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

  1. hdu 4452 Running Rabbits 模拟

    Running RabbitsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  2. 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 ...

  3. 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 ...

  4. 【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 ...

  5. [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) ...

  6. hdu-4452-Running Rabbits

    /* Running Rabbits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...

  7. 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> ...

  8. 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 ...

  9. 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

  1. 项目总结——深入浅出socket网络编程

    前言: 为什么会有如题的概念呢,我想对于没有主动听说过socket网络编程的人来说读到题目可能就已经蒙头了,为了很好的让大家进入场景,首先说一下一个需要用到这点东西的业务需求. 首先大家应该明确的是s ...

  2. wpf异常:指定的 Visual 不是此 Visual 的上级问题处理解析

    WPF在画线的时候,调用Control0.TransformToAncestor(Control1).Transform(new System.Windows.Point(0, 0))方法转换坐标的时 ...

  3. ·数据库基本内容回顾-day16.06.30

    一. 模式的定义和删除  ---创建了一个模式,就创建了一个数据库命名空间,一个框架.cascade.restrict create schema<模式名> authorization & ...

  4. Android UI WebView的使用:

    Android UI WebView的使用: /** * @author smiling * @date 2016/10 */ 布局: <?xml version="1.0" ...

  5. 关于ISAPI和CGI限制,这个要设为允许

    否则程序就报这个错误,注意,设置允许时不是在添加的网站上设置,而是在根iis,选择后右侧出现关于ISAPI和CGI限制,进去后选择相应版本,设置为允许就可以了

  6. eclipse - copy类的全名

    由于多次操作,感觉eclipse应该提供这个功能,网上搜一下,发现需要安装插件. 下载地址为 http://www.jave.de/eclipse/copyfully/copyfully_1.2.0. ...

  7. drop表,然后创建表,插入数据,并创建索引等内容。

    execute immediate 'drop table sjb_jhgl_ydjhtdsbb';   execute immediate 'create table dw_sc.sjb_jhgl_ ...

  8. 27 Java动态加载第三方jar包中的类

    我加载的方法是://参数fileName是jar包的路径,processorName 是业务类的包名+类名public static A load(String fileName, String pr ...

  9. bat命令中的变量声明及使用

    在bat文件中声明变量的方式如下: set xxx_variant_name=yyyyyyyyyyyy move D:\abc\efg\test.txt %xxx_variant_name%\test ...

  10. 如何使用sublime编辑器运行python程序

    现在越发喜欢sublime编辑器了,不仅界面友好美观.文艺,可扩展性还特别强. sublime本身是不具备运行python程序的能力的,需要做些设置才可以.以下是安装好sublime后设置的步骤: 点 ...