1740: [Usaco2005 mar]Yogurt factory 奶酪工厂
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 <= 10,000) 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.
Input
* Line 1: Two space-separated integers, N and S.
* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.
Output
* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.
Sample Input
88 200
89 400
97 300
91 500
Sample Output
第1周生产200单位奶酪并全部交付;第2周生产700单位,交付400单位,有300单位;第3周交
付300单位存货.第4周生产并交付500单位.
HINT
Source
题解:很经典的递推题(根本算不上DP,甚至有几分贪心的味道在里面),第五次见到了,不多说了,直接上代码(HansBug:难以想象这种题居然上usaco金组= =)
1 /************************************************************** 2 Problem: 1740 3 User: HansBug 4 Language: Pascal 5 Result: Accepted 6 Time:28 ms 7 Memory:224 kb 8 ****************************************************************/ 9 10 var 11 i,j,k,l,m,n:longint; 12 ans:int64; 13 function min(x,y:longint):longint; 14 begin 15 if x<y then min:=x else min:=y; 16 end; 17 begin 18 readln(n,m); 19 l:=maxlongint-m; 20 for i:=1 to n do 21 begin 22 readln(j,k); 23 l:=min(l+m,j); 24 ans:=ans+int64(l)*int64(k); 25 end; 26 writeln(ans); 27 readln; 28 end.
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
· 全程不用写代码,我用AI程序员写了一个飞机大战
· DeepSeek 开源周回顾「GitHub 热点速览」
· 记一次.NET内存居高不下排查解决与启示
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· .NET10 - 预览版1新功能体验(一)