HDU4452 Running Rabbits
涉及知识点:
1. direction数组。
2. 一一映射(哈希)。
Running Rabbits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1565 Accepted Submission(s): 1099
Problem Description
Rabbit Tom and rabbit Jerry are running in a field. The field is an N×N grid. Tom starts from the up-left cell and Jerry starts from the down-right cell. The coordinate of the up-left cell is (1,1) and the coordinate of the down-right cell is (N,N)。A 4×4 field and some coordinates of its cells are shown below:
The rabbits can run in four directions (north, south, west and east) and they run at certain speed measured by cells per hour. The rabbits can't get outside of the field. If a rabbit can't run ahead any more, it will turn around and keep running. For example, in a 5×5 grid, if a rabbit is heading west with a speed of 3 cells per hour, and it is in the (3, 2) cell now, then one hour later it will get to cell (3,3) and keep heading east. For example again, if a rabbit is in the (1,3) cell and it is heading north by speed 2,then a hour latter it will get to (3,3). The rabbits start running at 0 o'clock. If two rabbits meet in the same cell at k o'clock sharp( k can be any positive integer ), Tom will change his direction into Jerry's direction, and Jerry also will change his direction into Tom's original direction. This direction changing is before the judging of whether they should turn around.
The rabbits will turn left every certain hours. For example, if Tom turns left every 2 hours, then he will turn left at 2 o'clock , 4 o'clock, 6 o'clock..etc. But if a rabbit is just about to turn left when two rabbit meet, he will forget to turn this time. Given the initial speed and directions of the two rabbits, you should figure out where are they after some time.
Input
There are several test cases.
For each test case:
The first line is an integer N, meaning that the field is an N×N grid( 2≤N≤20).
The second line describes the situation of Tom. It is in format "c s t"。c is a letter indicating the initial running direction of Tom, and it can be 'W','E','N' or 'S' standing for west, east, north or south. s is Tom's speed( 1≤s<N). t means that Tom should turn left every t hours( 1≤ t ≤1000).
The third line is about Jerry and it's in the same format as the second line.
The last line is an integer K meaning that you should calculate the position of Tom and Jerry at K o'clock( 1 ≤ K ≤ 200).
The input ends with N = 0.
Output
For each test case, print Tom's position at K o'clock in a line, and then print Jerry's position in another line. The position is described by cell coordinate.
Sample Input
4
E 1 1
W 1 1
2
4
E 1 1
W 2 1
5
4
E 2 2
W 3 1
5
0
Sample Output
2 2
3 3
2 1
2 4
3 1
4 1
Source
2012 Asia JinHua Regional Contest
Recommend
zhuyuanchen520 | We have carefully selected several similar problems for you: 55665565556455635562
Statistic | Submit | Discuss | Note
#include<stdio.h> int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; int Map[256];
int Map1[256]; int main() {
Map['N'] = 0;
Map['W'] = 1;
Map['S'] = 2;
Map['E'] = 3;
Map1['N'] = 'W';
Map1['W'] = 'S';
Map1['S'] = 'E';
Map1['E'] = 'N';
int n;
while(~scanf("%d", &n), n) {
char d1[2], d2[2];
int s1, s2;
int t1, t2;
scanf("%s%d%d", d1, &s1, &t1);
scanf("%s%d%d", d2, &s2, &t2);
int k;
scanf("%d", &k);
int x1 = 1, y1 = 1, x2 = n, y2 = n;
for(int i = 0; i < k; i++) {
int id = Map[d1[0]];
if(!dir[id][0]) {
y1 += s1 * dir[id][1];
if(y1 <= 0) {
y1 = 2 - y1;
d1[0] = 'E';
}
if(y1 > n) {
y1 = 2 * n - y1;
d1[0] = 'W';
}
} else {
x1 += s1 * dir[id][0];
if(x1 <= 0) {
x1 = 2 - x1;
d1[0] = 'S';
}
if(x1 > n) {
x1 = 2 * n - x1;
d1[0] = 'N';
}
}
id = Map[d2[0]];
if(!dir[id][0]) {
y2 += s2 * dir[id][1];
if(y2 <= 0) {
y2 = 2 - y2;
d2[0] = 'E';
}
if(y2 > n) {
y2 = 2 * n - y2;
d2[0] = 'W';
}
} else {
x2 += s2 * dir[id][0];
if(x2 <= 0) {
x2 = 2 - x2;
d2[0] = 'S';
}
if(x2 > n) {
x2 = 2 * n - x2;
d2[0] = 'N';
}
}
if(x1 == x2 && y1 == y2) {
int tmp = d1[0];
d1[0] = d2[0];
d2[0] = tmp;
} else {
if(!((i + 1) % t1))
d1[0] = Map1[d1[0]];
if(!((i + 1) % t2))
d2[0] = Map1[d2[0]];
}
}
printf("%d %d\n%d %d\n", x1, y1, x2, y2);
}
return 0;
}
HDU4452 Running Rabbits的更多相关文章
- hdu 4452 Running Rabbits 模拟
Running RabbitsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
- 模拟 HDOJ 4552 Running Rabbits
题目传送门 /* 模拟:看懂题意,主要是碰壁后的转向,笔误2次 */ #include <cstdio> #include <algorithm> #include <c ...
