poj 1905Expanding Rods(二分法)（解题报告）_1905 expanding rods-优快云博客

本文链接：https://blog.youkuaiyun.com/CSUST_ACM/article/details/7952186

本文介绍了一种计算受热膨胀后薄杆中心位移的方法。通过数学推导结合二分查找算法，解决了薄杆受热膨胀形成圆弧段后的中心位移问题。输入包括初始长度、温度变化及材料膨胀系数。

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链接：http://poj.org/problem?id=1905

Expanding Rods

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.

When a thin rod of length L is heated n

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

思路：数学推论+二分查找

// 设 r为半径,&为角度 ,d为所求目标，L2=&r,sin(&/2)=L1/ (2.0*r),cos(&/2)=(r-d)/(r);

//且（cos&^2+sin&^2=1）则 r=（4*d*d+L1*L1)/(8*d), l=2*asin(l/(2.0*r))*r;比较l与L2的大小即可

#include<stdio.h>
#include<math.h>
int main()
{
    double L1,n,c;
    double r,l,d,high,low,mid,L2;
    while(scanf("%lf%lf%lf",&L1,&n,&c)!=EOF&&!(l<0&&n<0&&c<0))
    {
    L2=(1+n*c)*L1;
    high=L2;
    low=0;
    //printf("%.0lf\n",L);
    while(high-low>1e-6)//利用二分法查找目标 
    {
    mid=(high+low)/2.0;
    //printf("mid=%.0lf\n",mid);
    r=(4*mid*mid+L1*L1)/(8.0*mid);
    //printf("r=%.0lf\n",r);
    l=2*asin(l/(2.0*r))*r;
   // printf("L1=%.0lf\n",L1);
    if(l<L2)
    low=mid;
    else
    high=mid;
    }
    printf("%.3lf\n",high);
}
return 0;
    }