B2. Shave Beaver!   The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily! There…

A 开个数组记录一下 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #include<cmath> #include<map> using namespace std; #define LL long…

传送门:http://codeforces.com/contest/1108/problem/E2 E2. Array and Segments (Hard version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions i…

题目链接: http://codeforces.com/problemset/problem/522/D D. Closest Equals time limit per test3 secondsmemory limit per test256 megabytes 问题描述 You are given sequence a1, a2, ..., an and m queries lj, rj (1 ≤ lj ≤ rj ≤ n). For each query you need to print…

题目链接:http://codeforces.com/contest/799/problem/C 题目: 题意: 给你n种喷泉的价格和漂亮值,这n种喷泉题目指定用钻石或现金支付(分别用D和C表示),C和D之间不能相互转换.你现在需要修建两个喷泉,给你硬币数和现金数,问你怎样才能使修建的两个喷泉的总漂亮值最大. 思路: 易知,要修建的两个喷泉如果一个是用钻石支付,另一个用现金支付,那么只需找到小于给的钻石和现金的上限的漂亮值最大的两个温泉相加,此处遍历一边即可.对于这两个温泉都用同一种支付方式的情…

D. Closest Equals Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline/problem.php?id=3224 Description You are given sequence a1, a2, ..., an and m queries lj, rj (1 ≤ lj ≤ rj ≤ n). For each query you need to print the minimu…

Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them. Vasily decided to sort the cards. To do this,…

$ composer install Loading composer repositories with package information Updating dependencies (including require-dev) Your requirements could not be resolved to an installable set of packages. Problem 1 - phpunit/phpunit 5.7.5 requires php ^5.6 ||…

今天运行程序的时候出现了: You are using pip version 10.0.1, however version 18.1 is available.You should consider upgrading via the 'python -m pip install --upgrade pip' command. 对于很少用python的我,瞬间蒙了,尝试了很多方法之后才解决.希望分享给大家,我的解决方法吧 You are using pip version 10.0.1, h…

今天想用python代替shell做运维相关的事,写代码都是在本机,调试在服务器上 C:\Users\0>pip install psutilRequirement already satisfied: psutil in f:\programdata\anaconda3\lib\site-packages (5.4.5)distributed 1.21.8 requires msgpack, which is not installed.You are using pip version 10…

解决问题 You are using pip version 9.0.3, however version 10.0.1 is available.You should consider upgrading via the 'python -m pip install --upgrade pip' command. 解决方法 直接运行命令:python -m pip install --upgrade pip 完美!!!成功解决!!!…

使用pip命令安装或卸载第三方库时报You are using pip version 9.0.3, however version 18.0 is available.错误,一般情况下是pip版本过期需要更新pip版本. 使用如下方式可更新pip版本python -m pip install --upgrade pip…

pip安装selenium,pip install selenium 类型这样错误 1  原因可能不是以管理员身份运行cmd安装selenium 2  解决方式 也是要管理员身份运行 重点在最后一句 You are using pip version 9.0.1, however version 19.0.3 is available.You should consider upgrading via the 'python -m pip install --upgrade pip' comma…

在我们安装第三方库的时候会在结尾出现如下两行内容 You are using pip version 9.0.1, however version 18.0 is available. You should consider upgrading via the 'python -m pip install --upgrade pip' command. 解决 按照上面提示,执行下面代码(windows下要用管理员权限,linux下使用sudo) python -m pip install --u…

在安装第三方库时,出现如下提示: You are using pip version 10.0.1, however version 20.2.2 is available.You should consider upgrading via the 'python -m pip install --upgrade pip' command. 然后按照提示升级操作,又提示我已经是最新版本.还提示我添加上--trusted-host mirrors.aliyun.com命令试试 其实这样也解决不了问…

报错: You are using pip version 10.0.1, however version 18.0 is available. You should consider upgrading via the 'python -m pip install --upgrade pip' command. Win+R 输入 cmd 回车,运行了 python -m pip install --upgrade pip 即可.     虽然网络很慢,你看到了,6.8kB/s 在 Anacon…

我已经升级到了最新的版本 安装其他模块过程中出现下面提示,便说明你需要升级pip You are using pip version 10.0.1, however version 21.3.1 is available.You should consider upgrading via the 'python -m pip install --upgrade pip' command 当在下面出现Success,那么恭喜你安装成功. 失败的看这里: 失败的首先可以尝试更换升级指令,例如: py…

题意给定一个长度为n的序列,和m个区间.对一个区间的操作是:对整个区间的数-1可以选择任意个区间(可以为0个.每个区间最多被选择一次)进行操作后,要求最大化的序列极差(极差即最大值 - 最小值).easy version的范围是(1≤n≤300,0≤m≤300)hard version的范围是(1≤n≤1e5,0≤m≤300) 分析: 我们可以通过枚举最大或者最小值出现的位置 , 然后把不包括最大值位置或者包括最小值位置的区间选取 , 如果数据小我们是可以通过暴力的 , 但是这里的n有1e5 ,…

