.

Help Pikachu by finding the attacking powers of all Meowstic when he fights each of them for the last time, that is, for the $K^{th}$ round.

Note that, the influence of Helping Hand remains forever, and attacking powers DO NOT revert back after any round.

Example:

Input: n = 5, k = 3, arr = {1, 2, 3, 4, 5}
Output: 1 3 3 7 7

Approach

C++

#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;

int main()
{

    long long n = 5, k = 3;
    long long temp = 0, temp1;

    vector<long long> arr = {1, 2, 3, 4, 5};

    if (k > 100)
    {
        k = 100;
    }
    while (k--)
    {
        temp = 0;

        for (long long i = 0; i < n; i++)
        {
            temp1 = arr[i];
            arr[i] = arr[i] | temp;
            temp = temp1;
        }
    }
    for (auto i : arr)
    {
        cout << i << " ";
    }

    return 0;
}