POJ1905 Expanding Rods（计算几何推公式+二分）_计算几何推式子-优快云博客

本文链接：https://blog.youkuaiyun.com/Enjoying_Science/article/details/78598458

本文介绍了一种计算方法，用于求解当薄杆被加热时，由于热膨胀导致的杆中心位移。输入包括杆的初始长度、温度变化及材料的热膨胀系数，通过迭代计算得出精确到毫米级别的位移。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Link：http://poj.org/problem?id=1905

Expanding Rods

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17938		Accepted: 4800

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

AC code：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<map>
#include<set>
#define LL long long
#define INF 0xfffffff
#define PI acos(-1)
#define EPS 1e-6
using namespace std;


int main()
{
//	freopen("in.txt","r",stdin);
	double L,n,c,low,high,mid,s,r,L1;
	while(scanf("%lf%lf%lf",&L,&n,&c)!=EOF)
	{
		if(L<0 && n<0 && c<0)
		{
			break;
		}
		low=0;
		high=L/2;
		L1=(1+n*c)*L;
		while(high-low>EPS)
		{
			mid=(low+high)/2;
			r=(4*mid*mid+L*L)/(8*mid);
			s=2*asin(L/(2*r))*r;
			if(L1>s)
			{
				low=mid;
			}
			else
			{
				high=mid;
			}
		}
		printf("%.3f\n",mid);
	}
	return 0;
}