Product Array

There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1]. Solve it without division operator and in O(n).

#include "vector"
#include "iterator"
#include "algorithm"
#include "iostream"
#include "stdlib.h"
using namespace std;

/*
 * Pass 1
   --->
   1 2 3 4 5
       ^
   Pass 2
         <--
   1 2 3 4 5
       ^
*/

void ArrayProduct(vector& array)
{
    const int n = array.size();
    vector product(n);
    int prod = 1;
    for ( int i=0; i    {
        product[i] = prod;
        prod *= array[i];
    }
    prod = 1;
    for ( int i=n-1; i>=0; --i)
    {
        product[i] *= prod;
        prod *= array[i];
    }

    copy ( product.begin(), product.end(),
            ostream_iterator(cout, " ")
         );
}

int main ( int argc, char* argv[])
{
    const int n = atoi(argv[1]);
    vector array(n);
    for ( int i = 1; i<= n; ++i )
        array[i-1] = i;
    ArrayProduct(array);
    return 0;
}