El arreglo es palindromo (sin apuntadores).

#include <iostream>

int main () {

    int tam;
    std::cin >> tam;

    int arr [tam];

    for (int i = 0; i < tam ; ++i){
        std::cin >> arr [i];
    }

    int act = 0, cont = 0;
    while (act < tam / 2){
       if (arr[act] == arr[(tam - 1) - act]){
            ++cont;
        } else {
            break;
        }
        ++act;
    }

   std::cout << (cont == tam / 2 ? "Palindromo" : "No es palindromo\n");
}