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…
Reinforcement Learning for Profit July 17, 2016 Is RL being used in revenue generating systems today? Recently, one of my facebook friends, and alumni of the University of Alberta (with a PhD in Computing Science), Cosmin Paduraru posed a question:…
题目链接:uva 10581 - Partitioning for fun and profit 题目大意:给定m,n,k,将m分解成n份,然后依照每份的个数排定字典序,而且划分时要求ai−1≤ai,然后输出字典序排在k位的划分方法. 解题思路:由于有ai−1≤ai的条件.所以先记忆化搜索处理出组合情况dp[i][j][s]表示第i位为j.而且剩余的未划分数为s的总数为dp[i][j][s],然后就是枚举每一位上的值.推断序列的位置就可以. #include <cstdio> #include…
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…
We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job. Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at mo…
We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job. Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at mo…
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…
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…
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…
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…
原题链接在这里:https://leetcode.com/problems/most-profit-assigning-work/ 题目: We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job. Now we have some workers. worker[i] is the ability of the ith worker, whic…
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…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/most-profit-assigning-work/description/ 题目描述: We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job. Now we…
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],&…
本文是阅读 http://book.mixu.net/distsys/abstractions.html 的笔记. 第二章的题目是"Up and down the level of abstraction".这一章里面,作者主要介绍了分布式系统里面的一个重要概念:CAP理论. 什么是CAP理论呢?就是说在任何情况下,分布式系统只能满足下面三项中的两个: 一致性(Consistency),这里指的强一致性. 可用性(Availability). 对网络分割容错(Partition tol…
