Compute a partial sum of array using plus operator only

Problem:
There's an array of numbers A[1]...A[N]. Need to compute another array of numbers S[1]...S[N] such that each S[i] is a sum of all the elements A[1]...A[N] except the element A[i]. In other words:

S[i] = ƩA[j], 1 <= j <= N && j != i

Solution:
We need two passes over the array A[]. One to compute the sum in the forward direction (1..N) and assign the accumulated sum to each S[i] element. Another pass is performed in the backward direction (N..1) and adds the accumulated sum to the corresponding S[i] element. Each of the passes ensures that the sum accumulated in S[i] does not contain the value of A[i] itself:

1. take a variable say, Sum, initialize it as Sum = 0. 
2. For index i=1 to N do : 
      S[i] = Sum; 
      Sum+=A[i]; 
3. Sum = 0; 
4. For index i=N down to 1 do: 
      S[i]+=Sum; 
      Sum+=A[i]; 
5. Done.

Complexity:

time - O(n)
space - O(1)

Links and credits:
http://www.careercup.com/question?id=19301668