POJ 1905 Expanding Rods 浮点数二分

标签：c++   poj   二分   

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

Output

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

题解

又是二分答案的题目。题意是说，一个长为l的细杆会变长，如果固定两个端点，那么这个杆子就会拱起一个圆弧。求圆弧的高度。

如图：

我一开始想的比较简单，直接上手认为是一个可以搞得函数找零点的问题，但是手推函数之后发现一个问题。

对，这个函数是没有极限的，而且在不停地波动中，所以直接去计算零点就容易跪。所以就还是直接模拟去算了。代码挺简单的，主要还是精度问题。

代码示例

/****
	*@author    Shen
	*@title     poj 1905
	*/

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const double eps = 1e-5;
const double inf = 0xfffffff;
const double PI  = acos(1.0);

double l, n, c, lp;

inline bool test(double x)
{
    double angle = asin(l / x);
    double tmplp = x * angle;
    return tmplp > lp;
}

double Bsearch(double l, double r)
{
    while (r - l > eps)
    {
        double mid = (r + l) * 0.5;
        //printf("%16.4lf%16.4lf%16.4lf\n", l, mid, r);
        if (test(mid)) l = mid;
        else r = mid;
    }
    return r;
}

void solve()
{
    if (n * c < 0.000001)
        printf("0.000\n");
    else
    {
        lp = (1.0 + n * c) * l;
        lp /= 2; l /= 2;//都除二，减少运算
        double r = Bsearch(0.0, inf);
        double h = r - sqrt(r * r - l * l);
        //G++
            printf("%.3f\n", h);
        //C++
        //  printf("%.2lf\n", h);
    }
}

int main()
{
    while (~scanf("%lf%lf%lf", &l, &n, &c))
        if (l < 0) break;
        else solve();
    return 0;
}

POJ 1905 Expanding Rods 浮点数二分,布布扣,bubuko.com

POJ 1905 Expanding Rods 浮点数二分

标签：c++   poj   二分   

原文地址：http://blog.csdn.net/polossk/article/details/26378761