ACM-Expanding Rods

最新推荐文章于 2020-02-07 11:05:28 发布

weixin_30765319

最新推荐文章于 2020-02-07 11:05:28 发布

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原文链接：http://www.cnblogs.com/wejex/p/3218797.html

本文介绍了一个计算加热后杆形零件变形的程序实现方法。通过二分查找法求解加热后的杆形零件变成圆弧形状时中心位移的问题，并给出了具体的算法实现及样例输入输出。

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A - Expanding Rods

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

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.

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

集训的第一个题，我竟然连圆的几个公式都不记得了，好吧~

二分，综合两个公式，然后，精度弄高点，不然就wa了

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    double n,c,l;
    while(~scanf("%lf %lf%lf",&l,&n,&c))
    {

        if(l < 0 && n < 0 && c < 0)
            break;
        double ll = (1 + n * c) * l;
//        printf("%.3lf",l);
        double low, high ,mid , r;
        low  = 0;
        high = l / 2;

        while (high - low > 0.000001)
        {
            mid =(low  + high)/2;
            r  =((l * l)+(4*mid *mid ))/ (8 * mid );
            if ((2 * r * asin(l / (2*r)) < ll))
                low = mid;
            else
                high = mid;

        }
        printf ("%.3lf\n",mid);
    }
    return 0;
}