Codeforces Gym 100286G Giant Screen 水题
Problem G.Giant Screen
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/G
Description
You are working in Advanced Computer Monitors (ACM), Inc. The company is building and selling giant computer screens that are composed from multiple smaller screens. Your are responsible for design of the screens for your customers. Customers order screens of the specified horizontal and vertical resolution in pixels and a specified horizontal and vertical size in millimeters. Your task is to design a screen that has a required resolution in each dimension or more, and has required size in each dimension or more, with a minimal possible price. The giant screen is always built as a grid of monitors of the same type. The total resolution, size, and price of the resulting screen is simply the sum of resolutions, sizes, and prices of the screens it is built from. You have a choice of regular monitor types that you can order and you know their resolutions, sizes, and prices. The screens of each type can be mounted both vertically and horizontally, but the whole giant screen must be composed of the screens of the same type in the same orientation. You can use as many screens of the chosen type as you need
Input
The first line of the input file contains four integer numbers rh, rv, sh, and sv (all from 100 to 10 000 inclusive) — horizontal and vertical resolution and horizontal and vertical size of the screen you have to build, respectively. The next line contains a single integer number n (1 ≤ n ≤ 100) — the number of different screen types available to you. The next n lines contain descriptions of the available screen types. Each description occupies one line and consists of five integer numbers — rh,i, rv,i, sh,i, sv,i, pi (all from 100 to 10 000 inclusive), where first four numbers are horizontal and vertical resolution and horizontal and vertical size of i-th screen type, and pi is the price.
Output
Write to the output file a single integer — the minimal price of the specified giant screen.
Sample Input
1024 1024 300 300
3
1024 768 295 270 200
1280 1024 365 301 250
1280 800 350 270 210
Sample Output
250
HINT
题意
给你一个需要的屏幕分辨率和长宽
然后给你n个,让你挑选一种,让你花费最小的价格来达到规定的电视
题解:
水题,直接贪心除一下然后取min就吼了
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1) using namespace std;
const int maxn = + ;
int rh , rv , sh , sv , n;
typedef struct data
{
int rh , rv , sh , sv , p;
}; int Caculate(int x,int y)
{
int num = x / y;
if (y * num < x) num ++;
return num;
} data A[maxn]; int main(int argc,char *argv[])
{
freopen("giant.in","r",stdin);
freopen("giant.out","w",stdout);
scanf("%d%d%d%d%d",&rh,&rv,&sh,&sv,&n);
for(int i = ; i < n ; ++ i) scanf("%d%d%d%d%d",&A[i].rh,&A[i].rv,&A[i].sh,&A[i].sv,&A[i].p);
long long ans = 1LL << ;
for(int i = ; i < n ; ++ i)
{
int s1 = max(Caculate(rh,A[i].rh) , Caculate(sh,A[i].sh));
int s2 = max(Caculate(rv,A[i].rv),Caculate(sv,A[i].sv));
ans = min(ans ,1LL * (long long) s1 * (long long)s2 * (long long)A[i].p );
s1 = max(Caculate(rh,A[i].rv) , Caculate(sh,A[i].sv));
s2 = max(Caculate(rv,A[i].rh),Caculate(sv,A[i].sh));
ans = min(ans ,1LL * (long long) s1 * (long long)s2 * (long long)A[i].p );
}
cout << ans <<endl;
return ;
}
Codeforces Gym 100286G Giant Screen 水题的更多相关文章
- Codeforces gym 100685 C. Cinderella 水题
C. CinderellaTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/C ...
- Codeforces Gym 100431D Bubble Sort 水题乱搞
原题链接:http://codeforces.com/gym/100431/attachments/download/2421/20092010-winter-petrozavodsk-camp-an ...
- Codeforces GYM 100114 B. Island 水题
B. Island Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description O ...
- codeforces 577B B. Modulo Sum(水题)
题目链接: B. Modulo Sum time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #367 (Div. 2)---水题 | dp | 01字典树
A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #inclu ...
- codeforces 696A Lorenzo Von Matterhorn 水题
这题一眼看就是水题,map随便计 然后我之所以发这个题解,是因为我用了log2()这个函数判断在哪一层 我只能说我真是太傻逼了,这个函数以前听人说有精度问题,还慢,为了图快用的,没想到被坑惨了,以后尽 ...
- CodeForces 589I Lottery (暴力,水题)
题意:给定 n 和 k,然后是 n 个数,表示1-k的一个值,问你修改最少的数,使得所有的1-k的数目都等于n/k. 析:水题,只要用每个数减去n/k,然后取模,加起来除以2,就ok了. 代码如下: ...
- codeforces 710A A. King Moves(水题)
题目链接: A. King Moves 题意: 给出king的位置,问有几个可移动的位置; 思路: 水题,没有思路; AC代码: #include <iostream> #include ...
- codeforces 659A A. Round House(水题)
题目链接: A. Round House time limit per test 1 second memory limit per test 256 megabytes input standard ...
随机推荐
- Jquery插件写法及extentd函数
JQuery插件写法 JQuery插件又分为类扩展方法和对象扩展方法两种,类插件是定义在JQuery命令空间的全局函数,直接通过可调用,如可调用,如可调用,如.ajax():对象插件是扩展JQuery ...
- Android常用控件之FragmentTabHost的使用
最近在学TabHost时发现TabActivity在API level 13以后不用了,所以就去寻找它的替换类,找到FragmentActivity,可以把每个Fragment作为子tab添加到Fra ...
- Android 的实现TextView中文字链接的4种方法
Android 的实现TextView中文字链接的方式有很多种. 总结起来大概有4种: 1.当文字中出现URL.E-mail.电话号码等的时候,可以将TextView的android:autoLink ...
- PHP学习笔记--文件目录操作(文件上传实例)
文件操作是每个语言必须有的,不仅仅局限于PHP,这里我们就仅用PHP进行讲解 php的文件高级操作和文件上传实例我放在文章的最后部分.--以后我还会给大家写一个PHP类似于网盘操作的例子 注意:阅读此 ...
- 走向DBA[MSSQL篇] - 从SQL语句的角度提高数据库的访问性能(转)
最近公司来一个非常虎的DBA,10几年的经验,这里就称之为蔡老师吧,在征得我们蔡老同意的前提下 ,我们来分享一下蔡老给我们带来的宝贵财富,欢迎其他的DBA来拍砖. 目录 1.什么是执行计划?执行计划 ...
- memcpy、memmove、memset及strcpy函数实现和理解
memcpy.memmove.memset及strcpy函数实现和理解 关于memcpy memcpy是C和C++ 中的内存拷贝函数,在C中所需的头文件是#include<string.h> ...
- Vim小知识
在退出vim编辑的时候,强制退出是q! 感叹号在前,即!q,表示执行外部shell命令,感叹号在后,即q!,表示强制执行vi命令.
- android判断当前网络状态及跳转到设置界面
今天,想做这个跳转到网络设置界面, 刚开始用 intent = new Intent(Settings.ACTION_WIRELESS_SETTINGS); 不料老是出现settings.Wirele ...
- JavaScript——以简单的方式理解闭包
闭包,在一开始接触JavaScript的时候就听说过.首先明确一点,它理解起来确实不复杂,而且它也非常好用.那我们去理解闭包之前,要有什么基础呢?我个人认为最重要的便是作用域(lexical scop ...
- bitmap的实现方法
bitmap是一个十分有用的结构.所谓的Bit-map就是用一个bit位来标记某个元素对应的Value, 而Key即是该元素.由于采用了Bit为单位来存储数据,因此在存储空间方面,可以大大节省. 适用 ...