HDU 5753Permutation Bo

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 114    Accepted Submission(s): 67
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

 

 

很容易证明  中间一个比两遍大的概率是1/3  ，端点的概率1/2

乘一下就是期望了 

/* ***********************************************
Author        :guanjun
Created Time  :2016/7/26 16:00:08
File Name     :p302.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int c[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    while(cin>>n){
        for(int i=1;i<=n;i++){
            scanf("%d",&c[i]);
        }
        double ans=0;
        ans+=(c[1]+c[n])/2.0;
        for(int i=2;i<n;i++){
            ans+=c[i]/3.0;
        }
        printf("%.6f\n",ans);
    }
    return 0;
}