Codeforces Round #204 (Div. 2)

最新推荐文章于 2025-07-28 11:16:14 发布

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原文链接：http://www.cnblogs.com/BobHuang/p/7337854.html

Jeff has become friends with Furik. Now these two are going to play one quite amusing game.

At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes iand i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.

Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.

You can consider that the coin shows the heads (or tails) with the probability of 50 percent.

Input

The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.

Output

In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Examples

input

2
1 2

output

0.000000

input

5
3 5 2 4 1

output

13.000000

Note

In the first test the sequence is already sorted, so the answer is 0.

就是现在给出一个1~n的排列, Jeff和Furik分别轮流进行操作, Jeff先手, Jeff会选择相邻的两个数p[i], p[i + 1]交换位置, 然后轮到Furik, Furiki每次都会抛一个硬币, 出现正面就在序列中选取相邻的满足p[i] > p[i + 1]的两个数交换, 出现反面则选取任意一个相邻的满足p[i] < p[i + 1]的一对数进行交换, 操作时当这个序列变成递增序列的时候操作结束, 假设Jeff每一步都操作都最优(使得接下来剩余的操作次数最少)那么问一共需要的操作步数的期望是多少

其实这个期望就是逆序对数乘2？但是提交了并不对，所以是不是我想的不对啊， Jeff肯定会让逆序数-1，但是Furik是有0.5的可能让逆序数减1，所以如果逆序数是奇数的话，最后一次是Jeff啊，所以需要减1的，但是偶数就不用了

#include <bits/stdc++.h>
using namespace std;
int c[3005];
int n;
int lowbit(int i)
{
    return i&(-i);
}
int insert(int i,int x)
{
    while(i<=n)
    {
        c[i]+=x;
        i+=lowbit(i);
    }
    return 0;
}

int getsum(int i)
{
    int sum=0;
    while(i>0)
    {
        sum+=c[i];
        i-=lowbit(i);
    }
    return sum;
}
int main()
{
    while(cin>>n)
    {
        int ans=0;
        memset(c,0,sizeof(c));
        for(int i=1; i<=n; i++)
        {
            int a;
            cin>>a;
            insert(a,1);
            ans+=i-getsum(a);
        }
        if (ans&1) cout<<2*ans-1<<endl;
        else cout<<2*ans<<endl;
    }
    return 0;
}