VK Cup 2016 - Round 1 (Div. 2 Edition) A. Bear and Reverse Radewoosh 水题
A. Bear and Reverse Radewoosh
题目连接:
http://www.codeforces.com/contest/658/problem/A
Description
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be n problems. The i-th problem has initial score pi and it takes exactly ti minutes to solve it. Problems are sorted by difficulty — it's guaranteed that pi < pi + 1 and ti < ti + 1.
A constant c is given too, representing the speed of loosing points. Then, submitting the i-th problem at time x (x minutes after the start of the contest) gives max(0, pi - c·x) points.
Limak is going to solve problems in order 1, 2, ..., n (sorted increasingly by pi). Radewoosh is going to solve them in order n, n - 1, ..., 1 (sorted decreasingly by pi). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all ti. That means both Limak and Radewoosh will accept all n problems.
Input
The first line contains two integers n and c (1 ≤ n ≤ 50, 1 ≤ c ≤ 1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ 1000, pi < pi + 1) — initial scores.
The third line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000, ti < ti + 1) where ti denotes the number of minutes one needs to solve the i-th problem.
Output
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Sample Input
3 2
50 85 250
10 15 25
Sample Output
Limak
Hint
题意
有两个人在做CF,一个正着做,一个反着做
给你每道题的分值以及每道题做题的时间,问你谁的分数高
题解:
暴力模拟,扫一遍就好了
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 55;
long long p[maxn];
long long t[maxn];
int main()
{
int n,c;
scanf("%d%d",&n,&c);
for(int i=1;i<=n;i++)scanf("%lld",&p[i]);
for(int i=1;i<=n;i++)scanf("%lld",&t[i]);
long long p1 = 0,p2 = 0;
long long sum = 0;
for(int i=1;i<=n;i++)
{
sum+=t[i];
p1 = p1 + max(0LL,p[i]-c*sum);
}
sum=0;
for(int i=n;i>=1;i--)
{
sum+=t[i];
p2 = p2 + max(0LL,p[i]-c*sum);
}
if(p1>p2)cout<<"Limak"<<endl;
else if(p2>p1)cout<<"Radewoosh"<<endl;
else cout<<"Tie"<<endl;
}
VK Cup 2016 - Round 1 (Div. 2 Edition) A. Bear and Reverse Radewoosh 水题的更多相关文章
- VK Cup 2016 - Round 1 (Div. 2 Edition) A Bear and Reverse Radewoosh
A. Bear and Reverse Radewoosh time limit per test 2 seconds memory limit per test 256 megabytes inpu ...
- Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) A. Little Artem and Presents 水题
A. Little Artem and Presents 题目连接: http://www.codeforces.com/contest/669/problem/A Description Littl ...
- VK Cup 2016 - Round 1 (Div. 2 Edition) C. Bear and Forgotten Tree 3
C. Bear and Forgotten Tree 3 time limit per test 2 seconds memory limit per test 256 megabytes input ...
- Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D Bear and Two Paths
题目链接: http://codeforces.com/contest/673/problem/D 题意: 给四个不同点a,b,c,d,求是否能构造出两条哈密顿通路,一条a到b,一条c到d. 题解: ...
- Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C - Bear and Colors
题目链接: http://codeforces.com/contest/673/problem/C 题解: 枚举所有的区间,维护一下每种颜色出现的次数,记录一下出现最多且最小的就可以了. 暴力n*n. ...
- Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D. Bear and Two Paths 构造
D. Bear and Two Paths 题目连接: http://www.codeforces.com/contest/673/problem/D Description Bearland has ...
- Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C. Bear and Colors 暴力
C. Bear and Colors 题目连接: http://www.codeforces.com/contest/673/problem/C Description Bear Limak has ...
- Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) A. Bear and Game 水题
A. Bear and Game 题目连接: http://www.codeforces.com/contest/673/problem/A Description Bear Limak likes ...
- Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) B. Little Artem and Grasshopper 模拟题
B. Little Artem and Grasshopper 题目连接: http://www.codeforces.com/contest/669/problem/B Description Li ...
随机推荐
- python中随机数生成
1.random.random random.random()用于生成一个0到1的随机符小数: 0 <= n < 1.0 2.random.uniform random.uniform的函 ...
- docker安装总结 linux红帽系列
由于Docker限制分为两个版本CE和EE,所以之前yum里面的docker是老版本,需要先进行卸载,现在的包名叫做docker-ce yum remove docker docker-common ...
- 服务器部署之nginx的配置
nginx可作为Web和 反向代理 服务器,在高连接并发的情况下,Nginx是Apache服务器不错的替代品.下面记录一下自己对nginx的配置和使用. nginx的安装 环境:oracle-linu ...
- MySQL5.6 新特性之GTID【转】
转自 MySQL5.6 新特性之GTID - jyzhou - 博客园http://www.cnblogs.com/zhoujinyi/p/4717951.html 背景: MySQL5.6在5.5的 ...
- apache 各种配置
//apache 的网站配置文件 /usr/local/apache2/conf/extra/httpd-vhosts.conf -->在编辑这个文件前需要去httpd.conf把这个文件的注释 ...
- 垃圾回收算法与 JVM 垃圾回收器综述(转)
垃圾回收算法与 JVM 垃圾回收器综述 我们常说的垃圾回收算法可以分为两部分:对象的查找算法与真正的回收方法.不同回收器的实现细节各有不同,但总的来说基本所有的回收器都会关注如下两个方面:找出所有的存 ...
- 常见的 JavaScript 内存泄露
什么是内存泄露 指由于疏忽或错误造成程序未能释放已经不再使用的内存.内存泄漏并非指内存在物理上的消失, 而是应用程序分配某段内存后,由于设计错误,导致在释放该段内存之前就失去了对该段内存的控制,从而造 ...
- TGPPen 宽度的理解
procedure TForm4.Button1Click(Sender: TObject); var g: TGPGraphics; p: TGPPen; begin g := TGPGraphic ...
- HDU 1217 Arbitrage(Bellman-Ford判断负环+Floyd)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:问你是否可以通过转换货币从中获利 如下面这组样例: USDollar 0.5 Brit ...
- 配置OpenCV+VS2013环境
配置OpenCV+VS2013环境 准备工作 win7系统 下载opencv的windows编译版 安装vs2013 express 设定环境变量 按windows窗键输入path,选择第二个结果编辑 ...