H. Hit!
time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

"Hit!" is a popular game in ancient Byteland.

The very first version of the game is quite simple: each player picks up a stone and throws it at a circle drawn on the ground. A player wins if his/her stone lands inside the circle.

After 20 years of practice, Bitman, a young man living in ancient Byteland, has mastered the skill of throwing stones – he can throw a stone at any specific place he wants. With such skill, Bitman plays "Hit!" without losing a single game. He simply targets every stone at the center of the circle!

The King of Hackerland hears the story of Bitman and wants to challenge him with a harder, though still very simple, version of "Hit!".

In each game, two circles which share a positive common area are drawn on the ground. In order to win, the player must throw a stone at the common area of the two circles.

As Bitman had no idea how to target his stone at the common area, he asks for your help. Given the coordinates of the centers and radii of the two circles, please tell Bitman the coordinates of any point he can target at such that he can win the game.

For simplicity, you can consider the landing position of the stone as a single point.

Input

The input consists of two lines, each describes one circle drawn on the ground. Each line contains three integers xy and r, denoting respectively the x-coordinate, y-coordinate, and the radius of a circle.

All coordinates have their absolute value no more than 100, and 1 ≤ r ≤ 100 for both circles.

Output

Output two numbers, the x-coordinate and y-coordinate of a point where Bitman can throw his stone at to win the game.

Your answer will be accepted if for each of the two circles, the point lies inside the circle or that the distance between the point and the circle is not greater than 10 - 5.

Examples
input
0 0 3
3 4 3
output
1.5 2.5
input
-7 -9 3
-4 -4 5
output
-6 -7
Note

In the first sample, (1.5, 2.5) is a possible answer as it lies inside the common area of two circles drawn. Please note that there exists more than one possible answer in this case. For example, (2, 2), (1, 2) and (2.1, 1.87) are also possible answers.

思路:

水题,特判下两个圆的直径大于两圆心距离的情况,这时只要输出圆心就好了。还以为会卡精度,敲了半天,结果告诉队友特判,队友一顿乱敲没考虑精度直接过了。。mmp.

实现代码:

#include<bits/stdc++.h>
using namespace std; int main()
{
double x1,x2,y1,y2,z1,z2;
cin>>x1>>y1>>z1;
cin>>x2>>y2>>z2;
double len = pow((x2 - x1),) + pow((y2 - y1),);
len = sqrt(len);
//cout<<"len :"<<len<<endl;
double len1 = len - z1;
double len2 = len - z2;
//cout<<len1<<" "<<len2<<endl;
if(len1<=){
cout<<x2<<" "<<y2<<endl;
return ;
}
else if(len2<=){
cout<<x1<<" "<<y1<<endl;
return ;
}
double f1 = x2-x1;
double f2 = y2-y1;
double k = len2/len;
//cout<<k<<endl;
f1 *= k;
f2 *= k;
//printf("%.6lf\n",len2);
double ans;
while(){
ans = pow(f1,)+pow(f2,);
ans = sqrt(ans);
//printf("%.6lf\n",ans);
if(abs(ans - len2)<0.00001)
break;
else{
if(ans-len2<0.00001){
f1+=0.000001;f2+=0.000001;}
else if(ans-len2>0.00001){
f1-=0.000001;f2-=0.000001;}
}
}
printf("%.6lf %.6lf",x1+f1,y1+f2);
//cout<<pow((x1+f1),2)+pow((y1+f2),2)<<endl;
}

Gym - 101522H Hit!的更多相关文章

  1. Codeforces Gym 100803D Space Golf 物理题

    Space Golf 题目连接: http://codeforces.com/gym/100803/attachments Description You surely have never hear ...

  2. Codeforces Gym 100231F Solitaire 折半搜索

    Solitaire 题目连接: http://codeforces.com/gym/100231/ Description 给你一个8*8棋盘,里面有4个棋子,每个棋子可以做一下某个操作之一: 1.走 ...

