CodeForces-352D Jeff and Furik(递推)

D. Jeff and Furik

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 i and i + 1, for which an inequalitypi > 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

21 2

output

0.000000

input

53 5 2 4 1

output

13.000000

Note

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

题意：Jeff和Furik进行一场游戏，给出一个有n个数的序列，他们轮流进行一次操作，Jeff先进行操作：选取相邻两个递减的数交换顺序，然后Furik进行操作：一半的概率选取相邻两个递减的数交换顺序，一半的概率选取相邻两个递增的数交换顺序。求将这个序列变成上升序列需要的操作数的期望

题解：设初始序列的逆序数为cnt，a[i]是将i个逆序变成正序的期望，由于Jeff每次操作都会使逆序数-1，Furik每次操作一半的概率会使逆序数-1，一半的概率会使逆序数+1，因此每经过2次操作逆序数是不增加的，因此我们可以用递推求解。a[i]=0.5*a[i-1-1]+0.5*a[i-1+1]+2 => a[i]=a[i-2]+4

#in<span style="font-family: Arial, Helvetica, sans-serif;">clude<cstdio></span>

#include<algorithm>#include<string.h>#include<math.h>using namespace std;const int maxn = 3000 + 5;int a[maxn];double d[maxn*maxn];int main(){    int n,cnt;  //  freopen("in.txt","r",stdin);    while(~scanf("%d",&n)){        cnt=0;        for(int i=0;i<n;i++) scanf("%d",a+i);        for(int i=0;i<n;i++){            for(int j=i+1;j<n;j++){                if(a[i]>a[j]) cnt++;            }        }        d[0]=0;        d[1]=1;        /*        这个for其实可以省去，根据这个递推式可以知道cnt为偶数的时候答案就是2*cnt        cnt为奇数的时候答案就是cnt*2-1        */        for(int i=2;i<=cnt;i++) d[i]=4+d[i-2];        printf("%.6f\n",d[cnt]);    }    return 0;}