杭电 HDU ACM Milk
Milk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15697 Accepted Submission(s): 3947
Here are some rules:
1. Ignatius will never drink the milk which is produced 6 days ago or earlier. That means if the milk is produced 2005-1-1, Ignatius will never drink this bottle after 2005-1-6(inclusive).
2. Ignatius drinks 200mL milk everyday.
3. If the milk left in the bottle is less than 200mL, Ignatius will throw it away.
4. All the milk in the supermarket is just produced today.
Note that Ignatius only wants to buy one bottle of milk, so if the volumn of a bottle is smaller than 200mL, you should ignore it.
Given some information of milk, your task is to tell Ignatius which milk is the cheapest.
Each test case starts with a single integer N(1<=N<=100) which is the number of kinds of milk. Then N lines follow, each line contains a string S(the length will at most 100 characters) which indicate the brand of milk, then two integers for the brand: P(Yuan)
which is the price of a bottle, V(mL) which is the volume of a bottle.
2
2
Yili 10 500
Mengniu 20 1000
4
Yili 10 500
Mengniu 20 1000
Guangming 1 199
Yanpai 40 10000
Mengniu
MengniuHintIn the first case, milk Yili can be drunk for 2 days, it costs 10 Yuan. Milk Mengniu can be drunk for 5 days, it costs 20 Yuan. So Mengniu is the cheapest.In the second case,
milk Guangming should be ignored. Milk Yanpai can be drunk for 5 days, but it costs 40 Yuan. So Mengniu is the cheapest.这个题目当时acm入门的时候,怎么也解决不了 。大水题一直搁置到现在。现在闲着没事干,a它分分秒。看着曾经提交的code,代码质量令今天的我非常是感概。 又臭又长。而且回忆起曾经解这个题目时遇到的思路问题,感觉acm确实能给人带来思维上的强大进步,特别特别明显。 想问题也想的非常清楚。事实上这个题目就是简单的把问题划分清楚就ok了,同一时候注意点儿细节。#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
struct ls
{
string str;
int mo;
int ml;
double rate;
} gq[1000];
bool cmp(ls co,ls cn)
{
return co.rate<cn.rate;
}
int main()
{
int T,n;
cin>>T;
while(T--)
{
cin>>n;
for(int i=0; i<n; i++)
{
cin>>gq[i].str>>gq[i].mo>>gq[i].ml; if(gq[i].ml>=1200)
gq[i].rate=gq[i].mo/5;
else if((gq[i].ml>=200))
{
int x=(gq[i].ml)/200;
gq[i].rate=gq[i].mo/double(x*1.0);
}
else
gq[i].rate=-1;
}
double Min=0x1f1f1f1f;
int k;
for(int j=0; j<n; j++)
{
if(gq[j].rate!=-1)
{
if(gq[j].rate<Min)
{
Min=gq[j].rate;
k=j;
}
}
}
cout<<gq[k].str<<endl;
}
return 0;
}
/*
2
2
Yili 10 500
Mengniu 20 1000
4
Yili 10 600
Mengniu 20 1000
Guangming 1 199
Yanpai 40 10000
**/
杭电 HDU ACM Milk的更多相关文章
- 杭电 HDU ACM 2795 Billboard(线段树伪装版)
Billboard Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total ...
- 杭电 HDU ACM 1698 Just a Hook(线段树 区间更新 延迟标记)
欢迎"热爱编程"的高考少年--报考杭州电子科技大学计算机学院 Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memor ...
- ACM 杭电HDU 2084 数塔 [解题报告]
数塔 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submissi ...
- 杭电 HDU 4608 I-number
http://acm.hdu.edu.cn/showproblem.php?pid=4608 听说这个题是比赛的签到题......无语..... 问题:给你一个数x,求比它大的数y. y的要求: 1. ...
- 深搜基础题目 杭电 HDU 1241
HDU 1241 是深搜算法的入门题目,递归实现. 原题目传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1241 代码仅供参考,c++实现: #incl ...
- 杭电 HDU 1242 Rescue
http://acm.hdu.edu.cn/showproblem.php?pid=1242 问题:牢房里有墙(#),警卫(x)和道路( . ),天使被关在牢房里位置为a,你的位置在r处,杀死一个警卫 ...
- 杭电 HDU 1031 Design T-Shirt
Design T-Shirt Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- 杭电hdu 2089 数位dp
杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer). 杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以消除个别的士司机和乘客的心理障碍 ...
- 杭电 HDU 2717 Catch That Cow
Catch That Cow Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
随机推荐
- jquery之营销系统
// //////////////////////////优惠券开始//////////////////////////// // 给附加条件选择框添加事件 function inputFuJia() ...
- 《第一行代码》学习笔记35-服务Service(2)
1.Android的UI线程不安全,想要更新应用程序里的UI元素,则须在主线程中进行,否则会出现异常. 2.Android不允许在子线程里进行UI操作,对于该情况,Android提供了一套异步消息处理 ...
- IActiveView 接口 - 浅谈
IActiveView 和 IMap以其 Map, PageLayout之间的关系. 在ArcMap中, PageLayout 和 Map分别对应不同的视图: layout 和data view.在同 ...
- SQlserver表连接
连接是两元运算,可以对两个或多个表进行查询,结果通常是含有参加连接运算的两个表或多个表的指定列的表. 在T-SQL中,连接查询有两类: 第一类:符合SQL标准的连接谓词表示形式: 第二类:T-SQL扩 ...
- 【分享】JS图片滑动渐显渐隐插件-附使用方法。
前阵子总监要说做一个邀请函 效果 点击这里 鼠标拖拽进行浏览 它用的是Adobe edge软件生成的,代码量过大,冗余太多. 再加上我也没学过这个软件怎么使用,增加学习成本影响项目进度. 于是就自己写 ...
- vagrant 配置文件简析
#自定义box名字 config.vm.box = "website" #以ip192.168.33.10访问 config.vm.network "private_ne ...
- MySQL查询指定时间的数据
user_event :用户事件表 create_time :表中存储时间的字段 #获取当月数据 SELECT * FROM user_event WHERE DATE_FORMAT(create_t ...
- WF学习
1.添加argument 类的argument必须先在表达式里面实例化 argument和variable 没有区别啊??????? http://msdn.microsoft.com/en-us/l ...
- 配置TC2.0运行环境
一. 学习过程 下载TC2.0并打开如图: 这是一个集成的C语言环境包,包括TC.dosbox和一些编译工具,很明显这样我无法判断单个程序的功能,也无法区分哪些程序是必须的. 那么为了搞清楚哪些文件是 ...
- Ubuntu12.04 下配置tomcat
首先要配置好JDK,参考:http://www.cnblogs.com/yshyee/p/3352154.html 到http://tomcat.apache.org/download-60.cgi下 ...