题目链接：
Permutation Bo
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
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).
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).
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.
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
题意：
求这个函数值得期望;
思路：
发现首尾的概率为1/2,中间的为1/3;求一下就好;
AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef unsigned long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e7+10;
const int maxn=1000+10;
const double eps=1e-8;
int c[1100];
int main()
{
int n;
while(cin>>n)
{
For(i,1,n)
{
read(c[i]);
}
double ans=0;
for(int i=2;i<n;i++)
{
ans=ans+c[i]*1.0/3;
}
ans=ans+(c[1]+c[n])*1.0/2;
printf("%.6lf
",ans);
}
return 0;
}