Problem A: Expanding Rods

When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion.

When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.

Your task is to compute the distance by which the center of the rod is displaced.

The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision.

Sample input

1000 100 0.0001
15000 10 0.00006
10 0 0.001
-1 -1 -1

Output for sample input

61.329
225.020
0.000 二分法:二分高度H,计算出H对应的L弧长来推出答案 弧长根据补出一个园三角函数推出弧长即可
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cctype>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MAXN 10000+10
#define INF 1<<30
int MOD;
double l, n, c; double getL(double h){
double R = l*l/(8*h)+h/2;
return asin(l/(2*R))*(2*R);
} int main (){ while(scanf("%lf%lf%lf",&l, &n, &c) != EOF){
if(l < 0 && n < 0 && c < 0)
break;
double L = 0, R = l/2;
double tar = (1+n*c)*l;
for(int i = 0; i < 100; i++){
double mid = (L+R)/2;
if(getL(mid) < tar)
L = mid;
else
R = mid;
}
printf("%.3lf\n",L);
}
return 0;
}

  

UVA 10668 Expanding Rods的更多相关文章

  1. UVA 10668 - Expanding Rods(数学+二分)

    UVA 10668 - Expanding Rods 题目链接 题意:给定一个铁棒,如图中加热会变成一段圆弧,长度为L′=(1+nc)l,问这时和原来位置的高度之差 思路:画一下图能够非常easy推出 ...

  2. POJ 1905 Expanding Rods

                           Expanding Rods Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1 ...

  3. Expanding Rods(二分POJ1905)

    Expanding Rods Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13688   Accepted: 3527 D ...

  4. poj 1905 Expanding Rods(木杆的膨胀)【数学计算+二分枚举】

                                                                                                         ...

  5. POJ 1905:Expanding Rods 求函数的二分

    Expanding Rods Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13780   Accepted: 3563 D ...

  6. D - Expanding Rods POJ - 1905(二分)

    D - Expanding Rods POJ - 1905 When a thin rod of length L is heated n degrees, it expands to a new l ...

  7. POJ 1905 Expanding Rods(二分)

    Expanding Rods Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 20224 Accepted: 5412 Descr ...

  8. 1137 - Expanding Rods

    1137 - Expanding Rods    PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 M ...

  9. LightOj1137 - Expanding Rods(二分+数学)

    题目链接:http://lightoj.com/volume_showproblem.php?problem=1137 题意:有一根绳子的长度为l,在有温度的情况下会变形为一个圆弧,长度为 l1 = ...

随机推荐

  1. Python全栈 MongoDB 数据库(数据的查找)

      非关系型数据库和关系型数据库的区别? 不是以关系模型构建的,结构自由 非关系型数据库不保证数据一致性 非关系型数据库可以在处理高并发和海量数据时弥补关系数据库的不足 非关系型数据库在技术上没有关系 ...

  2. Mybatis 异常记录(1): Invalid bound statement (not found)

    错误信息: org.apache.ibatis.binding.BindingException: Invalid bound statement (not found): com.pingan.cr ...

  3. c# 对List<T> 某字段排序,取TOP条数据

    //排序的对象里的字段数据准备 try { cmr.v4 = Double.Parse(cmr.v3) - Double.Parse(cmr.v2); } catch (Exception e) { ...

  4. java调c# exe 程序,exe里写文件问题

    应用场景描述: java web程序,触发 调用c#写的后台exe程序,发现exe里写的文件找不到.单独在cmd命令行下执行exe没问题: 问题查找: 由于exe里获取文件路径错误导致: 解决方法: ...

  5. 学习bash——通配符与特殊符号

    一.通配符 这是bash操作环境中一个非常有用的功能,这让我们使用bash处理数据就更方便了. 常用通配符如下: 符号 意义 * 代表0个到无穷多个任意字符 ? 代表一个任意字符 [] 代表一定有一个 ...

  6. [译]在Python中如何使用额enumerate 和 zip 来迭代两个列表和它们的index?

    enumerate - 迭代一个列表的index和item <Python Cookbook>(Recipe 4.4)描述了如何使用enumerate迭代item和index. 例子如下: ...

  7. android多点触控自由对图片缩放

    在系统的相册中,观看相片就可以用多个手指进行缩放. 要实现这个功能,只需要这几步: 1.新建项目,在项目中新建一个ZoomImage.java public class ZoomImageView e ...

  8. Linux e1000e网卡驱动

    目录 识别网卡 命令行参数 附加配置 技术支持 一.识别网卡e1000e驱动支持Intel所有的GbE PCIe网卡,除了82575,82576,基于82580系列的网卡.提示:Intel(R) PR ...

  9. Flink源码解读之状态管理

    一.从何说起 State要能发挥作用,就需要持久化到可靠存储中,flink中持久化的动作就是checkpointing,那么从TM中执行的Task的基类StreamTask的checkpoint逻辑说 ...

  10. ScrollBarsEnabled的使用

    在WinForm中通过WebBrowser获取网页,我想把WebBrowser的ScollBar去掉,我的网页不需要滚动条. 设置方法如下:单击WebBrowser设计页面,在属性页面有一个Scrol ...