这里是一个比较简单的问题：考虑每个数对和的贡献。先考虑数列两端的值，两端的摆放的值总计有2种，比如左端：0，大，小；0，小，大；有1/2的贡献度。右端同理。
中间的书总计有6种可能。小，中，大。其中有两种对答案有贡献，即1/3的贡献度。加和计算可得到答案。
Permutation Bo
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 693 Accepted Submission(s): 421
Special Judge
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).
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).
If the absolute error between your answer and the standard answer is no more than 10−4, your solution will be accepted.
3 2 4 5
5
3 5 99 32 12
52.833333
#include<iostream>
#include<stdio.h>
using namespace std;
int a[];
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=; i<n; i++)
{
scanf("%d",a+i);
}
if(n==)
{
printf("%.6lf\n",(a[]+0.0));
continue;
}
if(n==)
{
printf("%.6lf\n",(a[]+0.0+a[])/);
continue;
}
double ans=0.0;
ans=(a[]+0.0+a[n-]+0.0);
ans/=;
double tmp=0.0;
for(int i=; i<n-; i++)
{
tmp+=(a[i]+0.0);
}
tmp/=;
ans+=tmp;
printf("%.6lf\n",ans); }
return ; }