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 <[email protected]> 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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