HDU 5120 (formula for finding the area of intersection of two circles)

S = a large b large-a large B small-a small B large + a small B small. (A is a ring, large represents a great circle, B is the same). Then directly set the template ,,,,

1#include <stdio.h>2#include <algorithm>3#include <string.h>4#include <cmath>5 using namespacestd;6 7 Const DoubleEPS = 1e-8;8 Const DoublePI = ACOs (-1.0);9 Ten intSgnDoublex) One { A     if(Fabs (x) < EPS)return 0; -     if(X <0)return-1; -     Else return 1; the } - struct Point - { -     Doublex, y; + Point () {} -Point (Double_x,Double_y) +     { Ax = _x; y =_y; at     } -Pointoperator-(ConstPoint &b)Const -     { -         returnPoint (X-b x, Y-b. y); -     } -  in     Double operator^ (ConstPoint &b)Const -     { to         returnX*b. y-y*b. x; +     } -  the     Double operator* (ConstPoint &b)Const *     { $         returnX*b. x + y*b. y;Panax Notoginseng     } -  the     voidTransxy (DoubleB) +     { A         Doubletx = X,ty =y; thex = tx* cos (B)-ty*sin (B); +y = tx* sin (B) + ty*cos (B); -     } $ }; $  - DoubleDist (Point A, point B) - { the     returnsqrt (A-B) * (Ab)); - }Wuyi  the DoubleAc (Point C1,DoubleR1, point C2,DoubleR2) - { Wu     DoubleD =Dist (C1,C2); -     if(R1 + R2 < D + EPS)return 0; About     if(D < Fabs (R1-R2) +EPS) $     { -         DoubleR =min (r1,r2); -         returnpi*r*R; -     } A     Doublex = (d*d + r1*r1-r2*r2)/(2*d); +     DoubleT1 = ACOs (x/R1); the     DoubleT2 = ACOs ((d-x)/r2); -     returnR1*R1*T1 + r2*r2*t2-d*r1*sin (t1); $ } the  the intMain () { the     intT; Point C1, C2; the     Doubleans, R, R, X1, y1, x2, y2; -scanf"%d", &T); in      for(intCAS =1; CAS <= T; ++CAs) { thescanf"%LF%LF%LF%LF%LF%LF", &r, &r, &x1, &y1, &AMP;X2, &y2); thec1.x = x1; C1.Y =Y1; Aboutc2.x = x2; C2.Y =Y2; theAns = AC (c1, R, C2, R)-AC (C1, R, C2, R)-Ac (C1, R, C2, R) the+Ac (C1, R, C2, r); theprintf"Case #%d:%.6lf\n", CAs, ans); +     } -     return 0; the}

HDU 5120 (formula for finding the area of intersection of two circles)