题目大意:给定$n$个数,第$i$个数为$a_i$,记这$n$个数的所有非空子集的和分别为$s_1,s_2,\dots,s_{2^n-1}$:求$s$的中位数. 题解:假设考虑的是所有子集,包括空子集. 令$sum=\sum\limits_{t=1}^n a_i$. 若有一个子集和为$x$,存在另一个子集和为$sum-x$. 因此不含空子集的中位数等价于所有出现过的数的中间两个的后一个. $0/1$背包就好了,可以用$bitset$优化 卡点:无 C++ Code: #include <cstd…

Median Sum You are given N integers A1, A2, ..., AN. Consider the sums of all non-empty subsequences of A. There are 2N−1 such sums, an odd number. Let the list of these sums in non-decreasing order be S1, S2, ..., S2N−1. Find the median of this list…

Problem Statement You are given N integers A1, A2, ..., AN. Consider the sums of all non-empty subsequences of A. There are 2N−1 such sums, an odd number. Let the list of these sums in non-decreasing order be S1, S2, ..., S2N−1. Find the median of th…

传送门 题意简述:输入一个数组an" role="presentation" style="position: relative;">anan. 对于所有2n−1" role="presentation" style="position: relative;">2n−12n−1个非空子集,每个子集的权值是包含的所有元素之和. 求这2n−1" role="presentatio…

高端操作qaq 又双叒叕读错题了= = 然后重新读题发现不会做了 于是瞅了一波题解 我靠要不要这么暴力呜呜呜 直接bitset O(n^3/w)QAQ 就是f[i]表示i是否能被搞出来 然后我们先不看2^n-1 补上空集就是2^n 然后这就两两对应分出组了 然后我们要找的就是比(sum+1)/2大的第一个元素 bitset强上= = 学习一发bitset正确优化姿势也海星 //Love and Freedom. #include<cstdio> #include<cstring>…

按照顺序来. Median Sum 大意: 给你一个集合,求其所有非空子集的权值的中位数. 某集合的权值即为其元素之和. 1 <= n <= 2000 解: 集合配对,每个集合都配对它的补集. 最大的那个没有配对,所以求(原集合的权值 + 1) >> 1,不小于这个的第一个即为所求. 用bitset实现可行性背包. #include <cstdio> #include <bitset> ; std::bitset<N * N> bt; int a…

Choosing Points 数学 Integers on a Tree 构造 Leftmost Ball 计数dp+组合数学 Painting Graphs with AtCoDeer tarjan+polya Building Cubes with AtCoDeer 枚举 AtCoDeer and Election Report 贪心 Snuke's Coloring 思维题 Snuke's Coloring 2 线段树+单调栈 Make Them Even 贪心 1D Reversi 模…

A - Move and Win 题解 看两个人相遇的时候谁先手即可,相遇之后第一个移动的人必输 代码 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #def…

1.pandas中的列的分位数 # 查看列的分位数 import pandas as pd # set columns type my_df['col'] = my_df['col'].astype(np.float64) # computations for 4 quantiles : quartiles bins_col = pd.qcut(my_df['col'], 4) bins_col_label = pd.qcut(my_df['col'], 4).labels 分位数 2.多重聚合…

---------------------------------------------------------------------------- 一题题目: 一题题解: 这个题目哪来入门再好不过了,支老板之前没有接触过这个东西,然后一点即通:就是把一个int(32位)拆成32个只放0或1的位置,然后这32个的单点操作或者32个一起操作的复杂度是O(1),所以长度位N的bitset的一次单点操作是O(1),整体操作是O(N/w),其中w=32.(long long 是64). 然后Bits…

1.DERIVATIVE()函数 作用:返回一个字段在一个series中的变化率. InfluxDB会计算按照时间进行排序的字段值之间的差异,并将这些结果转化为单位变化率.其中,单位可以指定,默认为1s. 语法: SELECT DERIVATIVE(<field_key>, [<unit>]) FROM <measurement_name> [WHERE <stuff>] 其中,unit取值可以为以下几种: u --microseconds s --seco…

A - Move and Win Time limit : 1sec / Memory limit : 512MB Score : 300 points Problem Statement A game is played on a strip consisting of N cells consecutively numbered from 1 to N. Alice has her token on cell A. Borys has his token on a different cel…

------------------------------------------聚合函数--------------------------------------------- --1: AVG(DISTINCT|ALL) ALL表示对所有的值求平均值,DISTINCT只对不同的值求平均值 SELECT AVG(SAL) FROM SCOTT.EMP; SELECT AVG(DISTINCT SAL) FROM SCOTT.EMP; --2: MAX(DISTINCT|ALL) 求最大值,…

