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 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

 

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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 0x3f3f3f3fusing 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);elseinsert(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;}elsel = mid + eps;}printf("%.6lf\n", ans);}}