HDU 5753 Permutation Bo

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 485    Accepted Submission(s): 286
Special Judge

Problem Description

There are two sequences h₁∼h_n and c₁∼c_n. h₁∼h_n is a permutation of 1∼n. particularly, h₀=h_n+1=0.

We define the expression [condition] is 1 when condition is True,is 0 when condition is False.

Define the function f(h)=∑n_i=c_i[h_i>h_i−1  and  h_i>h_i+1]

Bo have gotten the value of c₁∼c_n, 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 c_i(0≤c_i≤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

 

Source

2016 Multi-University Training Contest 3

 

 

 

解析：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。据此即可解答本题（注意特判n == 1的情况）

 

 

 

#include <bits/stdc++.h>

int n;
double c[1005];

int main()
{
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; ++i)
            scanf("%lf", &c[i]);
        if(n == 1){
            printf("%f
", c[1]);
            continue;
        }
        double ans = (c[1]+c[n])/2;
        for(int i = 2; i < n; ++i)
            ans += c[i]/3;
        printf("%f
", ans);
    }
    return 0;
}