前言: 最近想看看矢量中值滤波(Vector median filter, VMF)在GRB图像上的滤波效果,意外的是找了一大圈却发现网上没有现成的code,所以通过matab亲自实现了一个,需要学习的朋友可以拿过去用.本文的核心是VMF的matlab实现,最后通过在RGB图像上应用举例说明. VMF的数学表达: 含有N个矢量的集合{C1,C2,...CN},它的VMF结果如下所示: 其中,CVM1表示距离所有其他向量的距离和最小的那个向量.而距离可以自己定义,常用的欧氏距离,曼哈顿距离等等.…

D. Sum of Medians time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecu…

The geometric median of a discrete set of sample points in a Euclidean space is the point minimizing the sum of distances to the sample points. This generalizes the median, which has the property of minimizing the sum of distances for one-dimensional…

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examples: [2,3,4] , the median is 3 [2,3], the median is (2 + 3) / 2 = 2.5 Design a d…

传送门 D. Sum of Medians time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five con…

The L1 Median (Weber 1909) 链接网址 Derived from a transportation cost minimization problem, the L1 median is defined to be any point which minimizes the sum of Euclidean distances to all points in the data set (fig.2). As with most median definitions, t…

题目链接: Sum of Medians Time Limit:3000MSMemory Limit:262144KB 问题描述 In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is cal…

D. Sum of Medians time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecu…

一个人力资源咨询集团通过网络爬虫采集手段将多个知名招聘网站上发布的求职和招聘等信息准实时采集到自己的库里,形成一个数据量浩大的招聘信息库,跟踪全国招聘和求职的行业.工种.职位.待遇等信息,并通过商业智能系统,开展职业职位供求及趋势等相关统计分析.这家公司竟然用SSAS OLAP Cube多维数据集容纳如此数量级的数据,广告维成员包含了每一个广告条目.该商业智能团队的开发人员咨询如何用MDX求解薪水中位数.四分位数(Median,Quartile)等. 以下是对Median.Q1.Q3等问题的MD…

HDU 4981 Goffi and Median 思路:排序就能够得到中间数.然后总和和中间数*n比較一下就可以 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int N = 1005; int n, a[N], sum; int main() { while (~scanf("%d…

题目要求: Download the text file here. The goal of this problem is to implement the "Median Maintenance" algorithm (covered in the Week 5 lecture on heap applications). The text file contains a list of the integers from 1 to 10000 in unsorted order;…

Brown-Mood Median Test 对于两独立样本尺度中的位置参数(中位数)检验问题: \(H_0: med_x = med_y\)   \(H_1=med_x > med_y\) 在\(H_0\)假设下,两组数据具有相同的中位数 对于X,Y两组数据组成的两个数组,进行中位数比较. 核心是比较\(med_xy\) 和\(med_x\) \(med-y\)的关系. X Y sum >\(M_{xy}\) A B t <\(M_{xy}\) C D (m+n)-(A+B) \(\s…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). You may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = [1, 3]…

四年前第一次看到<100+ Times FasterWeighted Median Filter (WMF)>一文时,因为他附带了源代码,而且还是CVPR论文,因此,当时也对代码进行了一定的整理和解读,但是当时觉得这个算法虽然对原始速度有不少的提高,但是还是比较慢.因此,没有怎么在意,这几天有几位朋友又提到这篇文章,于是把当时的代码和论文又仔细的研读了一番,对论文的思想和其中的实现也有了一些新的新的,再次做个总结和分享. 这篇文章的官网地址是:http://www.cse.cuhk.edu.h…

E2 - Median on Segments (General Case Edition) 思路: 首先我们计算出solve(m):中位数大于等于m的方案数,那么最后答案就是solve(m) - solve(m+1) 那么怎么计算sovle(m)呢? 对于一个区间[l,r],如果它的中位数大于等于m,那么这个区间中 (大于等于m的数的个数) > (小于m的数的个数) 如果记a[i]大于等于m为+1,小于m 为 -1,即 sum(l, r)  > 0 我们枚举右端点 i ,并且同时计算sum(…

E1. Median on Segments (Permutations Edition) time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given a permutation p1,p2,…,pnp1,p2,…,pn. A permutation of length nn is a sequence suc…

median 中位数 odd 奇数 even 奇数 You are given an array aa of nn integers and an integer ss. It is guaranteed that nn is odd. In one operation you can either increase or decrease any single element by one. Calculate the minimum number of operations required…