Yaoge’s maximum profit Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5052 Problem Description Yaoge likes to eat chicken chops late at night. Yaoge has eaten too ma

咸鱼了好久...出来冒个泡_(:з」∠)_ 题目连接:1107G - Vasya and Maximum Profit 题目大意:给出\(n,a\)以及长度为\(n\)的数组\(c_i\)和长度为\(n\)的严格单调上升数组\(d_i\),求\(\max\limits_{1 \le l \le r \le n} (a\cdot(r-l+1)-\sum_{i=l}^{r}c_i-gap(l,r))\),其中\(gap(l, r) = \max\limits_{l \le i < r} (d_{i

https://github.com/Premiumlab/Python-for-Algorithms--Data-Structures--and-Interviews/blob/master/Mock%20Interviews/Large%20E-Commerce%20Company/E-Commerce%20Company%20-%20Interview%20Problems%20-%20SOLUTIONS/On-Site%20Question%201%20-%20SOLUTION.ipyn

Maximum Profit You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen. Write a

Codeforces 1107G 线段树最大子段和 + 单调栈 G. Vasya and Maximum Profit Description: Vasya got really tired of these credits (from problem F) and now wants to earn the money himself! He decided to make a contest to gain a profit. Vasya has \(n\) problems to choo

题目如下: We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i]. You're given the startTime , endTime and profit arrays, you need to output the maximum profit you can take such that there

题目大意: 给出一棵树.每一个点有商店.每一个商店都有一个价格,Yaoge每次从x走到y都能够在一个倒卖商品,从中得取利益.当然,买一顶要在卖之前.可是没次走过一条路,这条路上的全部商品都会添加一个v. 输出每次的最大利益. 思路分析: 非常easy想到树链剖分,但是关键在于怎样维护这样一个变量.使得每次都要让买的再卖的前面. 维护变量 ltr 和 rtl .表示从左去右和从右去左. 剖分熟练的时候.推断x 和 y的深度,一步一步往上爬. 然后维护区间最大值和最小值.爬的时候更新答案. ...4

典型的LCT操作,但是维护的是一个序列最左边减最右边的最小值,所以要维护左边减右边的最小值del[0]和一个右边减左边的最小值del[1](因为rev标记swap的时候对应的值也要交换).维护的时候del[0]可能是来自于左右儿子的del[0],也有可能是来自于左儿子的最小值减去右儿子及当前节点的值的最大值,还有就是当前节点减去右儿子的最大值,比赛的时候漏了当前节点减去右儿子的最大值因而WA了..- -0 #pragma warning(disable:4996) #include <iostr

意甲冠军: 特定n小点的树权. 以下n每一行给出了正确的一点点来表达一个销售点每只鸡价格的格 以下n-1行给出了树的侧 以下Q操作 Q行 u, v, val 从u走v,程中能够买一个鸡腿,然后到后面卖掉,输出max(0, 最大的收益) 然后给[u,v]路径上点点权+=val 思路: 树链剖分裸题.记录区间的最大最小值,→走的答案和←走的答案. 官方题解:点击打开链接 #pragma comment(linker, "/STACK:1024000000,1024000000") #inc

题目分析: 前缀和啥的模拟一下就行了. 代码: #include<bits/stdc++.h> using namespace std; ; int n,x,d[maxn],sta[maxn],top; long long minn[maxn],c[maxn],maxx[maxn]; void read(){ scanf("%d%d",&n,&x); ;i<=n;i++) scanf("%d%lld",&d[i],&

洛谷 Codeforces 我竟然能在有生之年踩标算. 思路 首先考虑暴力:枚举左右端点直接计算. 考虑记录\(sum_x=\sum_{i=1}^x c_i\),设选\([l,r]\)时那个奇怪东西的平方为\(f(l,r)\),使劲推式子: \[ ans_{l,r}=(r-l+1)\times a-sum_r+sum_{l-1}-f(l,r)\\ ans_{l,r}+l\times a-a-sum_{l-1}=r\times a-sum_r-f(l,r)\\ ans_{l,r}+l\times

题目传送门 题解: 枚举 r 的位置. 线段树每个叶子节点存的是对应的位置到当前位置的价值. 每次往右边移动一个r的话,那么改变的信息有2个信息: 1. sum(a-ci) 2.gap(l, r) 对于第一个东西,我们直接对[1,r]区间内的所有值都加上a-ci就好了. 对于第2个修改的值,首先要明白的是,这个值只会对[1,r-1]内的值存在影响. 并且可以发现,每一段区间的这个gap值每次更新完之后是从左到右递减的,所以我们可以用一个单调栈来维护这个gap值. 代码: /* code by:

Less Time, More profit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description The city planners plan to build N plants in the city which has M shops. Each shop needs products from some plants to make p

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5855 Description The city planners plan to build N plants in the city which has M shops. Each shop needs products from some plants to make profit of proi units. Building ith plant needs investment o

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5855 Less Time, More profit Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 The city planners plan to build N plants in the city which has M shops. Each shop needs p

The city planners plan to build N plants in the city which has M shops. Each shop needs products from some plants to make profit of proiproi units. Building ith plant needs investment of payipayi units and it takes titi days. Two or more plants can b

Less Time, More profit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 468    Accepted Submission(s): 172 Problem Description The city planners plan to build N plants in the city which has M sho

714. Best Time to Buy and Sell Stock with Transaction Fee - Medium Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee. You may c

from:http://www.embeddedlinux.org.cn/html/jishuzixun/201307/08-2594.html 段错误调试神器 - Core Dump详解 来源:互联网 作者:Alex 时间:2013-07-08 Tag:Linux   点击: 11670 一.前言: 有的程序可以通过编译, 但在运行时会出现Segment fault(段错误). 这通常都是指针错误引起的. 但这不像编译错误一样会提示到文件某一行, 而是没有任何信息, 使得我们的调试变得困难起来

http://blog.csdn.net/lanmolei814/article/details/45201693 ====================================== 1.Core文件简介 Core文件其实就是内存的映像,当程序崩溃时,存储内存的相应信息,主用用于对程序进行调试.当程序崩溃时便会产生core文件,其实准确的应该说是core dump 文件,默认生成位置与可执行程序位于同一目录下,文件名为core.***,其中***是某一数字. 2.开启或关闭Core文件