cf 462 E - Maximize!

 两个操作 处理和查询 分开做了 首先他要处理的那个中间数值  搞搞就好了，还有注意两个事情 地方开大点 应为开成了 1e5 挂了好多次 E. Maximize! time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

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 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 Copy 6 1 3 2 1 4 2 1 8 2 Output 0.0000000000 0.5000000000 3.0000000000 Input Copy 4 1 1 1 4 1 5 2 Output 2.0000000000

#include <cstdio> using namespace std; #define INF 0x7f const int maxn =1e6+6; typedef long long ll ; typedef double dl ; #define f(i,l,r) for(int i=l;i<=r;++i) #define g(i,l,r) for(int i=l;i>=r;--i) ll stack[maxn],s[maxn]; int index,op,n; dl dd; int l,r,x; int main() { scanf("%d",&n); while(n--) { scanf("%d",&op); if(op==1) { scanf("%d",&x); stack[++index]=x; s[index]=s[index-1]+x; if(index==1) continue; l =1;r=index-1; while(l!=r) { int mid = (l+r)>>1; if(s[mid]+x > 1ll*(mid+1)* stack[mid+1]) l=mid+1; else r=mid; } dd = dd>(dl)x-(dl)(s[l]+x)/(dl)(l+1) ?dd:(dl)x-(dl)(s[l]+x)/(dl)(l+1); } else printf("%.10lf\n",dd); } return 0; }未来的我一定会感谢现在正在成长的我