Treasure Hunt
Captain Bill the Hummingbird and his crew recieved an interesting challenge offer. Some stranger gave them a map, potion of teleportation and said that only this potion might help them to reach the treasure.
Bottle with potion has two values x and y written on it. These values define four moves which can be performed using the potion:
Map shows that the position of Captain Bill the Hummingbird is (x1, y1) and the position of the treasure is (x2, y2).
You task is to tell Captain Bill the Hummingbird whether he should accept this challenge or decline. If it is possible for Captain to reach the treasure using the potion then output "YES", otherwise "NO" (without quotes).
The potion can be used infinite amount of times.
The first line contains four integer numbers x1, y1, x2, y2 ( - 105 ≤ x1, y1, x2, y2 ≤ 105) — positions of Captain Bill the Hummingbird and treasure respectively.
The second line contains two integer numbers x, y (1 ≤ x, y ≤ 105) — values on the potion bottle.
Print "YES" if it is possible for Captain to reach the treasure using the potion, otherwise print "NO" (without quotes).
0 0 0 6
2 3
YES
1 1 3 6
1 5
NO
In the first example there exists such sequence of moves:
— the first type of move
— the third type of move
题解:
做了两次codeforces之后发现上面的题目很多没有看起来那么复杂,实际上只需要判断一下是否可以整除并且整除之后奇偶性是不是一样的就可以了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
using namespace std;
int x,x2,y,y2,a,b;
int main()
{
int i,j;
cin>>x>>y>>x2>>y2;
cin>>a>>b;
i=abs(x2-x);j=abs(y2-y);
if(i%a==&&j%b==&&((i/a)%==(j/b)%))cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
Treasure Hunt的更多相关文章
- zoj Treasure Hunt IV
Treasure Hunt IV Time Limit: 2 Seconds Memory Limit: 65536 KB Alice is exploring the wonderland ...
- POJ 1066 Treasure Hunt(线段相交判断)
Treasure Hunt Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4797 Accepted: 1998 Des ...
- ZOJ3629 Treasure Hunt IV(找到规律,按公式)
Treasure Hunt IV Time Limit: 2 Seconds Memory Limit: 65536 KB Alice is exploring the wonderland ...
- POJ 1066 Treasure Hunt(相交线段&&更改)
Treasure Hunt 大意:在一个矩形区域内.有n条线段,线段的端点是在矩形边上的,有一个特殊点,问从这个点到矩形边的最少经过的线段条数最少的书目,穿越仅仅能在中点穿越. 思路:须要巧妙的转换一 ...
- poj1066 Treasure Hunt【计算几何】
Treasure Hunt Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 8192 Accepted: 3376 Des ...
- zoj 3629 Treasure Hunt IV 打表找规律
H - Treasure Hunt IV Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu ...
- ZOJ 3626 Treasure Hunt I 树上DP
E - Treasure Hunt I Time Limit:2000MS Memory Limit:65536KB Description Akiba is a dangerous country ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)I - Piglet treasure hunt Series 1
题目描述 Once there was a pig, which was very fond of treasure hunting. The treasure hunt is risky, and ...
- 【树形dp】Treasure Hunt I
[ZOJ3626]Treasure Hunt I Time Limit: 2 Seconds Memory Limit: 65536 KB Akiba is a dangerous coun ...
随机推荐
- Ubuntu14.04双网卡主备配置
近日有个需求,交换机有两台,做了堆叠,服务器双网卡,每个分别连到一台交换机上.这样就需要将服务器的网卡做成主备模式,以增加安全性,使得当其中一个交换机不通的时候网卡能够自动切换. 整体配置不难,网上也 ...
- Flume总结(1)
一.日志采集:从网络端口接收数据,下沉到logger 文件netcat-logger.conf: # Name the components on this agent #给那三个组件取个名字 a1. ...
- 用php(session)实现留言板功能----2017-05-09
要实现留言功能,发送者和接受者必不可少,其次就是留言时间留言内容. 要实现的功能: 1.登录者只能查看自己和所有人的信息,并能够给好友留言 2.留言板页面,好友采取下拉列表,当留言信息为空时,显示提示 ...
- CentOS7.2静默安装oracle11g
http://www.centoscn.com/image-text/config/2015/0528/5552.html http://www.linuxidc.com/Linux/2016-04/ ...
- 下拉菜单制作——利用CSS实现的一个实例
本文实现了一个经典的下拉菜单的制作. 首先,写出Html部分: <!DOCTYPE html> <html> <head> <meta charset=&qu ...
- js获取宽高
document.body.clientWidth ==> BODY对象宽度 document.body.clientHeight ==> BODY对象高度 document.docume ...
- HDU4704Sum 费马小定理+大数取模
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4704 题目大意: 看似复杂,其实就是求整数n的划分数,4=1+1+2和4=1+2+1是不同的.因而可 ...
- (jquery+ajax)省市区三级联动(封装和不封装两种方式)-----2017-05-14
首先,要实现如下图效果, 1.要理清思路: 先做出三个下拉菜单----根据第一个下拉菜单的value值获取第二个下拉列表的内容,第三个同理. 2.用到的数据库表:Chinastates表 规律:根据国 ...
- 1-LPC1778建立工程
先来建立一个工程模板,,,要比32简单的多,假设32是用库开发的话,,,,因为还要把那些和库相关的文件加到工程里.... LPC呢就只需要把设置系统和总线的时钟文件(system_LPC177x_8x ...
- Java类加载和卸载的跟踪
博客搬家自https://my.oschina.net/itsyizu/blog/ 什么是类的加载和卸载 Java程序的运行离不开类的加载,为了更好地理解程序的执行,有时候需要知道系统加载了哪些类.一 ...