CodeForce 855B 暴力or线段树 题意 给你一串数,然后找出三个数,他们的前后关系和原来一样,可以相同,然后分别乘p,q,r,求他们积的和最大,并且输出这个数. 解题思路 这个可以使用线段树来做,找出区间内的最小值和最大值,如果x(代表pqr中的一个)小于零,就乘以这个区间的最小值,如果大于零,就乘以这个区间的最大值.然后\(j\)从1到n开始遍历. 或者可以暴力,不过这个暴力比一点重方法还要好,我看完就惊呆了,lxm大佬太强了. 代码实现 第一种: #include<cstdio>…

Codeforces Round #620 F2. Animal Observation (hard version) (dp + 线段树) 题目链接 题意 给定一个nm的矩阵,每行取2k的矩阵,求总共矩阵里的数的和最大值,重复取到的数不算 题解 dp[i]表示当前行从第i个数开始取矩阵的最大值 dp[i] = 上一行中最大数 + 当前行第i个数到第i+k-1个数的和 - 当前行重复的 + 下一行第i个数到第i+k-1个数的和 用线段树维护 上一行中最大数 + 当前行第i个数到第i+k-1个数的…

1304F2 - Animal Observation (hard version) 线段树or单调队列 +DP 题意 用摄像机观察动物,有两个摄像机,一个可以放在奇数天,一个可以放在偶数天.摄像机在同一天可以同时照到k个区域放下去可以持续两天.现在给出每一天每个区域的动物数量,问最多照到动物多少个.如果两个照相机同时照到一个动物只算一次.n<=50 k<=m<=2e4 思路 我们可以考虑只在一天的情况 那么就是一个简单的dp,状态为dp[i][j]第i天放置在区域j可以获得的最大数.那…

题目链接:https://codeforces.com/contest/1089/problem/K time limit per test: 2 seconds memory limit per test: 512 megabytes King Kog got annoyed of the usual laxity of his knights — they can break into his hall without prior notice! Thus, the King decided…

Block Art 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/block-art Description The NeoCubist artistic movement has a very distinctive approach to art. It starts with a rectangle which is divided into a number of squares. T…

http://codeforces.com/contest/760/problem/E 题目大意:现在对栈有m个操作,但是顺序是乱的,现在每输入一个操作要求你输出当前的栈顶, 注意,已有操作要按它们的时间顺序进行. 思路:线段树好题啊啊,我们把push当成+1, pop当成-1,按操作的位置建立线段树,那么如何 寻找栈顶呢,我们计算每个点的后缀,栈顶就是下标最大的>0的后缀,我们拿后缀建立线段树, 剩下的就是区间加减法,和求区间最大值啦. #include<bits/stdc++.h>…

有一些草,一开始高度都是0,它们的生长速率不同. 给你一些单增的日期,在这些日期要将>b的草的部分都割掉,问你每次割掉的部分有多少. 将草的生长速率从大到小排序,这样每次割掉的是一个后缀,而且不会影响它们生长速率的递增性. 就是三种操作,一种对一个后缀赋值,一种对整个数组作 + 另一个数组(d(i)-d(i-1))*a,一种求区间和. 可以通过打标记的线段树实现,标记下放通过预处理生长速率数组的前缀和可以实现. 队友的代码: #include <iostream> #include &l…

G. Raffles 题目连接: http://www.codeforces.com/contest/626/problem/G Description Johnny is at a carnival which has n raffles. Raffle i has a prize with value pi. Each participant can put tickets in whichever raffles they choose (they may have more than o…

 Buses                                                                                                                                                                                      time limit per test 2 seconds                                 …

Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular w mm  ×  h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what t…

传送门 题意:    有m个区间,n个a[ i ] , 选择若干个区间,使得整个数组中的最大值和最小值的差值最小.n<=1e5,m<=300; 思路: 可以知道每个i,如果一个区间包含这个点,就让这个区间发挥作用.枚举每个i,找到最大值即可. 当然这个复杂度不对,我们可以通过线段树保存数组的最大值和最小值,每次区间在左端点发挥作用,在右端点去掉作用. #include <algorithm> #include <iterator> #include <iostre…

原题地址:http://codeforces.com/problemset/problem/527/C Examples input H V V V output input H V V H V output 题意是给定一个矩形,不停地纵向或横向切割,问每次切割后,最大的矩形面积是多少. 最大矩形面积=最长的长*最宽的宽这题,长宽都是10^5,所以,用0 1序列表示每个点是否被切割,然后,最长的长就是长的最长连续0的数量+1最长的宽就是宽的最长连续0的数量+1于是用线段树维护最长连续零 问题转换…