HDU-5753 Permutation Bo(期望)（概率）

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 863    Accepted Submission(s): 517
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

 

h1-hn是n的1~n的全排列

让计算f(h)的期望，分开求单个的期望贡献，（不太懂期望mark一下），即ci和fi的概率乘积，

fi的概率在两端出现为大于两边的数的情况的概率为1/2，中间出现的概率为1/3

/*

*/
#include
#include
#include
#include
using namespace std;

const int N = 1005;
int n,c[N];
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 1;i <= n;i++)
            scanf("%d",&c[i]);
        double ans = 0.0;
        ans += (double)(c[1]+c[n]) / 2;
        for(int i = 2;i <= n-1;i++)
            ans += (double)c[i]/3;
        printf("%.6f\n",ans);
    }
    return 0;
}