2017 Multi-University Training Contest

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6070

题面:

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1599    Accepted Submission(s): 740
Special Judge

Problem Description

In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated in the following way. First let's ignore all the problems the team didn't pass, assume the team passed Xproblems during the contest, and submitted Y times for these problems, then the ''Dirt Ratio'' is measured as XY. If the ''Dirt Ratio'' of a team is too low, the team tends to cause more penalty, which is not a good performance.



Picture from MyICPC

Little Q is a coach, he is now staring at the submission list of a team. You can assume all the problems occurred in the list was solved by the team during the contest. Little Q calculated the team's low ''Dirt Ratio'', felt very angry. He wants to have a talk with them. To make the problem more serious, he wants to choose a continuous subsequence of the list, and then calculate the ''Dirt Ratio'' just based on that subsequence.

Please write a program to find such subsequence having the lowest ''Dirt Ratio''.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there is an integer n(1≤n≤60000) in the first line, denoting the length of the submission list.

In the next line, there are n positive integers a1,a2,...,an(1≤ai≤n), denoting the problem ID of each submission.

 

Output

For each test case, print a single line containing a floating number, denoting the lowest ''Dirt Ratio''. The answer must be printed with an absolute error not greater than 10−4.

 

Sample Input

1 5 1 2 1 2 3

 

Sample Output

0.5000000000

Hint

For every problem, you can assume its final submission is accepted.

 

思路：

　　分数规划的思想：二分答案，然后check。

　　check：dif(l,r)/(r-l+1)<=mid (dif(l,r)区间[l,r]的不同数个数）

　　这个形式很难在短时间内check，所以考虑等式变形。

　　变形成==》dif(l,r)+l*mid<=(r+1)*mid

　　在这种形式下可以通过枚举r，然后线段数维护左边的值。对于新加入的一个数，会在区间[pre[x],x]内贡献1，所以进行区间更新即可。

　　具体见代码：

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=6e4+7;
const int mod=1e9+7;

int n,a[K],pre[K],b[K];
double v[4*K],lz[4*K];
void push_down(int o)
{
    v[o<<1]+=lz[o],v[o<<1|1]+=lz[o];
    lz[o<<1]+=lz[o],lz[o<<1|1]+=lz[o];
    lz[o]=0;
}
double update(int o,int l,int r,int pos,double x)
{
    if(l==r) return v[o]=x;
    int mid=l+r>>1;
    push_down(o);
    if(pos<=mid) update(o<<1,l,mid,pos,x);
    else update(o<<1|1,mid+1,r,pos,x);
    v[o]=min(v[o<<1],v[o<<1|1]);
}
double update2(int o,int l,int r,int nl,int nr,double x)
{
    if(l==nl && r==nr) return v[o]+=x,lz[o]+=x;
    int mid=l+r>>1;
    push_down(o);
    if(nr<=mid) update2(o<<1,l,mid,nl,nr,x);
    else if(nl>mid) update2(o<<1|1,mid+1,r,nl,nr,x);
    else update2(o<<1,l,mid,nl,mid,x),update2(o<<1|1,mid+1,r,mid+1,nr,x);
    v[o]=min(v[o<<1],v[o<<1|1]);
}
bool check(double mid)
{
    for(int i=1,mx=n*4;i<=mx;i++) v[i]=1e9,lz[i]=0;
    for(int i=1;i<=n;i++)
    {
        update(1,1,n,i,i*mid);
        update2(1,1,n,pre[i]+1,i,1.0);
        if(v[1]<(i+1)*mid+eps)    return 1;
    }
    return 0;
}
int main(void)
{
    int t;cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        memset(b,0,sizeof b);
        for(int i=1;i<=n;i++) scanf("%d",a+i),pre[i]=b[a[i]],b[a[i]]=i;
        double l=0,r=1;
        for(int i=1;i<=14;i++)
        {
            double mid=(l+r)/2.0;
            if(check(mid)) r=mid;
            else l=mid;
        }
        printf("%.6f
",l);
    }
    return 0;
}