hdu-5753 Permutation Bo(概率期望)

题目链接：

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 131072/131072 K (Java/Others)

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

 

题意：

 

求这个函数值得期望;

 

思路：

 

发现首尾的概率为1/2,中间的为1/3;求一下就好;

 

AC代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  unsigned long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e7+10;
const int maxn=1000+10;
const double eps=1e-8;

int c[1100];

int main()
{
        int n;
        while(cin>>n)
        {
            For(i,1,n)
            {
                read(c[i]);
            }
            double ans=0;
            for(int i=2;i<n;i++)
            {
                ans=ans+c[i]*1.0/3;
            }
            ans=ans+(c[1]+c[n])*1.0/2;
            printf("%.6lf\n",ans);
        }

        return 0;
}