I think..
As like no are a,b,c,d,e
so sum will be
a+b,a+c,a+d,a+e,b+c,b+d,b+e,c+d,c+e,d+e;
so maximuum value will be d+e which is last element of array given

take last three value
1.c+d
2.c+e
3.d+e
eq(1)-eq(2)=d-e;
solving it with 3rd eq will give d and e
and with these value we can get other values





On Thu, Feb 24, 2011 at 2:52 AM, radha krishnan <
radhakrishnance...@gmail.com> wrote:

> This s a topcoder problem :)
>
> On Wed, Feb 23, 2011 at 7:16 PM, bittu <shashank7andr...@gmail.com> wrote:
> > If pairwise sums of 'n' numbers are given in non-decreasing order
> > identify the individual numbers. If the sum is corrupted print -1
> > Example:
> > i/p:
> > 4 5 7 10 12 13
> >
> > o/p:
> > 1 3 4 9
> >
> >
> > Thanks & Regards
> > Shashank
> >
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