  3. [LeetCode] Design Hit Counter 设计点击计数器

    Design a hit counter which counts the number of hits received in the past 5 minutes. Each function a ...

  4. ACM: Gym 101047M Removing coins in Kem Kadrãn - 暴力

     Gym 101047M Removing coins in Kem Kadrãn Time Limit:2000MS     Memory Limit:65536KB     64bit IO Fo ...

  5. ACM: Gym 101047K Training with Phuket's larvae - 思维题

     Gym 101047K Training with Phuket's larvae Time Limit:2000MS     Memory Limit:65536KB     64bit IO F ...

  6. ACM: Gym 101047E Escape from Ayutthaya - BFS

    Gym 101047E Escape from Ayutthaya Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I6 ...

  7. ACM: Gym 101047B Renzo and the palindromic decoration - 手速题

     Gym 101047B  Renzo and the palindromic decoration Time Limit:2000MS     Memory Limit:65536KB     64 ...

  8. Buffer cache hit ratio性能计数器真的可以作为内存瓶颈的判断指标吗?

    Buffer cache hit ratio官方是这么解释的:“指示在缓冲区高速缓存中找到而不需要从磁盘中读取的页的百分比.” Buffer cache hit ratio被很多人当做判断内存的性能指 ...

  9. LeetCode Design Hit Counter

    原题链接在这里:https://leetcode.com/problems/design-hit-counter/. 题目: Design a hit counter which counts the ...

随机推荐

  1. Beautifulsoap - request 网络爬虫 (转)

    http://www.cnblogs.com/jiayongji/p/7118939.html (转) python爬虫系列(2)—— requests和BeautifulSoup库的基本用法

  2. 使用Git进行协同开发

    用了一段时间github,一直想用时间来对git的使用来做一段笔记,前段时间比较忙,现在沉下心来学习也是极好的. 很多项目开发会采用git这一优秀的分布式版本管理工具来进行项目版本管理.因为git的使 ...

  3. 20155302《网络对抗》Exp6 信息收集与漏洞扫描

    20155302<网络对抗>Exp6 信息收集与漏洞扫描 实验内容 (1)各种搜索技巧的应用 (2)DNS IP注册信息的查询 (3)基本的扫描技术:主机发现.端口扫描.OS及服务版本探测 ...

  4. # 20155319 Exp3 免杀原理与实践

    20155319 Exp3 免杀原理与实践 基础问题 (1)杀软是如何检测出恶意代码的? 基于特征码的检测 启发式的恶意软件检测 基于行为的恶意软件检测 (2)免杀是做什么? 免杀,从字面进行理解,避 ...

  5. Android开发——断点续传原理以及实现

    0.  前言 在Android开发中,断点续传听起来挺容易,在下载一个文件时点击暂停任务暂停,点击开始会继续下载文件.但是真正实现起来知识点还是蛮多的,因此今天有时间实现了一下,并进行记录.本文原创, ...

  6. linux gz 解压缩

    Linux压缩保留源文件的方法:gzip –c filename > filename.gzLinux解压缩保留源文件的方法:gunzip –c filename.gz > filenam ...

  7. mfc CImageList和CListCtrl

    知识点: CImageList类的运用 CListCtrl添加图标 一.CImageList CImageList*SetImageList(CImageList*pImageList,int nIm ...

  8. 工具神器推荐 Vox 和 search everything

    工具神器推荐 Vox 和 search everything vox官网: http://www.wox.one/

  9. css修改select下拉列表的默认样式

    select的一些默认样式我们很难修改,比如图标的替换.接下来就说说如何修改这些默认样式: html代码: <div> <select name=""> & ...

  10. 【OpenCV学习笔记之一】图像加载,修改及保存

    加载图像(用cv::imread)imread功能是加载图像文件成为一个Mat对象 其中第一个参数表示图像文件名称第二个参数 表示加载的图像是什么类型 支持常见的三个参数值IMREAD_UNCHANG ...