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.
感觉现在的数据结构越来越不单纯了,看来还是得全面发展,这样一个东西我都没有想到:
sum指的是l到r之间有多少不相同的数,然后就for一遍r,居然不会tQAQ
#include<bits/stdc++.h>
using namespace std;
const int maxn=6e4+5;
double sum[maxn*4],flag[maxn*4];
int n;
void pushup(int root)
{
sum[root]=min(sum[root<<1],sum[root<<1|1]);
}
void pushdown(int root)
{
if(!flag[root])
return;
sum[root<<1]+=flag[root];
sum[root<<1|1]+=flag[root];
flag[root<<1]+=flag[root];
flag[root<<1|1]+=flag[root];
flag[root]=0;
}
void build(int l,int r,int root,double m)
{
flag[root]=sum[root]=0;
if(l==r)
{
sum[root]=(double)l*m;
return ;
}
int mid=l+r>>1;
build(l,mid,root<<1,m);
build(mid+1,r,root<<1|1,m);
pushup(root);
}
void update(int l,int r,int root,int ql,int qr,double val)
{
if(l>=ql&&r<=qr)
{
flag[root]+=val;
sum[root]+=val;
return ;
}
pushdown(root);
int mid=l+r>>1;
if(mid>=ql)
update(l,mid,root<<1,ql,qr,val);
if(mid<qr)
update(mid+1,r,root<<1|1,ql,qr,val);
pushup(root);
}
double query(int l,int r,int root,int ql,int qr)
{
if(l>=ql&&r<=qr)
return sum[root];
int mid=l+r>>1;
double ans=1e7;
if(mid>=ql)
ans=min(ans,query(l,mid,root<<1,ql,qr));
if(mid<qr)
ans=min(ans,query(mid+1,r,root<<1|1,ql,qr));
return ans;
}
int pre[maxn],last[maxn];
int check(double m)
{
build(1,n,1,m);
for(int i=1;i<=n;i++)
{
update(1,n,1,pre[i]+1,i,1);
//cout<<i<<" "<<query(1,n,1,1,i)<<endl;
if(query(1,n,1,1,i)<=(double)m*(i+1))
return 1;
}
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(last,0,sizeof(last));
int x;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
pre[i]=last[x];
last[x]=i;
}
double low=0,high=1,mid;
while(high-low>=1e-5)
{
mid=(low+high)/2;
if(check(mid)==1)
high=mid-1e-6;
else
low=mid+1e-6;
}
printf("%.10f\n",high);
}
return 0;
}