- HDU 4452 Running Rabbits (模拟题)
题意: 有两只兔子,一只在左上角,一只在右上角,两只兔子有自己的移动速度(每小时),和初始移动方向. 现在有3种可能让他们转向:撞墙:移动过程中撞墙,掉头走未完成的路. 相碰: 两只兔子在K点整(即处 ...
- 【HDU 4452 Running Rabbits】简单模拟
两只兔子Tom和Jerry在一个n*n的格子区域跑,分别起始于(1,1)和(n,n),有各自的速度speed(格/小时).初始方向dir(E.N.W.S)和左转周期turn(小时/次). 各自每小时往 ...
- [模拟] hdu 4452 Running Rabbits
意甲冠军: 两个人在一个人(1,1),一个人(N,N) 要人人搬家每秒的速度v.而一个s代表移动s左转方向秒 特别值得注意的是假设壁,反弹.改变方向 例如,在(1,1),采取的一个步骤,以左(1,0) ...
- hdu-4452-Running Rabbits
/* Running Rabbits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- 2012 Asia JinHua Regional Contest
Draw Something http://acm.hdu.edu.cn/showproblem.php?pid=4450 o(n)统计输入每个数的平方和. #include<cstdio> ...
- Crystal Clear Applied: The Seven Properties of Running an Agile Project (转载)
作者Alistair Cockburn, Crystal Clear的7个成功要素,写得挺好. 敏捷方法的关注点,大家可以参考,太激动所以转载了. 原文:http://www.informit.com ...
- Running Dubbo On Spring Boot
Dubbo(http://dubbo.io/) 是阿里的开源的一款分布式服务框架.而Spring Boot则是Spring社区这两年致力于打造的简化Java配置的微服务框架. 利用他们各自优势,配置到 ...
随机推荐
- 国内三大PT(Private Tracker)站分析
除这一行外,下面全部内容都是转载.出处不明. 国内三大PT(Private Tracker)站分析 先郑重的声明一下:本文以下的内容所有是复制粘贴的,不代表老夫的观点. 事实上内容我也没细致看. 贴这 ...
- Tomcat 加入windows 服务自启动设置
基于J2ee技术开发,可以运行在Tomcat.weblogic.websphere等J2ee应用服务器上,对于一般访问量不是很高的客户我们推荐使用Tomcat(开源免费),一般情况下Tomcat服务需 ...
- 画年利率 画图 自定义 View
使用 SlbSyView slbView; slbView = (SlbSyView) findViewById(R.id.slbView); slbView.setHeights(new float ...
- ssh注解开发
引入需要的jar包 @Entity public class Teacher { @Id @GeneratedValue(strategy=GenerationType.IDENTITY) priva ...
- java集合框架02
public class ListTest { public static void main(String[] args) { //创建几个新闻对象 News news1=new News(1, & ...
- jQuery中使用 .html() function在IE8和9中显示不正常源码中多出sizset和sizcache
错误原因:在引入jquery的时候,使用了html function,在IE8和IE9下面有可能会出现不兼容 解决办法:在html头部加一句 <meta http-equiv="X-U ...
- android系统360浏览器使用append动态添加元素出现黑色错乱背景
去掉样式 backface-visibility:hidden;
- html跳转到同一个页面的不同位置
<html> <body> <p><a href="#C4">查看 Chapter 4.</a></p> & ...
- C#Css/Js静态文件压缩--Yui.Compressor.Net
一.Asp.Net 自带静态文件压缩工具包 Microsoft.AspNet.Web.Optimization http://www.nuget.org/packages/Microsoft.AspN ...
- 华为S5300交换机配置基于接口的本地端口镜像
配置思路 1. 将Ethernet0/0/20接口配置为观察端口(监控端口) 2. 将Ethernet0/0/1----Ethernet0/0/10接口配置为镜像端口 配置步骤 1. 配置观察端 ...