C - Maximize! CodeForces - 939E （三分

題目來源：

https://vjudge.net/contest/293215#problem/C

http://codeforces.com/problemset/problem/939/E

You are given a multiset S consisting of positive integers (initially empty). There are two kind of queries:

1. Add a positive integer to S, the newly added integer is not less than any number in it.
2. Find a subset s of the set S such that the value is maximum possible. Here max(s) means maximum value of elements in s,  — the average value of numbers in s. Output this maximum possible value of .

Input

The first line contains a single integer Q (1 ≤ Q ≤ 5·105) — the number of queries.

Each of the next Q lines contains a description of query. For queries of type 1 two integers 1 and x are given, where x (1 ≤ x ≤ 109) is a number that you should add to S. It's guaranteed that x is not less than any number in S. For queries of type 2, a single integer 2 is given.

It's guaranteed that the first query has type 1, i. e. S is not empty when a query of type 2 comes.

Output

Output the answer for each query of the second type in the order these queries are given in input. Each number should be printed in separate line.

Your answer is considered correct, if each of your answers has absolute or relative error not greater than 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

Examples

Input

6
1 3
2
1 4
2
1 8
2

Output

0.0000000000
0.5000000000
3.0000000000

Input

4
1 1
1 4
1 5
2

Output

2.0000000000

題意：

第一行是一個Q，表示Q個操作。

操作有兩種：

操作一：在集合S中添加元素x，保證x不小於S中任何一個元素（保證S遞增）

操作二：查詢，s爲S子集，求函數max(s)-mean(s)的最大值，max(s)表示s中元素的最大值，mean(s)表示s中所有元素的平均數。對於每次操作二，輸出對應函數最大值，誤差小於10e-6

n爲S元素個數，由於S遞增，要使max(s)-mean(s)最大，只需要取S中的最後一個元素和S中前m(m<=n-1)個元素即可，而且由1到m函數max(s)-mean(s)的值先增加後減小

思路：三分最值+前綴和

三分原理：

lmid和rmid是l到r的兩個三等分點，如果lmid大於rmid，那麼最大值max一定在l與rmid之間，如果lmid小於rmid那麼最大值max一定在lmid與r之間。

三分講解具體參考：https://blog.csdn.net/forever_dreams/article/details/81093369

比賽的時候就想到是三分，但是不會實現，看見大佬們都沒ac所以直接放棄了嘗試，，，其實賊簡單，沒人做大概是題目沒讀懂吧（翻譯好評Orz

參考代碼：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pi pair<int,int>
#define Mp make_pair
const int inf = 0x3f3f3f3f;
const int N = 5e5+10;
#define ll long long
using namespace std;

ll a[N],sum[N],n,len;
double fun(int x)
{
    return (double)a[len]-((sum[x]+a[len])/(double)(x+1));
}
int main()
{
    ll x,l,r,lmid,rmid;
    double maxx;
    sum[0]=0;
    len=0;
    scanf("%lld",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&x);
        if(x==1)
        {
            len++;
            scanf("%lld",&a[len]);
            sum[len]=sum[len-1]+a[len];
        }
        else
        {
            r=len-1;
            l=1;
            while(r-l>2)
            {
                lmid=l+(r-l)/3;
                rmid=r-(r-l)/3;
                if(fun(lmid)>fun(rmid))
                    r=rmid;
                else if(fun(lmid)<fun(rmid))
                    l=lmid;
                else
                {
                    l=lmid;
                    r=rmid;
                }
            }
            maxx=0;
            for(int j=l; j<=r; j++)
                maxx=max(maxx,fun(j));
            printf("%.8lf\n",maxx);
        }
    }
    return 0;
}