【POJ1905】

最新推荐文章于 2020-05-05 22:18:26 发布

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本文介绍了一种计算两端固定的薄杆受热膨胀后形成圆弧段时，杆中心位移的方法。通过二分法找到合适的圆心角，并计算出最终的位移距离。

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Expanding Rods

Time Limit:500MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status

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. 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

1000 100 0.0001

150 10 0.00006

10 0 0.001

Sample Output

Case 1: 61.3289915

Case 2: 2.2502024857

Case 3: 0

大致题意：

一根两端固定在两面墙上的杆受热弯曲后变弯曲

求前后两个状态的杆的中点位置的距离

解法：弯曲后的长度：:s=2* θ*r, sinθ=(1/2)*l/r,联立得s=l*θ/sinθ; 实际上弯曲后的长度是l1=(1+n*c)*l;对角度θ进行二分；0到π/2； 将mid带入计算出s与l1比较

#include<cstdio>
#include<cmath>
#define P acos(-1.0)//表示π
using namespace std;
int main()
{
	int t,test=1;
	scanf("%d",&t);
	while(t--)
	{
		double l,n,c;
		scanf("%lf%lf%lf",&l,&n,&c);
		double l1=(1+n*c)*l;
		int Size=100;
		double k=0.0,g=P/2.0;
		while(Size--)
		{
		double mid=(k+g)/2.0;
		if(l*mid/sin(mid)>l1)
		g=mid;
		else
	k=mid;	
		}
		printf("Case %d: %.7lf\n",test++,0.5*l/sin(k)-0.5*l/tan(k));//半径-!!
	}
	return 0;
 }

设弦长为L0(即原长)，弧长为L1=(1+n*C)*l0，目标值为h，半径为R，弧所对圆心角为2θ(弧度制)。
可以得到以下方程组：
圆的弧长公式：L1=2θR
三角函数公式：L0=2*R*sinθ，变换得θ=arcsin(L0/(2*R))
勾股定理：R^2=(R-h)^2+(0.5*L0)^2，变换得L0^2+4*h^2=8*h*R
合并①②式得到
L1=2*R*arcsin(L0/(2*R))
半径R可以由③式变换得到
R=(L0^2+4*h^2)/(8*h)
可以用二分枚举h的值，计算出R和L1，与题目中L1进行比较。
化简得

r=（4h^2+l^2)/(8*h);

s=2*r*arcsin(l/(2r))

#include <stdio.h>
#include <math.h>
int main()
{
    double L0,L1,N,C;
    double high,low,h,r,L2;
    int t,test=1;
    scanf("%d",&t);
    while (t--)
    {
	scanf("%lf%lf%lf",&L0,&N,&C);
        L1=(1+N*C)*L0;
        high=0.5*L0;
        low=0.0;
        int Size=100;
        while (Size--)
        {
            h=(high+low)/2;
            r=(L0*L0+4*h*h)/(8*h);
            L2=2*r*asin(L0/(2*r));
            if (L1<L2) high=h;
            else low=h;
        }
        h=(high+low)/2;
        printf("Case %d: %.7f\n",test++,h);
    }
    return 0;
}