描述 There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following com…

Median Weight Bead Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3162 Accepted: 1630 Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, -, N.…

Median Weight Bead Time Limit: 1000ms Memory Limit: 30000KB This problem will be judged on PKU. Original ID: 197564-bit integer IO format: %lld      Java class name: Main   There are N beads which of the same shape and size, but with different weight…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…

链接:poj 1975 题意:n个珠子,给定它们之间的重量关系.按重量排序.求确定肯定不排在中间的珠子的个数 分析:由于n为奇数.中间为(n+1)/2,对于某个珠子.若有至少有(n+1)/2个珠子比它重或轻,则它肯定不排在中间 能够将能不能确定的权值初始化为0,能确定重量关系的权值设为1 #include<stdio.h> #include<string.h> int a[110][110]; int main() { int T,n,m,i,j,k,d,x,sum; scanf(…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…

Median Time Limit: 1 Second Memory Limit: 65536 KB Recall the definition of the median of elements where is odd: sort these elements and the median is the -th largest element. In this problem, the exact value of each element is not given, but relatio…

动态规划 动态规划 容易: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 不易: , , , , , , , , , , , , , , , , , , , , , , , , 推荐: , , , , , , , , , , , , , , , , , , , , , , , , Jury Compromise False co…

列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029, 2039, 2063, 20…

[1]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887,1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,…

列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029, 2039, 2063, 20…

]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887, 1926(马尔科夫矩阵,求平 衡), 1936,1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,2…

各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…

声明: 1.这份列表不是我原创的,放到这里便于自己浏览和查找题目. ※最近更新:Poj斜率优化题目 1180,2018,3709 列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 195…

动态规划 动态规划 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029, , 2063, 2081, 2082,…

题目描述 有n颗形状和大小都一致的珍珠,它们的重量都不相同.n为整数,所有的珍珠从1到n编号.你的任务是发现哪颗珍珠的重量刚好处于正中间,即在所有珍珠的重量中,该珍珠的重量列(n+1)/2位.下面给出将一对珍珠进行比较的办法:给你一架天平用来比较珍珠的重量,我们可以比出两个珍珠哪个更重一些,在作出一系列的比较后,我们可以将某些肯定不具备中间重量的珍珠拿走. 例如,下列给出对5颗珍珠进行四次比较的情况: 1.珍珠2比珍珠1重 2.珍珠4比珍珠3重 3.珍珠5比珍珠1重 4.珍珠4比珍珠2重 根据以…

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…

BEAD 源程序名 BEAD.???(PAS,C,CPP) 可执行文件名 BEAD.EXE 输入文件名 BEAD.IN 输出文件名 BEAD.OUT 时间限制 1S 内存限制 128MB 有n颗形状和大小都一致的珍珠,它们的重量都不相同.n为整数,所有的珍珠从1到n编号.你的任务是发现哪颗珍珠的重量刚好处于正中间,即在所有珍珠的重量中,该珍珠的重量列(n+1)/2位.下面给出将一对珍珠进行比较的办法: 给你一架天平用来比较珍珠的重量,我们可以比出两个珍珠哪个更重一些,在作出一系列的比较后,我们可…

文章目录 珍珠 题意 分析 增加限定条件 去掉限定条件 Code 珍珠 题意 一共n课珍珠,m颗悬挂,其余在桌子上.如图所示. 仆人每天从某一端"借"一颗珍珠珠.主人每天都会检查悬挂的珍珠数目. 每颗珍珠的摩擦系数都是kkk. 若有wh>kwtw_h \gt kw_twh​>kwt​,则珍珠落地,被主人发现. 珍珠iii的质量是wiw_iwi​,价格是cic_ici​. 问仆人最大可以借多少价钱的珍珠,并输出借的珍珠数目和一种"借"的方案. 分析 增加限…

问题: 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)). Example1:nums1 = [1, 3]nums2 = [2]The median is 2.0Example2:nums1 = [1, 2]n…

Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Different from the previous question where weight…

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…

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)). 这道题让我们求两个有序数组的中位数,而且限制了时间复杂度为O(log (m+n)),看到这个时间复杂度,自然而然的想到了应该使用二分查找法来求解.但是这道题…

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.…

题目简述: There are two sorted arrays A and B 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)). 解题思路: 这本身是个很简单的题目,但是题目要求他的复杂度为O(log(m+n)),就很有难度了.不过首先我们还是可以明确我们要用分治法.关键是怎么分治呢?我们…

B. Rebranding The name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding — an active marketing strategy, that includes a set of measures to change either the brand (both for the…

B. Pasha and PhonePasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits. Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, …

相信大家在日常开发中一定使用过weight这个属性,它的作用一个是权重,另一个就是渲染优先级,但是你真的能很6的使用它嘛?如果不是,那么请继续往下看!!! 我们知道,当weight起到不同作用的时候,weight属性的值可以影响控件的效果.但是具体的影响是怎么样的哪? 当起权重作用的时候,weight属性值越大,占据的空间越大.(要求weight属性必须和0dp一起使用) 当起渲染优先级作用的时候,weight属性值越大,越后渲染. 其实,所谓占据空间大小是根据下面这个公式算出来的: 实际宽(高…

本文地址:http://www.cnblogs.com/wuyudong/p/5898403.html,转载请注明源地址. 如果要实现如下图所示的将按钮居中显示,并且占据父视图的一半,无论屏幕是否旋转 合用weightSum属性和layout_weight属性 不同Android设备的尺寸往往是不同的.作为开发者,我们需要创建适用于不同尺寸屏幕的XML文件.硬编码是不可取的,因此需要其他方法来组织视图. 本文分析如何合用layout_weight和weightSum这两个属性来填充布局内部的任意…