这里是一个比较简单的问题：考虑每个数对和的贡献。先考虑数列两端的值，两端的摆放的值总计有2种，比如左端：0，大，小；0，小，大；有1/2的贡献度。右端同理。

中间的书总计有6种可能。小，中，大。其中有两种对答案有贡献，即1/3的贡献度。加和计算可得到答案。

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 693    Accepted Submission(s): 421
Special Judge

Problem Description

There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0.

We define the expression [condition] is 1 when condition is True,is 0 when condition is False.

Define the function f(h)=∑ni=1ci[hi>hi−1  and  hi>hi+1]

Bo have gotten the value of c1∼cn, and he wants to know the expected value of f(h).

Input

This problem has multi test cases(no more than 12).

For each test case, the first line contains a non-negative integer n(1≤n≤1000), second line contains n non-negative integer ci(0≤ci≤1000).

Output

For each test cases print a decimal - the expectation of f(h).

If the absolute error between your answer and the standard answer is no more than 10−4, your solution will be accepted.

Sample Input

4
3 2 4 5
5
3 5 99 32 12

Sample Output

6.000000
52.833333

Author

绍兴一中

#include<iostream>
#include<stdio.h>
using namespace std;
int a[10005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
        {
            scanf("%d",a+i);
        }
        if(n==1)
        {
            printf("%.6lf\n",(a[0]+0.0));
            continue;
        }
        if(n==2)
        {
            printf("%.6lf\n",(a[0]+0.0+a[1])/2);
            continue;
        }
        double ans=0.0;
        ans=(a[0]+0.0+a[n-1]+0.0);
        ans/=2;
        double tmp=0.0;
        for(int i=1; i<n-1; i++)
        {
            tmp+=(a[i]+0.0);
        }
        tmp/=3;
        ans+=tmp;
        printf("%.6lf\n",ans);

    }
    return 0;

}