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)/2;
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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