[BZOJ3834][Poi2014]Solar Panels Description Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the gold as within a few days  clients walked through his door. Each client has ordered a si…

题目描述 Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the gold as within a few days  clients walked through his door. Each client has ordered a single rectangular panel with specified w…

题意 询问两个区间[smin,smax],[wmin,smax]中是否存在k的倍数,使得k最大 分析 将其转化成\([\frac{smin-1}k,\frac{smax}k],[\frac{wmin-1}k,\frac{wmax}k]\) 用分块思想做,注意到这只有\(O(\sqrt{n})\)种取值,于是可以分段计算,做到\(O(\sqrt{n})\)每次询问 我的理解:每一块为[i,j],j=b/(b/i)表示该块的最右端,而i表示该块的最左端,b在[i,j]上的值b/i相同 trick 代…

3834: [Poi2014]Solar Panels Time Limit: 20 Sec  Memory Limit: 128 MBSubmit: 367  Solved: 285[Submit][Status][Discuss] Description Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the go…

题目描述 Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the gold as within a few days  clients walked through his door. Each client has ordered a single rectangular panel with specified w…

题目链接 BZOJ3834 题解 容易想到对于\(gcd(x,y) = D\),\(d\)的倍数一定存在于两个区间中 换言之 \[\lfloor \frac{a - 1}{D} \rfloor < \lfloor \frac{b}{D} \rfloor\] \[\lfloor \frac{c - 1}{D} \rfloor < \lfloor \frac{d}{D} \rfloor\] 整除分块即可做到\(O(n\sqrt{max\{b\}})\) #include<algorithm&…

问题相当于找到一个最大的k满足在$[x_1,x_2]$,$[y_1,y_2]$中都有k的倍数 等价于$\frac{x_2}{k}>\frac{x_1-1}{k}$且$\frac{y_2}{k}>\frac{y_1-1}{k}$ 注意到这只有$O(\sqrt{n})$种取值,于是可以分段计算,做到$O(\sqrt{n})$每次询问 #include<cstdio> int T,a,b,c,d,i,j,t; inline void up(int&a,int b){if(a>…

http://www.lydsy.com/JudgeOnline/problem.php?id=3834 题意:求$max\{(i,j)\}, smin<=i<=smax, wmin<=i<=wmax$,其中$smin<=smax<=10^9, wmin<=wmax<=10^9$,有$N<=1000$组数据 #include <bits/stdc++.h> using namespace std; int main() { int cs,…

整除分块枚举... 真的没有想到会这么简单. 要使一个数 \(p\) 满足 条件, 则 存在\(x, y\), \(a<=x \times p<=b\ \&\&\ c<=y \times p <=d\) 把\(p\) 除掉 则 \(\left\lceil\dfrac{a}{p}\right\rceil <=y <=\left\lfloor\dfrac{b}{p}\right\rfloor\) \(\left\lceil\dfrac{c}{p}\right…

题目大意: $T(T\le1000)$组询问,每次给出$A,B,C,D(A,B,C,D\le10^9)$,求满足$A\le x\le B,C\le y\le D$的最大的$\gcd(x,y)$. 思路: 令$n=\gcd(x,y)$,则若$n$为合法的答案,当且仅当$\lfloor\frac{A-1}n\rfloor<\lfloor\frac Bn\rfloor,\lfloor\frac{C-1}n\rfloor<\lfloor\frac Dn\rfloor$. 考虑数论分块,每次用块内最大值…

  Indian scientists have designed a new device they hope will solve one of the biggest problems with the use of solar energy. 印度科学家设计了一种新的设备,希望能够解决一个太阳能利用中的最大问题.   They call the device a solar tree. 他们将设备称之为太阳能树.   Solar trees have metal "branches&qu…

Skyscrapers Covered in Solar Panels An office tower on Miller Stree in Manchester is completely covered in solar panels. 曼彻斯特米勒大街上的一座办公大楼完全被太阳能电池板覆盖. They are used to create some of the energy used by the insurance company inside. 他们被用来制造大楼内部保险公司使用的部…

POI2014题解 [BZOJ3521][Poi2014]Salad Bar 把p当作\(1\),把j当作\(-1\),然后做一遍前缀和. 一个合法区间\([l,r]\)要满足条件就需要满足所有前缀和\(\ge 0\),所有后缀和\(\ge 0\),也就是\(\forall i\in[l,r],sum_i-sum_{l-1}\ge 0,sum_r-sum_{i-1}\ge 0\). 也就是说\(sum_{l-1}\)要是\([l-1,r]\)内的最小值,\(sum_r\)要是\([l-1,r]\…

Nexiq 125032 usb link is Diesel Truck diagnostic Interface. Nexiq truck scanner can compatible with 17 software. Below auto diagnostic obd sharing Tips of Nexiq wireless adapter Nexiq 125032 usb link update. Many individuals will probably be interest…

