CodeForces - 846F Random Query(期望)

You are given an array a consisting of n positive integers. You pick two integer numbers l and r from 1 to n, inclusive (numbers are picked randomly, equiprobably and independently). If l > r, then you swap values of l and r. You have to calculate the expected value of the number of unique elements in segment of the array from index l to index r, inclusive (1-indexed).

Input

The first line contains one integer number n (1 ≤ n ≤ 10⁶). The second line containsn integer numbers a₁, a₂, ... a_n (1 ≤ a_i ≤ 10⁶) — elements of the array.

Output

Print one number — the expected number of unique elements in chosen segment.

Your answer will be considered correct if its absolute or relative error doesn't exceed 10^- 4 — formally, the answer is correct if , where xis jury's answer, and y is your answer.

Example

Input

2
1 2

Output

1.500000

Input

2
2 2

Output

1.000000

题意：给你一个序列a，随机生成l,r有可能l>r，则看做r,l处理。将权值定义为l-r中不同数字的个数，求期望。

题解：第一道期望题！首先肯定是想暴力，枚举每一种l,r，扫l-r中不同的数的个数，复杂度（n^3），考虑优化，对于每个值k，他能对前一个值为k的位置到如今的所有l和之后的所有r做贡献，贡献为（i-pre[k]）*(n-i+1),因为l,r可互换，所以有两种可能性,所以贡献为（i-pre[k]）*(n-i+1)。
因为l=r的情况只算一种，但之前算了两遍，所以应该减去。

代码如下：

#include<cstdio>
#include<algorithm>
using namespace std;

long long pre[1000010],n,i,j,ans;

int main()
{
    scanf("%lld",&n);
    ans-=n;
    for(i=1;i<=n;i++)
    {
        long long x;
        scanf("%lld",&x);
        ans+=(i-pre[x])*(n-i+1)*2;
        pre[x]=i;
    }
    double hehe=(double)ans/(n*n);
    printf("%.6lf",hehe);
}

 

 

 

 

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