HDU6070-Dirt Ratio

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1087 Accepted Submission(s): 475
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

题目大意： 给出一个序列，问所有连续子序列中，种类/长度最小为多少？
解题思路：线段树+二分

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define lson rt<<1#define rson rt<<1|1using namespace std;typedef long long LL;const int INF=0x3f3f3f3f;const double eps=1e-10;const int MAXN=1e5+5;const int MOD=1e9+7;int a[MAXN],last[MAXN],pre[MAXN];int n;struct segment_tree{    double val[MAXN*4],lazy[MAXN*4];    void pushup(int rt)    {        val[rt]=min(val[lson],val[rson]);    }    void build(int l,int r,int rt,double M)    {        lazy[rt]=0;        if(l==r) {val[rt]=l*M;return;}        int m=(l+r)>>1;        build(l,m,lson,M);        build(m+1,r,rson,M);        pushup(rt);    }    void pushdown(int rt)    {        lazy[lson]+=lazy[rt];lazy[rson]+=lazy[rt];        val[lson]+=lazy[rt];val[rson]+=lazy[rt];        lazy[rt]=0;    }    void update(int L,int R,int c,int l,int r,int rt)    {        if(L<=l&&R>=r)        {            lazy[rt]+=c;val[rt]+=c;return;        }        if(lazy[rt]!=0) pushdown(rt);        int m=(l+r)>>1;        if(L<=m) update(L,R,c,l,m,lson);        if(R>m) update(L,R,c,m+1,r,rson);        pushup(rt);    }    double query(int L,int R,int l,int r,int rt)    {        if(L<=l&&R>=r) return val[rt];        if(lazy[rt]!=0) pushdown(rt);        int m=(l+r)>>1;        if(R<=m) return query(L,R,l,m,lson);        else if(L>m) return query(L,R,m+1,r,rson);        else return min(query(L,m,l,m,lson),query(m+1,R,m+1,r,rson));        pushup(rt);    }    bool check(double M)    {        build(1,n,1,M);        for(int i=1;i<=n;++i)        {            update(pre[i]+1,i,1,1,n,1);            if(query(1,i,1,n,1)<=M*(i+1)) return true;        }        return false;    }};segment_tree st;//val=size(l,r)+mid*l//val<=mid(r+1)int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        memset(pre,0,(n+5)*sizeof(int));        memset(last,0,(n+5)*sizeof(int));        for(int i=1;i<=n;++i)        {            scanf("%d",&a[i]);            pre[i]=last[a[i]];            last[a[i]]=i;        }        double L=0,R=1,M;        for(int i=0;i<20;++i)        {            M=(L+R)/2;            if(st.check(M)) R=M;            else L=M;        }        printf("%.10f\n",L);    }    return 0;}