Lucky Winners

Question

Passing arrays to functions

A local supermarket is running a promotion: each Nth customer will receive one item for free. Customers names are sequentially given as array of strings (see template).
Write a function that receives the array of customers, its size, and the N number as arguments, and prints the names of the lucky customers each in a new line.

Sample Input
3

Sample output
Rayan
Bruce
Richard
Mary
Remember to pass the size of the array as argument to the function.

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

//complete the function
void winners(string arr[], int size, int n) {
for (int i=0;i>size;i++) {

}
}

int main() {

string customers[] = {"Alice", "Bob", "Rayan", "Emma", "Ann", "Bruce", "Synthia", "Daniel", "Richard", "Sam", "Nick", "Mary", "Paul"};

//getting the lucky number as input
int n;
cin >> n;
//call function
winners(customers, 13, n);

return 0;
}

I am unsure how to represent the variable n which defines the lucky number in my function.

Answer 1

void winners(string arr[], int size, int n) {
for (int i=0; i < size; i++) {
if((i + 1) % n == 0)
cout << arr[i] << endl;
}
}

Answer 2

While (i <size-n){
i+=n;
cout <<customers [i]<<endl;

}

Answer 3

Question is missing some details surely...
But i guess, nth customer means multiple 3 customers, n, n*2, n*3...
Like for 3 => 3,6,9... Customers and index starts from 3 so ith index value when i+1%3==0 is true.
You have i>size infinite, use i<size

Answer 4

..or
void winners(string arr[], int size, int n) {
for (int i = n - 1; i < size; i+=n) {
cout << arr[i] << endl;
}
}

... or
void winners(string arr[], int size, int n) {
for (int i = n; i < size; i+=n) {
cout << arr[i - 1] << endl;
}
}