hdu 6070 Dirt Ratio

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 384    Accepted Submission(s): 120
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  X problems 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.

   
    
  

 

Source

2017 Multi-University Training Contest - Team 4

 

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Statistic |  Submit |  Discuss |  Note 题意：x为区间数字的种数，y为区间长度，求x/y的最小值。

思路：赛后看官方题解秒懂。。当时有想过二分答案但没细想下去gg。思路和官方一模一样就不细说了。直接上代码：

//#pragma comment(linker, "/STACK:102400000,102400000") 
#include<iostream>  
#include<cmath>  
#include<queue>  
#include<cstdio>  
#include<queue>  
#include<algorithm>  
#include<cstring>  
#include<string>  
#include<utility>
#include<set>
#include<map>
#include<stack>
#include<vector>
#define maxn 60005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
const double eps = 1e-5;
const int mod = 1e6 + 3;
int n, a[maxn], Left[maxn], vis[maxn], mark[maxn << 2];
double minnum[maxn << 2];
void insert(int l, int r, int index, int now, double x){
	if (l == r){
		minnum[now] = x*l;
		return;
	}
	int mid = l + r >> 1;
	if (mark[now]){
		minnum[now << 1] += mark[now];
		mark[now << 1] += mark[now];
		minnum[now << 1 | 1] += mark[now];
		mark[now << 1 | 1] += mark[now];
		mark[now] = 0;
	}
	if (index <= mid)
		insert(l, mid, index, now << 1, x);
	else
		insert(mid + 1, r, index, now << 1 | 1, x);
	minnum[now] = min(minnum[now << 1], minnum[now << 1 | 1]);
}
void update(int l1, int r1, int l2, int r2, int now){
	if (l2 <= l1&&r2 >= r1){
		minnum[now] += 1;
		mark[now]++;
		return;
	}
	int mid = l1 + r1 >> 1;
	if (mark[now]){
		minnum[now << 1] += mark[now];
		mark[now << 1] += mark[now];
		minnum[now << 1 | 1] += mark[now];
		mark[now << 1 | 1] += mark[now];
		mark[now] = 0;
	}
	if (l2 <= mid)
		update(l1, mid, l2, r2, now << 1);
	if (r2 > mid)
		update(mid + 1, r1, l2, r2, now << 1 | 1);
	minnum[now] = min(minnum[now << 1], minnum[now << 1 | 1]);
}
bool solve(double x){
	memset(mark, 0, sizeof(mark));
	for (int i = 1; i < maxn << 2; i++)
		minnum[i] = inf;
	for (int i = 1; i <= n; i++){
		insert(1, n, i, 1, x);
		update(1, n, Left[i] + 1, i, 1);
		if (minnum[1] - x * (i + 1) < eps)
			return true;
	}
	return false;
}
int main(){
	int t;
	scanf("%d", &t);
	while (t--){
		scanf("%d", &n);
		memset(vis, 0, sizeof(vis));
		for (int i = 1; i <= n; i++){
			scanf("%d", &a[i]);
			Left[i] = vis[a[i]];
			vis[a[i]] = i;
		}
		double l = 0, r = 1;
		double ans = 0;
		while (l - r < eps){
			double mid = (l + r) / 2;
			if (solve(mid)){
				ans = mid;
				r = mid - eps;
			}
			else
				l = mid + eps;
		}
		printf("%.6lf\n", ans);
	}
}