POJ1905-Expanding Rods - 二分法。

最新推荐文章于 2019-01-28 17:39:43 发布

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最新推荐文章于 2019-01-28 17:39:43 发布

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分类专栏：二分和三分

本文链接：https://blog.youkuaiyun.com/qq_23168773/article/details/47030015

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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、温度变化及材料膨胀系数，利用循环和数学公式计算出杆形变后的中心位移。文章提供了完整的C++实现代码，并注意到了循环次数和精度控制的重要性。

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Expanding Rods

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13354		Accepted: 3442

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

方程容易求的。

最后直接枚举答案。（SL为起始长度， L为之后长度）

不过本题有两个坑点。

1、Input data guarantee that no rod expands by more than one half of its original length.（如果R取太大了，答案肯定是错了的， R应取SL / 4， 保险也可取 SL / 2）。

2、二分循环当循环量很大时， 用浮点数可能有误差。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <vector>
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define mem(a) memset(a, 0, sizeof(a))
using namespace std;


//typedef __int LL;
typedef long long LL;
const double epx = 1e-7;





int main()
{
    double sl, l, r, n, c;
    while (scanf("%lf %lf %lf", &sl, &n, &c), sl >= 0 && n >= 0 && c >= 0) {
        l = (1 + n * c) * sl;


        double L = 0.0, R = sl / 4;
        FOR(i, 0, 1000) {               //这里是重点。
            double mid = (L + R) / 2;
            r = mid / 2 + (sl * sl) / (8 * mid);

            if (r >= (l / (2 * asin(sl / (2 * r))))) R = mid;
            else L = mid;
        }


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