100 atcoder#ABC138C. [ABC138C] Alchemist

[ABC138C] Alchemist

Score : 300300 points

Problem Statement

You have a pot and NN ingredients. Each ingredient has a real number parameter called value, and the value of the ii-th ingredient (1iN)(1 \leq i \leq N) is viv_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x+y)/2(x + y) / 2 where xx and yy are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N1N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

  • 2N502 \leq N \leq 50
  • 1vi10001 \leq v_i \leq 1000
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN

v1v_1 v2v_2 \ldots vNv_N

Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10510^{-5}.

2
3 4
3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 33 and 44 is (3+4)/2=3.5(3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

3
500 300 200
375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

  • Use the ingredients of values 500500 and 300300 to produce an ingredient of value (500+300)/2=400(500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200200, resulting in an ingredient of value (400+200)/2=300(400 + 200) / 2 = 300.
  • Use the ingredients of values 500500 and 200200 to produce an ingredient of value (500+200)/2=350(500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300300, resulting in an ingredient of value (350+300)/2=325(350 + 300) / 2 = 325.
  • Use the ingredients of values 300300 and 200200 to produce an ingredient of value (300+200)/2=250(300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500500, resulting in an ingredient of value (250+500)/2=375(250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375375.

Printing 375.0 and so on will also be accepted.

5
138 138 138 138 138
138