Last three values could be:
1.b+e
2.c+e
3.d+e

On Thu, Feb 24, 2011 at 5:09 PM, ashish agarwal <
ashish.cooldude...@gmail.com> wrote:

> 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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