The Designer

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 761    Accepted Submission(s): 142

Problem Description
Nowadays, little haha got a problem from his teacher.His teacher wants to design a big logo for the campus with some circles tangent with each other. And now, here comes the problem. The teacher want to draw the logo on a big plane. You could see the example of the graph in the Figure1

At first, haha's teacher gives him two big circles, which are tangent with each other. And, then, he wants to add more small circles in the area where is outside of the small circle, but on the other hand, inside the bigger one (you may understand this easily if you look carefully at the Figure1.

Each small circles are added by the following principles.
* you should add the small circles in the order like Figure1.
* every time you add a small circle, you should make sure that it is tangented with the other circles (2 or 3 circles) like Figure1.
    
The teacher wants to know the total amount of pigment he would use when he creates his master piece.haha doesn't know how to answer the question, so he comes to you.

Task
The teacher would give you the number of small circles he want to add in the figure. You are supposed to write a program to calculate the total area of all the small circles.

 
Input
The first line contains a integer t(1≤t≤1200), which means the number of the test cases. For each test case, the first line insist of two integers R1 and R2 separated by a space (1≤R≤100), which are the radius of the two big circles. You could assume that the two circles are internally tangented. The second line have a simple integer N (1≤N≤10 000 000), which is the number of small circles the teacher want to add.
 
Output
For each test case: 
Contains a number in a single line, which shows the total area of the small circles. You should out put your answer with exactly 5 digits after the decimal point (NO SPJ).
 
Sample Input
2
5 4
1
4 5
1
 
Sample Output
3.14159
3.14159
 
Source
 
 #include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
#define mod 1000000007
typedef long long ll;
int t;
int r1,r2,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&r1,&r2,&n);
if(r1<r2) swap(r1,r2);
double k1,k2,k3,k4,ans;
k1=-1.0/r1;
k2=1.0/r2;
k3=1.0/(r1-r2);
k4=k1+k2+k3;
ans=(r1-r2)*(r1-r2);
n--;
for(int i=; i<=n; i+=)
{
double r4=1.0/k4;
if(r4*r4<1e-)
break;
ans+=r4*r4;
if(i+<=n) ans+=r4*r4;
double k5=*(k1+k2+k4)-k3;
k3=k4;
k4=k5;
}
printf("%.5f\n",ans*acos(-1.0));
}
return ;
}

HDU 6158 笛卡尔定理+韦达定理的更多相关文章

  1. HDU 6158 笛卡尔定理 几何

    LINK 题意:一个大圆中内切两个圆,三个圆两两相切,再不断往上加新的相切圆,问加上的圆的面积和.具体切法看图 思路:笛卡尔定理: 若平面上四个半径为r1.r2.r3.r4的圆两两相切于不同点,则其半 ...

  2. The Designer (笛卡尔定理+韦达定理 || 圆的反演)

    Nowadays, little haha got a problem from his teacher.His teacher wants to design a big logo for the ...

  3. CF77E Martian Food(圆的反演or 笛卡尔定理+韦达定理)

    题面 传送门 这题有两种方法(然而两种我都想不到) 方法一 前置芝士 笛卡尔定理 我们定义一个圆的曲率为\(k=\pm {1\over r}\),其中\(r\)是圆的半径 若在平面上有两两相切,且六个 ...

  4. 爆炸几何之 CCPC网络赛 I - The Designer (笛卡尔定理)

    本文版权归BobHuang和博客园共有,不得转载.如想转载,请联系作者,并注明出处.   Nowadays, little hahahaha got a problem from his teache ...

  5. 2018 Multi-University Training Contest 1 H - RMQ Similar Sequence(HDU - 6305 笛卡尔树)

    题意: 对于一个序列a,构造一个序列b,使得两个序列,对于任意的区间 [l, r] 的区间最靠近左端点的那个最大值的位置,并且序列 b 满足 0 < bi < 1. 给定一个序列 a ,求 ...

  6. HDU - 6158 The Designer

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6158 本题是一个计算几何题——四圆相切. 平面上的一对内切圆,半径分别为R和r.现在这一对内切圆之间,按 ...

  7. HDU - 6305 RMQ Similar Sequence(笛卡尔树)

    http://acm.hdu.edu.cn/showproblem.php?pid=6305 题目 对于A,B两个序列,任意的l,r,如果RMQ(A,l,r)=RMQ(B,l,r),B序列里的数为[0 ...

  8. hdu 6305 RMQ Similar Sequence——概率方面的思路+笛卡尔树

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=6305 看题解,得知: 0~1内随机取实数,取到两个相同的数的概率是0,所以认为 b 序列是一个排列. 两个 ...

  9. hdu 1506 Largest Rectangle in a Histogram——笛卡尔树

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=1506 关于笛卡尔树的构建:https://www.cnblogs.com/reverymoon/p/952 ...

随机推荐

  1. POJ 2388&&2299

    排序(水题)专题,毕竟如果只排序不进行任何操作都是极其简单的. 事实上,排序算法十分常用,在各类高级的算法中往往扮演着一个辅助的部分. 它看上去很普通,但实际的作用却很大.许多算法在失去排序后将会无法 ...

  2. linux下ipython无法保存历史记录

    在Centos7下使用ipython时,发现有个warning,提示无法保存历史记录 [root@localhost pip-]# ipython /usr/local/lib/python3./si ...

  3. PowerBI开发 第六章:数据网管

    Power BI的本地数据网管(On-Premises Data Gateway)是运行在组织内部的软件,用于管控外部用户访问内部(on-premises)数据的权限.PowerBI的网管像是一个尽职 ...

  4. 如何解决markdown中图片上传的问题

    1.第一种方式(图床) 1.1 google中的插件-新浪微博图床 2.第二种方式,操作流程如下 2.1 下载一个有道云笔记客户端 2.2 然后把图片通过有道云笔记分享出来,见下动态图 3.总结一下 ...

  5. 巧用cheerio重构grunt-inline

    grunt-inline是楼主之前写的一个插件,主要作用是把页面带了__inline标记的资源内嵌到html页面去.比如下面的这个script标签. <script src="main ...

  6. Kubernetes采用CoreDNS

    参考文档: kubernetes插件:https://github.com/kubernetes/kubernetes/tree/master/cluster/addons/dns/coredns 自 ...

  7. Python机器学习/LinearRegression(线性回归模型)(附源码)

    LinearRegression(线性回归) 2019-02-20  20:25:47 1.线性回归简介 线性回归定义: 百科中解释 我个人的理解就是:线性回归算法就是一个使用线性函数作为模型框架($ ...

  8. c#版flappybird 未完全实现

    这些天开始在深圳找工作,想着把从前有些淡忘的技术再温故下.看到尊敬的<传智播客>有一期公开课,讲的是用c#编写flappybird小游戏,也就自己搜了下游戏资源,也来试试看. 其实用到的技 ...

  9. LINUX基础实验报告

    实验一:主要是介绍Linux系统概况,无运行代码. 实验二:Linux的基本操作 重要知识点 [Tab] 使用Tab键来进行命令补全,Tab键一般键盘是在字母Q旁边,这个技巧给你带来的最大的好处就是当 ...

  10. Linux第五周学习总结——扒开系统调用的三层皮(下

    Linux第五周学习总结--扒开系统调用的三层皮(下) 作者:刘浩晨 [原创作品转载请注明出处] <Linux内核分析>MOOC课程http://mooc.study.163.com/co ...