Hello Ashish, try with this 2-5-6-8-11
so, if we do this then pair wise sum will be
(7-8-10-13) | (11-13-16) | (14 17) | (19)
So the last three elements will be (16-17-19)
Here, 16 is b+e
So,it is not necessary that the last three elements will be exactly
1.c+d
2.c+e
3.d+e
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the problem is of which srm
On Feb 23, 6:46 pm, bittu 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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You recei
http://www.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm182
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@ above
cn u please explain your logic.
On Fri, Jan 21, 2011 at 2:02 PM, juver++ wrote:
> Here is solution from Igor Naverniouk(Google):
> http://shygypsy.com/tools/pairsums.cpp
>
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> To pos
Here is solution from Igor
Naverniouk(Google): http://shygypsy.com/tools/pairsums.cpp
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