Codeforces 600D area of Circles ' intersection

Separation, inclusion, want to make three kinds of situations to discuss.

is mainly the accuracy and data range of the problem, the first data with a long double, can be judged by the integer type should not use floating-point type.

The coordinates given in the title, the radius of the integer type, there is a card floating point to judge the situation is relatively small.

At the end of the Triangle area of the time do not use Helen Formula, there are four connected multiply very easy to explode data range loss accuracy, even if 22 times to consider the symbol

(Taking the logarithm is also a good way). (I don't know how the accuracy of sqrt and Cos,sin compare

#include <bits/stdc++.h>using namespaceStd;typedefLong DoubleLd;typedefLong Longll;ld x[2],y[2],r[2];inline ld SQR (ld x) {returnx*x;} inline ll SQRL (ll x) {returnx*x;} Inline LD Fcos (ld A, LD B, ld c) {returnACOSL ((a*a+b*b-c*c)/(2*a*b));} Inline LD cut (LD ang, ld R) {ld S1= ang*r*R; LD S2= Sinl (ANG) *cosl (ANG) *r*R; returnS1-S2;}ConstLD pi = ACOSL (-1);Doublesolve () {if(r[0] > r[1]) {Swap (r[0],r[1]); Swap (x[0],x[1]); Swap (y[0],y[1]); } LL DX= x[0]-x[1], dy = y[0]-y[1]; LL DS= SQRL (dx) +sqrl (DY); if(DS >= SQRL (r[0]+r[1]))return 0; LD D=Sqrtl (DS); if(d+r[0] <= r[1])returnSQR (r[0])*Pi; LD ang[2]; ang[0] = Fcos (d,r[0],r[1]); ang[1] = Fcos (d,r[1],r[0]); /*WA-ld area = ANG[1]*SQR (r[1]) + ANG[0]*SQR (r[0]);    LD s = (d+r[0]+r[1])/2; Area-= Sqrtl (s* (s-d) * (S-r[0]) * (s-r[1]) * *;    O (n^4) Split the sign problem logarithm may be feasible return area; */    returnCut (ang[0],r[0]) + Cut (ang[1],r[1]);}//#define LOCALintMain () {#ifdef LOCAL freopen ("In.txt","R", stdin);#endifCin>>x[0]>>y[0]>>r[0]>>x[1]>>y[1]>>r[1]; printf ("%.10f\n", Solve ()); return 0;}

Codeforces 600D area of Circles ' intersection