That is exactly what my solution is doing.

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

> There must be another good solution..please let me know .
> Thanks
>
> 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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>>> >
>>> >
>>>
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>>
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