2017多校联合四1004/hdu6070Dirt Ratio [二分+线段树]

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1104    Accepted Submission(s): 486
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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题意：给n个数，设x/y为区间内不同数的个数除以区间长度，问x/y的最小值。

题解：二分答案x/y，当我们去检验mid的时候，会得到如下表达式（size表示区间不同数的个数）

之后我们对表达式进行转换

这样我们就可以根据如上表达式判断mid是否合法。我们每次枚举右端点，当新加入一个点的时候，我们需要对不包含这个数的最右的连续区间进行整体加1，这时候我们就可以，用线段树来维护区间最小值，线段树上的每一个叶子节点表示，size（l,r）+mid*l 。

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)#define eps 1e-8//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;using namespace std;//维护size(l,r)+mid*l<=mid*(r+1)const int maxn=60006;double tree[maxn<<2],add[maxn<<2];int last[maxn];int a[maxn];void pushdown(int root){    tree[root<<1]+=add[root];    tree[root<<1|1]+=add[root];    add[root<<1]+=add[root];    add[root<<1|1]+=add[root];    add[root]=0;}void update(int l,int r,int L,int R,int root,double k){    if(l<=L && r>=R)    {        tree[root]+=k;        add[root]+=k;        return ;    }    if(add[root]>eps)        pushdown(root);    int mid=(L+R)/2;    if(r<=mid) update(l,r,L,mid,root<<1,k);    else if(l>mid) update(l,r,mid+1,R,root<<1|1,k);    else    {        update(l,mid,L,mid,root<<1,k);        update(mid+1,r,mid+1,R,root<<1|1,k);    }    tree[root]=min(tree[root<<1],tree[root<<1|1]);}double query(int l,int r,int L,int R,int root){    if(l<=L && r>=R) return tree[root];    if(add[root]>eps) pushdown(root);    int mid=(L+R)/2;    if(r<=mid)return query(l,r,L,mid,root<<1);    else if(l>mid)return query(l,r,mid+1,R,root<<1|1);    else return min(query(l,mid,L,mid,root<<1),query(mid+1,r,mid+1,R,root<<1|1));    tree[root]=min(tree[root<<1],tree[root<<1|1]);}int main(){    int T;    cin>>T;    while(T--)    {        int n;        cin>>n;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        double l=0,r=1;        for(int bin=0;bin<20;bin++)        {            double mid=(l+r)/2.0;            memset(tree,0,sizeof(tree));            memset(add,0,sizeof(add));            memset(last,0,sizeof(last));            int flag=0;            for(int i=1;i<=n;i++)            {                update(last[a[i]]+1,i,1,n,1,1);                update(i,i,1,n,1,mid*i);                last[a[i]]=i;                double k=query(1,i,1,n,1);                if(k<=(double)mid*(i+1))                {                    flag=1;                    break;                }            }            if(!flag) l=mid;            else  r=mid;        }        printf("%.4f\n",r);    }    return 0;}