hdu5753Permutation Bo+数学期望

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

Source

2016 Multi-University Training Contest 3

1 2 3

1 2 3 用到了第3位

1 3 2 用到了第2位

2 1 3 用到了第1,3位

2 3 1 用到了第2位

3 1 2 用到了第1,3位

3 2 1 用到了第1位

总共有6种，第一位出现的3次，第二位出现了2次，第三位出现了3次。。

找个大点的排列直接打个表，观察每一位出现的频率，可观察得到第一位和最后一位出现频率为0.5，其他都是1/3；

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define LL long long

int main(){
int n;
int a[1005];
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
if(n==1){
printf("%d\n",a[1]);
}
else{
double ans=0;
ans+=a[1]*0.5;
ans+=a
*0.5;
for(int i=2;i<n;i++) ans+=a[i]*(1.0/3);
printf("%f\n",ans);
}
}
return 0;
}