Score : $500$ points

问题描述

设有0号、1号、$\ldots$以及（N-1）号人士，他们按照逆时针方向均匀地围绕在一个转盘周围。第$p_i$道菜位于第i号人士面前的桌子上。

你可以执行以下操作零次或多次：

• 将转盘逆时针旋转 $\frac{1}{N}$ 圈。在旋转前位于第i号人士面前的那道菜现在会出现在第 $(i+1) \bmod N$ 号人士面前。

完成所有操作后，第i号人士会增加 $k$ 点沮丧值，其中$k$是最小整数，使得第i号菜此时位于第 $(i-k) \bmod N$ 号或第 $(i+k) \bmod N$ 号人士面前。

求最终这N位人士沮丧值之和的最小可能值。

什么是$a \bmod m$？对于一个整数$a$和一个正整数$m$，$a \bmod m$表示的是在$0$到$(m-1)$（包含两端点）之间的一个整数$x$，满足$(a-x)$是$m$的倍数。（可以证明这样的$x$是唯一的。）

以上为通义千问 qwen-max 翻译，仅供参考。

Problem Statement

Person $0$, Person $1$, $\ldots$, and Person $(N-1)$ are sitting around a turntable in counterclockwise order, evenly spaced. Dish $p_i$ is in front of Person $i$ on the table.
You may perform the following operation $0$ or more times:

• Rotate the turntable by one $N$-th of a counterclockwise turn. The dish that was in front of Person $i$ right before the rotation is now in front of Person $(i+1) \bmod N$.

When you are finished, Person $i$ gains frustration of $k$, where $k$ is the minimum integer such that Dish $i$ is in front of either Person $(i-k) \bmod N$ or Person $(i+k) \bmod N$.
Find the minimum possible sum of frustration of the $N$ people.

What is $a \bmod m$? For an integer $a$ and a positive integer $m$, $a \bmod m$ denotes the integer $x$ between $0$ and $(m-1)$ (inclusive) such that $(a-x)$ is a multiple of $m$. (It can be proved that such $x$ is unique.)

Constraints

• $3 \leq N \leq 2 \times 10^5$
• $0 \leq p_i \leq N-1$
• $p_i \neq p_j$ if $i \neq j$.
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$

$p_0$ $\ldots$ $p_{N-1}$

Output

Print the answer.

Sample Input 1

4
1 2 0 3

Sample Output 1

2

The figure below shows the table after one operation.

Here, the sum of their frustration is $2$ because:

• Person $0$ gains a frustration of $1$ since Dish $0$ is in front of Person $3\ (=(0-1) \bmod 4)$;
• Person $1$ gains a frustration of $0$ since Dish $1$ is in front of Person $1\ (=(1+0) \bmod 4)$;
• Person $2$ gains a frustration of $0$ since Dish $2$ is in front of Person $2\ (=(2+0) \bmod 4)$;
• Person $3$ gains a frustration of $1$ since Dish $3$ is in front of Person $0\ (=(3+1) \bmod 4)$.

We cannot make the sum of their frustration less than $2$, so the answer is $2$.

Sample Input 2

3
0 1 2

Sample Output 2

0

Sample Input 3

10
3 9 6 1 7 2 8 0 5 4

Sample Output 3

20

update @ 2024/3/10 11:18:43