接下来的N(1≤N10000)星期中,奶酪工厂在第i个星期要花C_i分来生产一个单位的奶酪.约克奶酪工厂拥有一个无限大的仓库,每个星期生产的多余的奶酪都会放在这里.而且每个星期存放一个单位的奶酪要花费S分.工厂最近收到了客戶N个星期的订单,第i个星期要向客戶提供Y_i 个单位的奶酪.当然这些奶酪可以在第i个星期时生产,也可以从仓库中拿取.采用怎样的生产策略约克奶酪工厂的花费最小呢?Input第一行两个整数:N和S:接下来的N行中,第i行的两个数表示:C_i和Y_i.Output仅一行,即工厂生产

1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 119  Solved: 100[Submit][Status][Discuss] Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10

bzoj1680[Usaco2005 Mar]Yogurt factory bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂 题意: n个月,每月有一个酸奶需求量(吨)和酸奶成本(元每吨).酸奶可以保存,费用为S(元每月每吨),求最小总费用.n≤10000 题解: 第i月每吨酸奶的成本为Cj+s*(i-j),j∈[1,i],化简得Cj-s*j+s*i,因为s*i只和当前相关,所以维护一个最小的Cj-s*j即可.注意开long long. 代码: #includ

Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) c

n<=10000天每天Ci块生产一东西,S块保存一天,每天要交Yi件东西,求最少花多少钱. 这个我都不知道归哪类了.. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> //#include<iostream> using namespace std; int n,m; #define maxn 10011 int a,b; #define

Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) c

Yogurt factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14771   Accepted: 7437 Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk

2393:Yogurt factory 总时间限制:  1000ms 内存限制:  65536kB 描述 The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost th

http://poj.org/problem?id=2393 Yogurt factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7341   Accepted: 3757 Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10

Factory method工厂方法模式是一种实现了“工厂”概念的面向对象设计模式.就像其他创建型模式一样,它也是处理在不指定对象具体类型的情况下创建对象的问题.工厂方法模式的实质是“定义一个创建对象的接口,但让实现这个接口的类来决定实例化哪个类.工厂方法让类的实例化推迟到子类中进行.” 创建一个对象常常需要复杂的过程,所以不适合包含在一个复合对象中.创建对象可能会导致大量的重复代码,可能会需要复合对象访问不到的信息,也可能提供不了足够级别的抽象,还可能并不是复合对象概念的一部分.工厂方法模式通

1680: [Usaco2005 Mar]Yogurt factory Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 106  Solved: 74[Submit][Status][Discuss] Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000)

Yogurt factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8205   Accepted: 4197 Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to prod

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Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) c

http://www.lydsy.com/JudgeOnline/problem.php?id=1680 看不懂英文.. 题意是有n天,第i天生产的费用是c[i],要生产y[i]个产品,可以用当天的也可以用以前的(多生产的).每单位产品保存一天的费用是s.求最小费用 显然贪心,每次查找之前有没有哪一天保存到现在的价值最小,然后比较更新.. #include <cstdio> #include <cstring> #include <cmath> #include <

Factory Methord 工厂方法模式(创建型模式) 从耦合关系谈起耦合关系直接决定着软件面对变化时的行为 -模块与模块之间的紧耦合使得软件面对变化时,相关的模块都要随之变更 -模块与模块之间的松耦合使得软件面对变化时,一些模块更容易被替换或者更改,但其他模块保持不变 对代码的关注要在理解了设计模式之后,对于不同的代码但是解决的是同一类问题,他们就是同一种设计模式.解决了哪一类的问题就是哪一类的设计模式. 软件需求的变化是软件工程的一部分,是我们要解决的问题. 把模块分为主模块.次模块主模

Abstract Factory 抽象工厂(创建型模式) 常见的对象创建方法:    //创建一个Road对象    Road road=new Road();    new的问题:    -实现依赖,不能应对“具体实例化类型”的变化    解决思路:    -封装变化点——那里变化,封装那里    -潜台词:如果没有变化,当然不需要额外的封装    工厂模式的缘起    变化点在“对象创建”,因此就封装“对象创建”    面向接口编程——依赖接口,而非依赖实现    最简单的解决方法:    

Yogurt factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6821   Accepted: 3488 Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk

Yogurt factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16669   Accepted: 8176 Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk