Codeforces Round 526 (Div. 2) |
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Finished |
The Fair Nut lives in $$$n$$$ story house. $$$a_i$$$ people live on the $$$i$$$-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening.
It was decided that elevator, when it is not used, will stay on the $$$x$$$-th floor, but $$$x$$$ hasn't been chosen yet. When a person needs to get from floor $$$a$$$ to floor $$$b$$$, elevator follows the simple algorithm:
Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the $$$x$$$-th floor. Don't forget than elevator initially stays on the $$$x$$$-th floor.
The first line contains one integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of floors.
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the number of people on each floor.
In a single line, print the answer to the problem — the minimum number of electricity units.
3
0 2 1
16
2
1 1
4
In the first example, the answer can be achieved by choosing the second floor as the $$$x$$$-th floor. Each person from the second floor (there are two of them) would spend $$$4$$$ units of electricity per day ($$$2$$$ to get down and $$$2$$$ to get up), and one person from the third would spend $$$8$$$ units of electricity per day ($$$4$$$ to get down and $$$4$$$ to get up). $$$4 \cdot 2 + 8 \cdot 1 = 16$$$.
In the second example, the answer can be achieved by choosing the first floor as the $$$x$$$-th floor.
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