bench: search multiple candidates

[akiradeveloper]

Let

• n: number of nodes
• k: choose k nodes

and assume each node has the same cap.

The number of searching segment table will be

sum (n/(n-i)) for i = [0,k)

for example, if n=10 and k=3 it will be 1+10/9+10/8 ~ 3

Write benchmark to check if the actual execution time can be explained by the math.

[akiradeveloper]

done.