Zoj 3806 incircle and circumcircle [] [Special Judge]

Question link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3806.

Calculate the outer circle radius and incircle radius of a triangle and the three sides of the triangle that meet the given conditions.

The example given by the question is a bit difficult... it is easy to mislead people.

Sample Input

 
1 22 59 9
 

Sample ouput

 
3.464101615137754587 3.464101615137754587 3.4641016151377545876 8 10no solution!
 

The answer to the two and five examples is three integers. So it may be misleading.

When I first saw the question, I was misled by the example. I had no idea, but I found a special case after drawing it on paper for a long time.

The center of the incircle and the incircle given by YY is above a straight line, and then the relationship between length x and angle α can be obtained based on the sine theorem and a series of deformation.

Then the mathematical formula shows that the formula of angle α and X is

X = (1 + 1/sin (α)/(2-2 * sin (α ))

According to the formula binary sin (α)

The value range of sin (α) is 0 ~ 1. Compare the size of input x and re-determine the upper and lower bounds. (This function is incremental)

After sin (α) is obtained, three sides of the triangle can be obtained based on the x size and α angle.

Note that the radius of the incircle is 1 by default after the transformation. Therefore, the three sides of the triangle must be multiplied by R/R.

# Include <iostream> # include <math. h> # include <cstring> # include <stdio. h >#include <algorithm> using namespace STD; # define EPS 1e-11double Er (Double X) {double L = 0, r = 1, mid; while (L <R) {mid = (L + r)/2; // printf ("%. 9lf %. 9lf \ n ", l, R); double temp = (1.0 + 1.0/Mid)/(2.0-2.0 * (mid * mid )); if (temp-x> EPS) L = mid; else if (temp + EPS <X) r = mid; else break;} // printf ("%. 13lf \ n ", mid); return mid;} int main () {doubl E n, m; while (~ Scanf ("% lf", & N, & M) {double Bi = M/N; If (Bi-2) <-EPS) {printf ("no solution! \ N "); continue;} double SiNx = ER (BI); double cosx = SQRT (1-sinx * SiNx); Double Y = (1.0 + 1.0/SiNx)/cosx; double T = (1.0 + SiNx)/cosx; printf ("%. 18lf %. 18lf %. 18lf \ n ", y * n, y * n, 2 * T * n );}}
 

Ignore the definition of my function name and the appearance of an ER (BI )........

zoj 3806 incircle and circumcircle [ry] [Special Judge]