ACM: 二分法 数论题 poj 1905

最新推荐文章于 2022-07-09 18:05:13 发布

Smile_Benson

最新推荐文章于 2022-07-09 18:05:13 发布

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探讨了利用二分法解决物理问题——热胀冷缩导致的位移计算。通过数学模型和编程实现，精确计算出加热后杆件中心的位移。

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Expanding Rods

Description

ACM: <wbr>二分法 <wbr>数论题 <wbr>poj <wbr>1905

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

题意: 一个棍子热胀冷缩, 求热涨之后棍子升高的距离.

解题思路:

设升高的距离为: mid. 原来棍子长度length

1. 范围题,用二分逼近吧.

2. 要求出圆的半径为: (mid*mid + length*length/4) / (2*mid)

3. 逼近法:

当 r * asin( length/ (2*r) ) < (length*(1+n*c)) / 2

mid = left

否则

mid = right

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const double eps = 1e-5;

double length, c, n;

int main()
{
double left, right, mid;
double r;
double length2;
// freopen("input.txt","r",stdin);
while(scanf("%lf %lf %lf",&length,&n,&c) != EOF)
{
if(length == -1 && n == -1 && c == -1)
break;

  left = 0;
  right = length/2;
  length2 = (1.0+n*c)*length;
  while(right - left > eps)
  {
   mid = (left+right) / 2;
   r = (mid*mid + length*length/4) / (mid*2);
   if(r * asin( length/ (2*r) ) < length2 / 2 )
    left = mid;
   else
    right = mid;
  }
  printf("%.3lf\n",mid);
}
return 0;
}