uva_375 - Inscribed Circles and Isosceles Triangles

最新推荐文章于 2022-02-03 15:56:25 发布

原创最新推荐文章于 2022-02-03 15:56:25 发布 · 562 阅读

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本文介绍了一种计算等腰三角形中一系列内接圆周长之和的方法，通过递归方式累加直到最小圆半径达到指定阈值。利用三角函数计算每个内接圆的半径，并给出完整的C++实现代码。

Given two real numbers

the width of the base of an isosceles triangle in inches

the altitude of the same isosceles triangle in inches

Compute to six significant decimal places

the sum of the circumferences of a series of inscribed circles stacked one on top of another from the base to the peak; such that the lowest inscribed circle is tangent to the base and the two sides and the next higher inscribed circle is tangent to the lowest inscribed circle and the two sides, etc. In order to keep the time required to compute the result within reasonable bounds, you may limit the radius of the smallest inscribed circle in the stack to a single precision floating point value of 0.000001.

For those whose geometry and trigonometry are a bit rusty, the center of an inscribed circle is at the point of intersection of the three angular bisectors.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input will be a single line of text containing two positive single precision real numbers (B H) separated by spaces.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The output should be a single real number with twelve significant digits, six of which follow the decimal point. The decimal point must be printed in column 7.

Sample Input

1

0.263451 0.263451

Sample Output

     0.827648

不断的累加内接圆面积直至半径r<0.000001. 刚开始用等面积法三角形内接圆半径一直wa，可能是用到开根号算出来有些微小误差。。。

后来改用cmath里的三角函数求就ac了

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;



double circum ( double b, double h)
{
    double theta = atan(2*h/b);     //腰与高的夹角
    double r = b/2*tan(theta/2);   
    if ( r < 0.000001) return 0;

    return 2*M_PI*r + circum( (1-2*r/h)*b, h-2*r);

}

int main()
{
    int t;
    double b, h;

    cin >> t;
    while( t--){
        cin >> b >> h;
        cout << fixed << setprecision(6);
        cout << setw(13) << circum( b, h) << endl;

        if( t) cout << endl;
    }
    return 0;
}