UVa 375 Inscribed Circles and Isosceles Triangles

UVa 375 Inscribed Circles and Isosceles Triangles (等腰三角形內切圓&規律)

375 - Inscribed Circles and Isosceles Triangles

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=101&page=show_problem&problem=311

Given two real numbers

B

the width of the base of an isosceles triangle in inches
H

the altitude of the same isosceles triangle in inches

Compute to six significant decimal places

C

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

查看本欄目更多精彩內容：http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

1. 如何求內切圓半徑？——過內心向三條邊做垂線，再連結內心與三個頂點，由面積相等得0.5*(a+b+c)*r = 三角形面積S，即r=2*S/(a+b+c)

2. 下一個圓的半徑呢？把公式中的高H和底B換成H-2r和B*(H-2r)/H即可——進而發現序列r成了一個等比序列。(但O(1)的公式法實現會存在精度不準的問題，不可行)

完整代碼：

/*0.069s*/    #include<cstdio>  #include<cmath>  const double pi = 2 * acos(0);      int main()  {      int n;      double B, H, r, k, R;      scanf("%d", &n);      while (n--)      {          scanf("%lf%lf", &B, &H);          r = B * H / (B + sqrt(B * B + 4 * H * H));          k = 1 - 2 * r / H;          R = 0;          while (r >= 1e-6)          {              R += r;              r *= k;          }           printf("%13.6f\n", pi * R * 2);///注意是13          if (n) putchar(10);      }      return 0;  }