CodeForces - D - Maximize!

最新推荐文章于 2024-08-11 19:08:12 发布

原创最新推荐文章于 2024-08-11 19:08:12 发布 · 237 阅读

0 ·

CC 4.0 BY-SA版权

本文探讨了一种涉及正整数多集合的查询问题，主要关注如何通过特定类型的查询操作来寻找集合中最大子集的平均值。文章提供了一个算法解决方案，并通过示例输入输出展示了该算法的有效性和实现细节。

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

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

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 $\text{[math]}$ .

Example

Input

Output

0.0000000000
0.5000000000
3.0000000000

Input

Output

2.0000000000

题解：

可以发现：平均值需要尽可能小，多组数据猜测采取的子集的数据应该取小的加上最后一个数据，而且平均值是一个随个数而变化的单峰极值函数，就是一个三分；

Code：

#include <queue>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int M = 5e5 + 233;
const int Mod = 1e9 + 7;
const int INF = 0x3f3f3f;
typedef long long ll;
ll a[M], s[M], cnt;
inline double Cal(int x) {  return (s[x] + a[cnt]) * 1.0 / (x+ 1);}
int main()
{
     int n, c;
     ll x;
     scanf("%d",&n);
     for(int i=0;i<n;i++)
     {
         scanf("%d",&c);
         if(c == 1) {
                scanf("%lld",&x);
                a[++cnt] = x;
                s[cnt] = s[cnt-1] + x;
        }
         else{
            int l = 1, r = cnt;
            double ans = Cal(1);
            while(l<r)
            {
                int mid = l+r>>1;
                int mmid = mid+r>>1;
                if(Cal(mid) > Cal(mmid)) l = mid;
                else r = mmid;
                ans  = min(ans,min(Cal(mid),Cal(mmid)));
            }
            printf("%.9f\n", a[cnt] - ans );
         }
     }
}
// 3 3  aaa 1 2 2 3 3 1