@poised:
i think this:
double round(double num)
{ return (int)(num+0.5)
}
works for all...
http://www.ideone.com/WfEIw
On Aug 3, 5:45 pm, Gene <[email protected]> wrote:
> Your solution to 1 works fine. I hope you get the job. But it needs
> O(N) additional storage for the stack. You can also do with constant
> additional storage.
>
> #include <stdio.h>
> int main(void)
> {
> #define N (sizeof a / sizeof a[0])
> int a[] = {7, 9, 4, 8, 2};
> int result[N], i, product;
> for (i = 0, product = 1; i < N; product *= a[i++])
> result[i] = product;
> for (i = N - 1, product = 1; i >= 0; product *= a[i--])
> result[i] *= product;
> for (i = 0; i < N; i++) printf("%d ", result[i]);
> printf("\n");
> return 0;
>
> }
>
> On Aug 3, 7:08 am, Poised~ <[email protected]> wrote:
>
>
>
>
>
>
>
> > I am not looking for answer. Just sharing these Section 2 questions:
>
> > 1. Given an array arr[] of n integers, construct a Product Array prod[] (of
> > same size) such that prod[i] is equal to the product of all the elements of
> > arr[] except arr[i]. Solve it without division operator. Give an efficient
> > code.
> > (if you are interested, here is my solution:http://ideone.com/EaTUF,
> > developed at the test time itself).
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