UVa 10250 The Other Two Trees
还是读了很长时间的题,不过题本身很简单。
可以把四棵树想象成正方形的四个顶点,已知两个相对顶点的坐标,求另外两个坐标。
不过,原题可没直接这么说,中间需要一些小证明。
题中说有一个平行四边形然后分别以四条边为边长向外作正方形,四棵树就在四个正方形中心的位置。
这是我用几何画板画的图。
下面证△FOE≌△HGO
设CD=2a,BC=2b
∴EF=OH=a,
OF=HG=b
易知∠DFO=∠OHB
∴∠EFO=∠OHG=∠DFO+90°
∴△FOE≌△HGO(边角边)
∴∠E=∠GOH
∠EOG=∠EOF+∠DFO+∠E=90° (因为再加上∠DFE就等于内角和180°了)
综上,OE=OG且OE⊥OG
由于对称性,四棵树就构成了个正方形。
下面说说算法,利用旋转。我是用复数做的。
比如说一个复平面的向量a+bi乘上i所得向量就是逆时针旋转90°所得向量,乘上-i就是顺时针。
可以计算出中点O的坐标,然后向量OG=Gx-Ox+(Gy-Oy)i,乘上单位向量i得Oy-Gy+(Gx-Ox)i。
在加上O的坐标就得到E的坐标(Ox+Oy-Gy,Oy-Ox+Gx)。
同理也可以很容易求出顺时针转90°后点的坐标。
10250
Problem E
The Other TwoTrees
Input: standard input
Output: standard output
Time Limit: 2 seconds
You have a quadrilateral(四边形) shaped land whose opposite fences are ofequal length(四边形对边相等,也就是平行四边形). You have four neighborswhose lands are exactly adjacent(邻近的) to your four fences, that means you have acommon fence with all of them. For example if you have a fence of length d inone side, this fence of length d is also the fence of theadjacent neighbor on that side. The adjacent neighbors have no fence in commonamong themselves and their lands also don’t intersect(相交). The main difference between their land andyour land is that their lands are all square(正方形) shaped. All your neighbors have a tree at the center oftheir lands. Given the Cartesian coordinates(笛卡尔坐标) of trees of two opposite neighbors, you will have tofind the Cartesian coordinates of the other two trees.
Input
The input file contains several lines ofinput. Each line contains four floating point or integer numbers x1,y1, x2, y2, where (x1, y1), (x2, y2) are thecoordinates of the trees of two opposite neighbors. Input is terminated by endof file.
Output
For each line of input produce one line ofoutput which contains the line “Impossible.” without thequotes, if you cannot determine the coordinates of the other two trees.Otherwise, print four floating point numbers separated by a single space withten digits after the decimal point ax1, ay1, ax2, ay2, where (ax1,ay1) and (ax2, ay2) are the coordinates of the othertwo trees. The output will be checked with special judge program, so don’tworry about the ordering of the points or small precision errors. The sampleoutput will make it clear.
Sample Input
10 0 -10 0
10 0 -10 0
10 0 -10 0
Sample Output
0.000000000010.0000000000 0.0000000000 -10.0000000000
0.000000000010.0000000000 -0.0000000000 -10.0000000000
0.0000000000-10.0000000000 0.0000000000 10.0000000000
(World FinalWarm-up Contest, Problem Setter: Shahriar Manzoor)
AC代码:
//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std; int main(void)
{
#ifdef LOCAL
freopen("10250in.txt", "r", stdin);
#endif double x[], y[], midx, midy;
while(scanf("%lf %lf %lf %lf", &x[], &y[], &x[], &y[]) == )
{
midx = (x[] + x[]) / ;
midy = (y[] + y[]) / ;
x[] = midx + midy - y[];
y[] = x[] - midx + midy;
x[] = midx - midy + y[];
y[] = midx + midy - x[];
printf("%.10lf %.10lf %.10lf %.10lf\n", x[], y[], x[], y[]);
}
return ;
}
代码君
UVa 10250 The Other Two Trees的更多相关文章
- 【例题 6-17 UVa 10562】Undraw the Trees
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 模拟+递归 [代码] #include <bits/stdc++.h> using namespace std; con ...
- UVA题目分类
题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...
- [LintCode]——目录
Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (20 ...
- (Step1-500题)UVaOJ+算法竞赛入门经典+挑战编程+USACO
http://www.cnblogs.com/sxiszero/p/3618737.html 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年 ...
- ACM训练计划step 1 [非原创]
(Step1-500题)UVaOJ+算法竞赛入门经典+挑战编程+USACO 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成 ...
- 算法竞赛入门经典+挑战编程+USACO
下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinej ...
- Uva 10007 / HDU 1131 - Count the Trees (卡特兰数)
Count the Trees Another common social inability is known as ACM (Abnormally Compulsive Meditation) ...
- UVA.122 Trees on the level(二叉树 BFS)
UVA.122 Trees on the level(二叉树 BFS) 题意分析 给出节点的关系,按照层序遍历一次输出节点的值,若树不完整,则输出not complete 代码总览 #include ...
- UVa 10562 Undraw the Trees 看图写树
转载请注明: 仰望高端玩家的小清新 http://www.cnblogs.com/luruiyuan/ 题目大意: 题目传送门:UVa 10562Undraw the Trees 给定字符拼成的树,将 ...
随机推荐
- 批量安装操作系统之cobbler
Cobbler 部署文档 服务端配置 操作系统:Centos6.4 关闭防火墙及 selinux 安装cobbler软件 添加yum源 rpm -Uvh https://dl.fedoraprojec ...
- 本地安装xssing
本地安装xssing 环境为apache+mysql+php,linux下和Windows均进行了尝试. 步骤: (1)Mysql中创建xing数据库,将xssing.sql中的语句复制并在xing数 ...
- Set Matrix Zeroes
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. click ...
- 【面试题】Google of Greater China Test for New Grads of 2014总结
2014年Google中国校园招聘采用在线技术笔试,在Code Jam平台上,14号9点到15号9点开放测试题,帮助大家熟悉环境.这个周末也有够忙的,当时就上去看了一下,把输入文件下了一下,今天才把题 ...
- HDU1724 Ellipse(数值积分)
补一下一些小盲区,譬如simpson这种数值积分的方法虽然一直知道,但是从未实现过,做一道例题存一个模板. #pragma warning(disable:4996) #include<iost ...
- codeforces 439C Devu and Partitioning of the Array(烦死人的多情况的模拟)
题目 //这是一道有n多情况的烦死人的让我错了n遍的模拟题 #include<iostream> #include<algorithm> #include<stdio.h ...
- (4)用opengl读入off文件生成可执行文件把模型显示出来(未完待续)
·找了好几个程序,好像都达不到我的要求,去教程里看看吧! 在往上抛出了这问题,好几天才有人回答,我已经找到程序了 正好的他的分析对我分解程序很有用 这是一个难度比较高的 首先你要分析.off文件结构, ...
- ECharts案例教程1
原文:http://blog.csdn.net/whqet/article/details/42703973 简介 ECharts,缩写来自Enterprise Charts,商业级数据图表,是百度的 ...
- Textures
LPDIRECT3DVERTEXBUFFER9 g_VertexBuffer=NULL; //顶点缓存 LPDIRECT3DTEXTURE9 g_Texture=NULL;//纹理对象 bool In ...
- 欧拉工程第51题:Prime digit replacements
题目链接 题目: 通过置换*3的第一位得到的9个数中,有六个是质数:13,23,43,53,73和83. 通过用同样的数字置换56**3的第三位和第四位,这个五位数是第一个能够得到七个质数的数字,得到 ...