Problem N
Description
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 produce one unit of yogurt in week i.
Yucky's factory, being well-designed, can produce arbitrarily many
units of yogurt each week.
Yucky Yogurt owns a warehouse that can store unused yogurt at a
constant fee of S (1 <= S <= 100) cents per unit of yogurt
per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's
warehouse is enormous, so it can hold arbitrarily many units of
yogurt.
Yucky wants to find a way to make weekly deliveries of Y_i (0 <=
Y_i <= 10,000) units of yogurt to its clientele (Y_i is the
delivery quantity in week i). Help Yucky minimize its costs over
the entire N-week period. Yogurt produced in week i, as well as any
yogurt already in storage, can be used to meet Yucky's demand for
that week.
integers, N and S.< br>< br>* Lines 2..N+1: Line i+1
contains two space-separated integers: C_i and Y_i.
integer: the minimum total cost to satisfy the yogurt schedule.
Note that the total might be too large for a 32-bit integer.
200
400
300
500
#include
#define maxn 10000
using namespace std;
int main()
{
//freopen("in.txt", "r", stdin);
int
n,s,c[maxn],y[maxn];
long long
money=0,daymoney=0;
while(scanf("%d%d",&n,&s)!=EOF)
{
money=daymoney=0;
for(int i=0;i
scanf("%d%d",&c[i],&y[i]);
money=c[0]*y[0];
for(int i=1;i
{
daymoney=min(c[i]*y[i],(c[i-1]+s)*y[i]);
//printf("c[i]=%d (c[i-1]+s)=%d\n",c[i],(c[i-1]+s));
//printf("daymoney=%d\n",daymoney);
money+=daymoney;
}
printf("%lld\n",money);
}
}
Problem N的更多相关文章
- 1199 Problem B: 大小关系
求有限集传递闭包的 Floyd Warshall 算法(矩阵实现) 其实就三重循环.zzuoj 1199 题 链接 http://acm.zzu.edu.cn:8000/problem.php?id= ...
- No-args constructor for class X does not exist. Register an InstanceCreator with Gson for this type to fix this problem.
Gson解析JSON字符串时出现了下面的错误: No-args constructor for class X does not exist. Register an InstanceCreator ...
- C - NP-Hard Problem(二分图判定-染色法)
C - NP-Hard Problem Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144 ...
- Time Consume Problem
I joined the NodeJS online Course three weeks ago, but now I'm late about 2 weeks. I pay the codesch ...
- Programming Contest Problem Types
Programming Contest Problem Types Hal Burch conducted an analysis over spring break of 1999 and ...
- hdu1032 Train Problem II (卡特兰数)
题意: 给你一个数n,表示有n辆火车,编号从1到n,入站,问你有多少种出站的可能. (题于文末) 知识点: ps:百度百科的卡特兰数讲的不错,注意看其参考的博客. 卡特兰数(Catalan):前 ...
- BZOJ2301: [HAOI2011]Problem b[莫比乌斯反演 容斥原理]【学习笔记】
2301: [HAOI2011]Problem b Time Limit: 50 Sec Memory Limit: 256 MBSubmit: 4032 Solved: 1817[Submit] ...
- [LeetCode] Water and Jug Problem 水罐问题
You are given two jugs with capacities x and y litres. There is an infinite amount of water supply a ...
- [LeetCode] The Skyline Problem 天际线问题
A city's skyline is the outer contour of the silhouette formed by all the buildings in that city whe ...
- PHP curl报错“Problem (2) in the Chunked-Encoded data”解决方案
$s = curl_init(); curl_setopt($s, CURLOPT_POST, true); curl_setopt($s, CURLOPT_POSTFIELDS, $queryStr ...
随机推荐
- ubuntu下程序员常用命令大全
一.ubuntu下用命令查询系统版本 1.在终端中执行下列指令: cat /etc/issue 该命令可查看当前正在运行的ubuntu的版本号. 效果如图: 2.使用 lsb_release 命令也可 ...
- 每天学点SpringMVC-异常处理
1. 第一步先写个Hello World 1.1 编写一个抛出异常的目标方法 @RequestMapping("/testException.do") public String ...
- crontab的两大坑:百分号和环境变量
今天想给服务器加个自动备份mysql数据库的功能(别怪我这么久才加,阿里云每天全盘备份的,不怕丢数据库),本以为只要5分钟就能搞定的,结果入了两个大坑. 我的crontab是这样写的: * * * m ...
- web版的tty
1.wetty Wetty是使用Node.js和websockets开发的一个开源`Web-based SSH` 2.环境配置 2.1.配置epel源 [epel] name=epel baseu ...
- Happy 2006 poj2773
Happy 2006 Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 9049 Accepted: 3031 Descri ...
- JavaWeb(三)JSP概述
一.JSP概述 1.1.JSP简介 一种动态网页开发技术.它使用JSP标签在HTML网页中插入Java代码.标签通常以<%开头以%>结束.JSP是一种Java servlet,主要用于实现 ...
- Java 浅拷贝和深拷贝的理解和实现方式
Java中的对象拷贝(Object Copy)指的是将一个对象的所有属性(成员变量)拷贝到另一个有着相同类类型的对象中去.举例说明:比如,对象A和对象B都属于类S,具有属性a和b.那么对对象A进行拷贝 ...
- jquery系列教程3-DOM操作全解
全栈工程师开发手册 (作者:栾鹏) 快捷链接: jquery系列教程1-选择器全解 jquery系列教程2-style样式操作全解 jquery系列教程3-DOM操作全解 jquery系列教程4-事件 ...
- DevOps之归纳总结
唠叨话 关于德语关我屁事的知识点,仅提供精华汇总,具体知识点细节,参考教程网址,如需帮助,请留言. DevOps归纳总结 <DevOps功能与性能>浏览器(饼干Cookie.会话Sessi ...
- Spring MVC Ajax 嵌套表单数据的提交
概述 在一些场景里,某个大表单里常常嵌套着一个或若干个小逻辑块,比如以下表单里"设计预审"中包括了一个子模块表单"拟定款项". 在这种情况下该怎么去设计实体类以 ...