spoj#RNDORDER. The Least Number
The Least Number
You are given n symbols a1, a2,..., an. You are told that there is a total ordering of the symbols. That is, there is a permutation [P1, P2,..., Pn] of [1,2,...,n] such that aP1< aP2 <...< aPn. You are trying to figure out the order by doing comparisons. The process you follow for determining the order is as follows:
- Compare [a1, a2]
- Compare [a2, a3], [a1, a3]
- Compare [a3, a4], [a2, a4], [a1, a4]
- ....
- ....
- Compare [an-1,an], [an-2,an],..., [a1, an]
Note that you compare in the order specified. That is you compare [a2, a3], then and only then do you compare [a1, a3].
Definition of Compare[ai, aj] (i < j)
- If Compare [ai, aj] = 1, it means ai > aj. If Compare[ai, aj] = -1, it means ai < aj.
- Compare is consistent. Suppose, that you queried [a2, a6] and it was already established [a2 < a6] (because for example a2 < a5 and a5 < a6 - since both of these comparisons happen earlier), then [a2, a6] returns -1.
- If no relationship is known between ai and aj, Compare[ai, aj] = 1 with probablity 1/2 and -1 with probability 1/2
Your task is to output the probability that a1 is the smallest element of the final ordering so obtained.
Input
First line contains T, the number of test cases
Each of the next T lines contains one number each, n(1 <= n <= 1000).
Output
Output T lines in total, one per test case: Probability that a1 is indeed the smallest element at the end of the comparisons. Your output will be judged correct if it differs by no more than 10-9 to the reference answer.
Example
Input:
3
1
2
3</p>Output:
1
0.500
0.3750000
Explanation:
n = 1 is trivial
For n = 2, only comparison is [a1, a2]. a1 is lower with probability 1/2.
For n = 3, a1 is not the least element if either (a1 > a2) or (a1 < a2 and a3 < a2 and a3 < a1).
So, probability that a1 is not the least element = 1/2 + 1/8 = 5/8. Probability that a1 is the least = 3/8 = 0.375.