POJ-1905 Expanding Rods 解题报告

最新推荐文章于 2021-02-23 11:15:02 发布

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#input #numbers #distance #output #each #算法

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本文介绍了一种通过二分法或直接使用公式来计算因热膨胀导致的细长杆中心位移的方法。输入包括杆的初始长度、温度变化及材料的热膨胀系数，输出为精确到三位小数的位移值。

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

题目链接：http://poj.org/problem?id=1905

解法类型：二分或直接公式求解

解题思路：很简单的二分题目，注意精度就可以了，直接对d进行二分，精度只要达到1e-4 ^_^ 其实，这题可以直接算出求解公式来的，为什么偏要归结于二分题呢？或许是精度不够高吧。

算法实现：

//STATUS:C++_AC_0MS_148K
#include<stdio.h>
#include<math.h>
#define esp 1e-4    //精度要控制好
int main()
{
	//freopen("in.txt","r",stdin);
    double high,mid,low,r,L1,L2,lt,N,C;
    while(scanf("%lf%lf%lf",&L1,&N,&C)&&L1!=-1)
	{
		L2=(1+N*C)*L1;
        low=0 ,high=L1/2+1; 

        for(mid=(low+high)/2;high-low>esp;mid=(low+high)/2)    //二分~
		{
            r=(L1*L1)/(4*mid)+mid;
            lt=2*r*atan(2*mid/L1);
			if(lt>L2)high=mid;
			else low=mid;
		}
		printf("%.3lf\n",mid);
	}
	return 0;
}