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）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string>
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 typedef long double ld;
10 #define MAX 1005
11
12 int n;
13 int c[MAX];
14 double ans;
15 void solve()
16 {
17     for(int i=1;i<=n;i++)
18         cin >> c[i];
19     if(n==1)
20     {
21         ans = c[1];
22         return;
23     }
24     if(n==2)
25     {
26         ans = (c[1]+c[2])/2.0;
27         return;
28     }
29     ans = 0;
30     for(int i=2;i<n;i++)
31         ans += (c[i])/3.0;
32     ans+=c[1]/2.0;
33     ans+=c[n]/2.0;
34 }
35 int main()
36 {
37     while(~scanf("%d",&n))
38     {
39         solve();
40         printf("%.6f\n",ans);
41     }
42     return 0;
43 }