LightOJ 1137 Expanding Rods（加热变化的路，二分）

本文链接：https://blog.youkuaiyun.com/shyazhut/article/details/52049768

本文介绍了一种计算受热膨胀后薄杆中心位移的方法。通过输入初始长度、温度变化及热膨胀系数，采用逼近法计算出薄杆受热膨胀后的形状变化及中心位移。

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1137 - Expanding Rods（进入题目）

PDF (English)

Statistics

Forum

Time Limit: 0.5 second(s)

Memory Limit: 32 MB

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. That means you have to calculate h as in the picture.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case contains three non-negative real numbers: the initial length of the rod in millimeters L, the temperature change in degrees n and the coefficient of heat expansion of the material C. Input data guarantee that no rod expands by more than one half of its original length. All the numbers will be between 0 and 1000 and there can be at most 5 digits after the decimal point.

Output

For each case, print the case number and the displacement of the center of the rod in single line. Errors less than 10^-6 will be ignored.

Sample Input

Output for Sample Input

1000 100 0.0001

150 10 0.00006

10 0 0.001

Case 1: 61.3289915

Case 2: 2.2502024857

Case 3: 0

SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET, PICTURES)

题意：

已知一个原来长度 l 的路，加热后变化（如图演示）

代码：

#include<stdio.h>
#include<math.h>
#define PI 3.1415926

int main() {
	int t,v=1;
	scanf("%d",&t);
	double L,n,c;//原来长度：L ;加热的温度：n ;热膨胀系数：c
	while(t--) {
		scanf("%lf%lf%lf",&L,&n,&c);
		double L_change=(1.0+n*c)*L;//加热变化后的长度
		double left=0,right=PI/2.0;//这里逼近是转化到圆心角大小
		int dd=64;
		while(dd--) {
			double mid=(left+right)/2.0;//圆心角 
			double R=(L/2)/sin(mid);//半圆弧所在圆半径
			double new_L=2*R*mid;//新区间逼近的弧长 
			if(new_L>L_change) {
				right=mid;
			} else {
				left=mid;
			}
		}
		double ans=0.5*L/sin(left)-0.5*L/tan(left);//计算差值 h 
		printf("Case %d: %.7lf\n",v,ans);
		v++;
	}
	return 0;
}