Codeforces Round #413 A. Carrot Cakes
1 second
256 megabytes
In some game by Playrix it takes t minutes for an oven to bake k carrot cakes, all cakes are ready at the same moment t minutes after they started baking. Arkady needs at least n cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take d minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get n cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
The only line contains four integers n, t, k, d (1 ≤ n, t, k, d ≤ 1 000) — the number of cakes needed, the time needed for one oven to bake k cakes, the number of cakes baked at the same time, the time needed to build the second oven.
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
8 6 4 5
YES
8 6 4 6
NO
10 3 11 4
NO
4 2 1 4
YES
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.
题目大意:
有一个面包机每t分钟可以完成一次任务,每次任务可以制作k个蛋糕。
现在有n个蛋糕需要被制作,为了提高效率他可以在花费d分钟再制作一个面包机
在制作新机器的同时,旧机器依旧可以运作。新机器制作完成之后两台机器可以同时运转
他最多可以制作一台新面包机。
问重新制作一台面包机能否提高效率? 可以 YES 、 否NO
解题思路:
大致思路就是分别计算出制作一台新机器完成任务所花费的总时间和只使用旧机器完成任务的总时间,然后比较。
因为在制作新机器的同时,旧机器依旧可以运转,因为不好判断到底是新机器完成最后一次任务
还是旧机器完成最后一次任务。所以分两种情况计算,找出最优情况(即花费时间最少的)
AC代码:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#define ll long long using namespace std; int main ()
{
int n,t,k,d;
int i,a,j;
while (scanf("%d %d %d %d",&n,&t,&k,&d)!=EOF)
{
if (k >= n){ // 新机器建造完成之前 蛋糕就能全部完成
printf("NO\n"); continue;
}
int len = (n+k-)/k;
i = d/t; // 建造新机器的过程中旧机器完成了几次任务(包括正在完成的)
if (i == ) i == ; // 正在完成的
if (d%t) i ++; j = len - i; // 还剩余几次任务
a = min(max(i*t+t*(j/),d+t*(j-j/)),max(i*t+t*(j-j/),d+t*(j*))); // 有可能是新机器最后完成 也有可能是旧机器最后完成 if (a >= len*t)
printf("NO\n");
else
printf("YES\n");
}
return ;
}
思路比较麻烦,仅供参考哈!
Codeforces Round #413 A. Carrot Cakes的更多相关文章
- Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)(A.暴力,B.优先队列,C.dp乱搞)
A. Carrot Cakes time limit per test:1 second memory limit per test:256 megabytes input:standard inpu ...
- Codeforces Round#413 Div.2
A. Carrot Cakes 题面 In some game by Playrix it takes t minutes for an oven to bake k carrot cakes, al ...
- Codeforces Round#413 Problem A - C
Problem#A Carrot Cakes vjudge链接[here] (偷个懒,cf链接就不给了) 题目大意是说,烤面包,给出一段时间内可以考的面包数,建第二个炉子的时间,需要达到的面包数,问建 ...
- Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) 一夜回到小学生
我从来没想过自己可以被支配的这么惨,大神讲这个场不容易掉分的啊 A. Carrot Cakes time limit per test 1 second memory limit per test 2 ...
- Codeforces Round #413(Div. 1 + Div. 2, combined)——ABCD
题目在这里 A.Carrot Cakes 乱七八糟算出两个时间比较一下就行了 又臭又长仅供参考 #include <bits/stdc++.h> #define rep(i, j, k) ...
- Codeforces Round #542 B Two Cakes
B. Two Cakes time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- Codeforces Round #413, rated, Div. 1 + Div. 2 C. Fountains(贪心 or 树状数组)
http://codeforces.com/contest/799/problem/C 题意: 有n做花园,有人有c个硬币,d个钻石 (2 ≤ n ≤ 100 000, 0 ≤ c, d ≤ 100 ...
- C.Fountains(Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)+线段树+RMQ)
题目链接:http://codeforces.com/contest/799/problem/C 题目: 题意: 给你n种喷泉的价格和漂亮值,这n种喷泉题目指定用钻石或现金支付(分别用D和C表示),C ...
- Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) C. Fountains 【树状数组维护区间最大值】
题目传送门:http://codeforces.com/contest/799/problem/C C. Fountains time limit per test 2 seconds memory ...
随机推荐
- Java的定时调度
一般在web开发中定时调度比较有用,因为要维护一个容器不关闭才可以一直定时操作下去. 定时调度:每当一段时间之后,程序就会自动执行,就称为定时调度.如果要使用定时调动,则必须要保证程序要始终运行着,也 ...
- Android Fragment原理及应用
1.Fragment简介 Fragment(片段) 表示 Activity 中的行为或用户界面部分.您可以将多个片段组合在一个 Activity 中来构建多窗格 UI,以及在多个 Activity 中 ...
- 使用IntelliJ IDEA 前最好修改的配置
目录 1.下载 2.破解 3.修改配置 下载 详见我的另外一篇博客:<软件开发资源下载>中的[IDE]->[IntelliJ IDEA] 破解 详见我的另外一篇博客:<最新版I ...
- POJ_1990 MooFest 【树状数组】
一.题面 POJ1990 二.分析 一个简单的树状数组运用.首先要把样例分析清楚,凑出57,理解一下.然后可以发现,如果每次取最大的v就可以肆无忌惮的直接去乘以坐标差值就可以了,写代码的时候是反着来的 ...
- python 面向过程和面向对象比较
面向过程 VS 面向对象 面向过程的程序设计:核心是过程二字,过程指的是解决问题的步骤,即先干什么再干什么......面向过程的设计就好比精心设计好一条流水线,是一种机械式的思维方式. 优点是:复杂度 ...
- Vue项目中使用HighChart
记:初次在Vue项目中使用 HighChart 的时候要走的坑 感谢这位哥们的思路 传送门 Vue-cli项目使用 npm install highcharts --save 让我们看看 highch ...
- Linux 远程登录ssh服务器
1.安装ssh服务器 sudo apt-get install openssh-server 2.在另一端输入ssh IP及密码(或ssh 用户名@IP)就可以远程登录到IP所在计算机
- Python多线程&进程
一.线程&进程 对于操作系统来说,一个任务就是一个进程(Process),比如打开一个浏览器就是启动一个浏览器进程,打开一个记事本就启动了一个记事本进程,打开两个记事本就启动了两个记事本进程, ...
- Git学习系列之 Git 、CVS、SVN的比较
Git .CVS.SVN比较 项目源代码的版本管理工具中,比较常用的主要有:CVS.SVN.Git 和 Mercurial (其中,关于SVN,请参见我的博客:SVN学习系列) 目前Google C ...
- Request笔记
1 Request 的简介和运行环境 1.HttpServletRequest 概述 我们在创建 Servlet 时会覆盖 service()方法,或 doGet()/doPost(),这些方法都有两 ...