Manoeuvring a satellite in orbit usually requires thrusters. Sometimes the thrust is provided by a fuel-burning rocket motor. Sometimes it comes from electrically heated gas. Both methods, though, add weight in the form of propellant, thus reducing l…

https://www.ted.com/talks/tabetha_boyajian_the_most_mysterious_star_in_the_universe/transcript00:12Extraordinary claims require extraordinary evidence, and it is my job, my responsibility, as an astronomer to remind people that alien hypotheses[haɪˈp…

This is the view from the instrument deployment camera of InSight, America’s latest probe to Mars, which landed safely on November 26th. InSight joins one, or possibly two, other missions now operating on the Martian surface (an American rover called…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

1. 主页:http://log4cpp.sourceforge.net“Log4cpp is library of C++ classes for flexible logging to files, syslog, IDSA and other destinations. It is modeled after the Log4j Java library, staying as close to their API as is reasonable.” api文档地址:http://log…

In particular embodiments, a method includes, from an indexer in a sensor network, accessing a set of sensor data that includes sensor data aggregated together from sensors in the sensor network, one or more time stamps for the sensor data, and metad…

很多觉得自己英语成绩还行的同学经常在自己的文章里用一些浮夸或者是难度特别大的词语,以显示自己的英语水平.当然了,成功了倒也无可厚非,要是有些词语用错了那可是会被别人笑掉大牙的.因此英语中的精准用词就成了留学生大问题了.下面就和小编来看看要怎么增强用词的准确性. 先来看个例子: Enter Stregato the self-named assidious low-profile who had a late start of English-learning odyssey where mult…

很多海外留学生在Essay写作时往往不善于对单词进行变化,不能将同一个意思用不同的方式表达出来,使得Essay显得单调乏味最终拿不到高分.小编建议大家应该尽量减少Essay写作中的重复用词.本文将为大家介绍这几种方法减少重复用词,实现用词多样化: 方法一:同义词替换 这是最简单的一种替换方法,当句子或段落中一个词出现太多次时,我们可以用单词的同义词来替换.举两个<经济学人>的例子: Just as drones can make up for poor roads,the theory goe…

A. Radio Prize All boring tree-shaped lands are alike, while all exciting tree-shaped lands are exciting in their own special ways.What makes Treeland more exciting than the other tree-shaped lands are the raddest radio hosts in the local area: Roota…

背景 在网站开发中,文件上传是很常见的一个功能.相信很多人都会遇到这种情况,想传一个文件上去,然后网页提示"该文件过大".因为一般情况下,我们都需要对上传的文件大小做限制,防止出现意外的情况. 但是在有些业务场景中,大文件上传又是必须的,比如邮箱附件,或者内部OA等等. 问题 服务端为什么不能直接传大文件?跟php.ini里面的几个配置有关 upload_max_filesize = 2M //PHP最大能接受的文件大小 post_max_size = 8M //PHP能收到的最大PO…

传送:主席树做法http://www.cnblogs.com/candy99/p/6160704.html 做那倒带修改的主席树时就发现分块可以做,然后就试了试 思想和教主的魔法差不多,只不过那个是求>=v的有几个 既然一个数v的名次可以求,我们二分这个数就行了啊 然而...... 首先,你二分到的这个数不一定在区间里出现过 比如 1 2 5 8 9 4和5的名次都是3 于是,我修改了某个区间名次的定义: “如果一个数的名次是x,但是区间中没有次数,那么他的名次为x-1” 实现上只需要find里…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4467 题意:给定n个点m条边的无向图,点被染色(黑0/白1),边带边权.然后q个询问.询问分为两种: Change u:把点u的颜色反转(黑变白,白变黑),Asksum a b(a,b的值为0/1):统计所以边的权值和,边的两个点满足一个点的颜色为a,一个点的颜色为b. 思路:考虑暴力的做法.修改可以做法O(1),但是查询就得O(m).所以总复杂度为O(m*q)会TLE.然后考虑图分块.参考HDU…

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 3192    Accepted Submission(s): 371 Problem Description You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negat…

题意: 求一个序列中顺序的长度为3的等差数列. SOL: 对于这种计数问题都是用个数的卷积来进行统计.然而对于这个题有顺序的限制,不好直接统计,于是竟然可以分块?惊为天人... 考虑分块以后的序列: 一个块内直接枚举统计三个或两个在块内的. 只有一个在当前块我们假设它是中间那个,对左右其它块做卷积. 但是还是感觉复杂度有点玄学啊... 我比较傻逼...一开始块内统计根本没有想清楚...最后做卷积硬生生把复杂度变成了 $\sqrt{N}*N*log(N)$... 改了一个晚上终于没忍住看标程...…

据说是道lct求深度的题 但是在小猫大的指点下用分块就n^1.5水过了 = =数据忘记加强系列 代码极其不美观,原因是一开始是听小猫大讲的题意,还以为是弹到最前面... #include <cstdio> #include <cmath> using namespace std; int n,m,p,q,k; ],b[],c[]; int main() { scanf("%d",&n); ;i<=n;i++) scanf(]); int N=(in…

二分法(必须要保证数据是有序排列的):   分块查找(数据有如下特点:块间有序,块内无序):    …