nice approach Ashish....thkz

On Fri, Jul 9, 2010 at 11:14 AM, Ashish Goel <ashg...@gmail.com> wrote:

> a minor correction
>
>> nC2 possibilities
>>
>> so n can be found using nC2=say6 then n=4
>>
>>
>> a+b=p0
>> a+c=p1
>> a+d=p2
>> b+c=p3
>> b+d=p4
>> c+d=p5
>>
>> 3a+b+c+d=po+p1+p2
>> b+c+d=(p3+p4+p5)/2
>>
>> so a= (2(p0+p1+p2)-p3-p4-p5)/6
>>
>>
>>
>> take case of 5 numbers {1,2,3,4,5}
>> P={3,4,5,6,5,6,7,7,8,9}
>> 5C2=10 so n=5
>>
>> //b+c+b+d+b+e+c+d+c+e+d+e=3b+3c+3d+3e
>> for (int i=0;i<n-1;i++)
>> {sumX+=p[i];}
>>
>> for (int j=i+1;j<nC2;j++)
>> sumY+=p[j];
>> sumY=sumY/(n-2);
>> a[0]=(sumX-sumY)/(n-1);
>>
>> using this rest of the numbers can be found
>>
>>
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Fri, Jul 9, 2010 at 9:21 PM, amit <amitjaspal...@gmail.com> wrote:
>>
>>> Given a list of numbers, A = {a0, a1, ..., an-1}, its pairwise sums P
>>> are defined to be all numbers of the form ai + aj for 0 <= i < j < n.
>>> For example, if A = {1,2,3,4}, then
>>> P = {1+2, 1+3, 1+4, 2+3, 2+4, 3+4} = {3, 4, 5, 5, 6, 7}.
>>> Now give you P, design an algorithm to find all possible A.
>>>
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>>
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