HDU5753 Permutation Bo（2016多校训练）

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 777    Accepted Submission(s): 468
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

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5753

 

根据题意，可以考虑每个位置对期望的贡献。当i不在排列两端的时候，有3! = 6种大小关系，其中有2种对期望有贡献，因此贡献为c_i/3；当i在排列两端时，有2! = 2种大小关系，其中有1种对期望有贡献，因此贡献为c_i/2。（注意特判n == 1的情况）

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define MAX 1005

int n;
int c[MAX];
double ans;
void solve()
{
    for(int i=1;i<=n;i++)
        cin >> c[i];
    if(n==1)
    {
        ans = c[1];
        return;
    }
    if(n==2)
    {
        ans = (c[1]+c[2])/2.0;
        return;
    }
    ans = 0;
    for(int i=2;i<n;i++)
        ans += (c[i])/3.0;
    ans+=c[1]/2.0;
    ans+=c[n]/2.0;
}
int main()
{
    while(~scanf("%d",&n))
    {
        solve();
        printf("%.6f\n",ans);
    }
    return 0;
}