zoukankan      html  css  js  c++  java
  • POJ 1905, Expanding Rods

    Time Limit: 1000MS  Memory Limit: 30000K
    Total Submissions: 4863  Accepted: 1058


    Description
    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.

     

    Input
    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.

    Output
    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

    Sample Output
    61.329
    225.020
    0.000

    Source
    Waterloo local 2004.06.12


    // POJ1905.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <cmath>
    #include 
    <algorithm>
    using namespace std;

    int main(int argc, char* argv[])
    {
        
    double N,C,L;
        
    while(scanf("%lf%lf%lf",&L,&N,&C)&&N>=0&&C>=0&&L>=0)
        {
            
    if(N==0||L==0||C==0)
            {
                printf(
    "0.000\n");
                
    continue;
            }

            
    double minv = 0,maxv = acos(-1.0), midv;
            
    double L2 = ( 1 + N * C ) * L;
            
    while(maxv - minv > 1e-12)    
            {
                midv 
    = (minv + maxv) / 2;
                
    if2 * L2 / L > midv / sin(midv / 2))
                    minv 
    = midv;
                
    else
                    maxv 
    = midv;
            }
            printf(
    "%.3lf\n",L2 / midv * (1-cos(midv / 2)));
        }
        
    return 0;
    }

  • 相关阅读:
    activeMQ
    读写xml
    PLSQL
    oracle语法
    cxf远程调用服务
    FastDFS在linux下的安装和整合nginx实现上传图片和url访问
    dubbo和zookeeper的应用
    solr和Lucene的配置方式和应用
    win10 下安装 MongoDB 数据库支持模块(python)
    nodeJs 对 Mysql 数据库的 curd
  • 原文地址:https://www.cnblogs.com/asuran/p/1592697.html
Copyright © 2011-2022 走看看