Expanding Rods

最新推荐文章于 2020-05-02 19:02:25 发布

Yjonben

最新推荐文章于 2020-05-02 19:02:25 发布

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分类专栏：二分文章标签：二分

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探讨了当薄木棍受热膨胀时，如何计算其弯曲成圆弧形状后中心点的位移距离。通过二分法枚举计算，确保精度达到题目要求。

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题面：

Expanding Rods

Time Limit: 1000MS
Memory Limit: 30000K
Total Submissions: 19870
Accepted: 5311

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. Alt
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

题面描述：

这根木棍会受热膨胀并向上弯曲，给出了受热膨胀后长度的公式和温度，要求膨胀后弯曲的木棍中点距离原始状态的中点的距离。输入的数据保证杆的长度不会超过原来长度的一半。

题目分析：

正常想数学算法好像是求不出杆中心位移距离的，这时候要用二分一个个枚举，控制精度比题目的1e-3小就行，当二分的左右都在这个精度内随便输出哪个都行。

代码：

#include<algorithm>
#include<stdio.h> 
#include<math.h>
#include<iostream>
using namespace std;
double l,n,c;

int ch(double mid){
	double nlen,nrad,rlen;
	rlen=(1+n*c)*l,nrad=(mid*mid+l*l/4)/(2*mid),nlen=asin((l/2)/nrad)*nrad*2;//rl弧长公式 
	if(rlen>nlen) return 1;
	else return 0;
}


int main(){
	double al,ar,mid;
	while(scanf("%lf %lf %lf",&l,&n,&c)&&l>=0&&n>=0&&c>=0){
		al=0,ar=l/2;
		while(ar-al>0.0001){
			mid=(al+ar)/2;
			if(ch(mid)) al=mid;
			else ar=mid;
		}
		printf("%.3lf\n",al);
	}
}