HDU-6070 Dirt Ratio

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1117    Accepted Submission(s): 493
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

151 2 1 2 3

 

Sample Output

0.5000000000
Hint
 For every problem, you can assume its final submission is accepted.
 

 

Source

2017 Multi-University Training Contest - Team 4

题意：

简单理解就是 在连续子区间内

比值 = 元素种类个数 / 连续子区间长度 

求比值的最小值

题解：

1，首先二分答案，来逼近答案； 
2，列出求解式子：

size(l,r)/(r-l+1)<=mid 
size(l,r)为区间(l,r)中不同的元素个数； 
mid是二分的结果（即比值）； 
接下来式子可以转化为：size(l,r)+mid*l<=(r+1)*mid; 
然后，枚举区间右端点从1到n 
从线段树中去维护size(l,r)+mid*l的最小值； 

#include<stdio.h>#include<string.h>#include<algorithm>#define ll long longusing namespace std;const int MAX = 1e5 + 5;const ll INF = 1e10;int last[MAX],pre[MAX],n;double add[MAX*4],sum[MAX*4];void UP(int rt){sum[rt] = min(sum[rt*2],sum[rt*2+1]);}void Down(int rt){add[rt*2] += add[rt];add[rt*2+1] += add[rt];sum[rt*2] += add[rt];sum[rt*2+1] += add[rt];add[rt] = 0;}void build(int l,int r,int rt){sum[rt] = add[rt] = 0;if(l == r)return ;int m = (l + r) >> 1;build(l,m,rt*2);build(m+1,r,rt*2+1);UP(rt);}void update(int L,int R,double c,int l,int r,int rt){if(L <= l && R >= r){add[rt] += c;sum[rt] += c;return ;}Down(rt);int m = (l + r) >> 1;if(L <= m)update(L,R,c,l,m,rt*2);if(R > m)update(L,R,c,m+1,r,rt*2+1);UP(rt);}double query(int L,int R,int l,int r,int rt){if(L <= l && R >= r)return sum[rt];Down(rt);int m = (l + r) >> 1;double ans = n;if(L <= m)ans = min(ans,query(L,R,l,m,rt*2));if(R > m)ans = min(ans,query(L,R,m+1,r,rt*2+1));UP(rt);return ans;}bool solve(double k){build(1,n,1);for(int i = 1;i <= n;i++){update(pre[i] + 1,i,1,1,n,1);update(1,i,-k,1,n,1);if(query(1,i,1,n,1) <= 0)return 1;}return 0;}int main(){int T,x;scanf("%d",&T);while(T--){scanf("%d",&n);memset(last,0,sizeof(last));memset(pre,0,sizeof(pre));for(int i = 1;i <= n;i++){scanf("%d",&x);pre[i] = last[x];last[x] = i;}double l = 0,r = 1;for(int i = 0;i < 20;i++){double m = (l + r) / 2;if(solve(m))r = m;elsel = m;}printf("%.9lf\n",r);}return 0;}