HDU5753 Permutation Bo

題目連結：HDU5753

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 451    Accepted Submission(s): 269

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
        

  題意：2個序列h1~hn,c1~cn，第一個是1到n的一個排列，ans是c1到cn的連加，如果ci對應的hi比它周圍2個數都大的話就去ci的值，否則不取。問結果的期望是多少。

題目分析：先求每個點取的機率，兩邊的點隻與1個數比較，是以機率1／2，中間的點與2個數比較，機率1／3；之後求期望即可。注意對1的特判。

//
//  main.cpp
//  Permutation Bo
//
//  Created by teddywang on 2016/7/27.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    int n;
    double ans,buf;
    while(cin>>n)
    {
        ans=0;
        if(n==1)
        {
            cin>>buf;
            printf("%.6lf\n",buf);
        }
        else
        {
            for(int i=0;i<n;i++)
            {
                cin>>buf;
                if(i==0||i==n-1)
                {
                    ans+=buf/2;
                }
                else ans+=buf/3;
            }
            printf("%.6lf\n",ans);
        }
    }
}