hdu5753(2016多校第三场，数学题)

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 593    Accepted Submission(s): 360
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 off(h).

If the absolute error between your answer and the standard answer is no more than10−4, your solution will be accepted.

 

Sample Input

43 2 4 553 5 99 32 12

 

Sample Output

6.00000052.833333

题意不多描述了，很容易理解。

思路：对每个位置求总共出现的次数，然后和n！求一个比值，最终的结果就是第i个数乘以第i个比值，可是很巧妙的发现是第一个位置和最后一个位置总是1/2，中间的总是1/3

。所以当大于等于3的时候直接把两端乘以0.5，其他数乘以0.33333333就好，然后就是只有一个数的时候单独判断一下就好……

这份代码是推的公式算的，其实最终的结果都是两边是1/2，中间是1/3.

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <vector>using namespace std;int n;int a[100005];int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        double quan=0;        if(n==1)        {            printf("%.6f\n",a[0]*1.0);        }        else if(n==2)        {            double sun=0;            for(int i=0;i<2;i++)            {                sun+=a[i]/2.0;            }            printf("%.6f\n",sun);        }        else        {            quan=0;            for(int i=2;i<n;i++)                quan+=i*(i-1);            quan/=n*(n-1)*(n-2);            double sum=0;            sum+=a[0]/2.0;            sum+=a[n-1]/2.0;            for(int i=1;i<n-1;i++)                sum+=a[i]*quan;            printf("%.6lf\n",sum);            //cout<<"--quan:  "<<quan<<endl;        }    }    return 0;}