2017 Multi-University Training Contest - Team 4 hdu6070 Dirt Ratio

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-;
 const double pi=acos(-1.0);
 const int K=6e4+;
 const int mod=1e9+;

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